3 by 3 system word problems

Video transcript

Solve the following application
problem using three equations with three unknowns. And they tell us the
second angle of a triangle is 50 degrees less than
four times the first angle. The third angle is 40
degrees less than the first. Find the measures
of the three angles. So let’s draw ourselves
a triangle here. And let’s call the first angle
a, the second angle b, and then the third angle c. And before we even look
at these constraints, one property we
know of triangles is that the sum of their
angles must be 180 degrees. So we know that a plus b plus
c must be equal to 180 degrees. Now, with that out of
the way, let’s look at these other constraints. So they tell us the second
angle of a triangle— let me do that in
another color— they tell us the second
angle of a triangle is 50 degrees less than
four times the first angle. So we’re saying b
is the second angle. So they’re saying, the
second angle of a triangle is 50 degrees less than
four times the first angle. So 4 times the first
angle would be 4a, we’re calling a the first angle. So 4 times the
first angle is 4a. But it’s 50 degrees less
than that, so minus 50. Now the next constraint they
give us, the third angle is 40 degrees less
than the first. So the third angle is 40
degrees less than the first. So the first angle, a, is
going to be 40 degrees less than that. So we have three equations
with three unknowns. And so we just have
to solve for it. And let’s see, what’s a
good first variable to try to eliminate? And just to try to visualize
that a little bit better, I’m going to bring these a’s
onto the left hand side of each of these equations over here. So I’m going to rewrite
the first equation. We have a plus b plus
c is equal to 180. And then this equation,
if we subtract 4a from both sides of this
equation, if we subtract 4a we have negative 4a plus
b is equal to negative 50. And then this equation
right over here, if we subtract a
from both sides we get negative a plus c
is equal to negative 40. I just subtract a
from both sides. So we now want to
eliminate variables. And we already have—
this third equation here is only in
terms of a and c. This is only in
terms of a and b. And this first one is
in terms of a, b, and c. So if we could— let’s see, this
is already in terms of a and c. If we could turn these
first two equations, if we can use the information
in these first two equations to end up with an
equation that’s only in terms of
a and c, then we could use whatever
we end up with along with this third equation
right over here, and we’ll have a system of two
equations with two unknowns. So let’s do that. So if we wanted to just end up
with an equation only in terms of a and c using
these first two, we would want to
eliminate the b’s. So we could multiply
one of these equations times negative 1, and one
of these positive b’s will turn into a negative b. So let’s do that. Let’s multiply this first
equation over here, let’s multiply it times negative 1. So it’ll become
negative a minus b minus c is equal
to negative 180. And then we have this
green equation right over here, which is really just
this equation, just rearranged. So we have negative 4a plus
b is equal to negative 50. And now we can add
these two equations. Actually, let me do
that in the other color, just so you see where
that’s coming from. I’ll do it in that green color. So this is negative 4a plus
b is equal to negative 50. We can add these two up now. And we get negative a
minus 4a is negative 5a. The b’s cancel out. We have a minus c is equal
to negative 180 minus 50 is negative 230. So now using these
top two equations we have an equation only
in terms of a and c. We have another equation
only in terms of a and c. And it looks like if we
add them together their c’s will cancel out. So let me just rewrite
this equation over here. And you have to be
careful that you’re using all of the equations. Otherwise you’ll do
a circular argument. You have to be careful that over
here this first equation came from these two over here. Now I want to combine that
with this third constraint, s constraint that’s not already
baked into this equation right over here. So we have negative a plus
c is equal to negative 40. And we add these two equations. Negative 5a minus
a is negative 6a. The c’s cancel out. And then you have
negative 230 minus 40. This is equal to negative 270. We can divide both
sides by negative 6. And we get a is equal
to negative 270 over 6. Let me see how many times—
let me see something. 270 is divisible
by both 3 and 2. So it should be divisible by 6. So let me just divide it. The negative signs
obviously will— a negative divided by a negative is
going to be a positive. And if we take 6 into 270,
6 goes into 27 4 times. 4 times 6 is 24. We subtract. We get 3, bring down the 0. 6 goes into 30 five times. So we get a is equal to 45. Now let’s look at
the other ones. We can substitute back
in to solve for c. c is equal to a
minus 40 degrees. So that is equal to— let
me write it right over here in yellow— so c
is equal to 45 minus 40, which is equal to 5 degrees. So, so far we have a
is equal to 45 degrees, c is equal to 5 degrees. And then you could
substitute into either one of these other ones
to figure out b. We could use this one
right over here in green. B is equal to 4a minus 50. So b is going to be equal to
4 times 45 is— let’s see, 2 times 45 is 90. So 4 times 45 is 180. So it’s going to
be 180 minus 50, by this equation
right over here, which is equal to 130 degrees. So we get b is equal
to 130 degrees. And then we can— let
me write over here. So a is equal to 45. So if I wanted to
draw this triangle it would actually look
something like this. a is a 45 degree angle, b is a
130 degree angle, and c is 5. So it’ll look
something like this. It will look
something like this, where this is a at 45
degrees, b is 135 degrees, and then c is 5 degrees. And you can verify
that it works. One, you can just
add up the angles. 45 plus 5 is 50. Oh, sorry, this isn’t 135. It’s 130. We solved it right over here. It’s 130 and this is 5. So when you add them
all up— 45 plus 130 plus 5— that does
indeed equal 180 degrees. 45 plus 5 is 50 plus 130. So this does
definitely equal 180. So it meets our
first constraint, Then on our second
constraint, b needs to be equal to 4a minus 50. Well, 4 times a is 180
minus 50 is 130 degrees. So it meets our
second constraint. And then our third constraint,
c is a minus 40 degrees. Well, a is 45, c is 5. So if you subtract 40 from
45 you get 5, which is c. So it meets all of our
constraints and we are done.

Related Pages
Systems of Equations — Graphical Method
Solving Equations
More Algebra Lessons
More Lessons for Grade 9 Math
Math Worksheets

Systems of Equations — Word Problems

Example:
A rental car company charges a flat daily fee plus a charge for each mile driven. A car rented for 5 days and driven for 300 miles costs $178, while a car rented for four days and driven for 550 miles costs $197. Find the daily fee and mileage charge.

How to solve word problem using a system of equations with 2 variables?

Example:
At a baseball game, 212 cups of coffee were sold, and $489 was collected. If small coffees sold for $2 each and large coffees sold for $3 each, how many of each size was sold?

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Systems of Equations word problems

Example:
As a birthday gift, Zoey gave her niece an electronic piggy bank that displays the total amount of money in the bank as well as the total number of coins.
After depositing some number of nickels and quarters only, the display read:
Money $2.00
Number of coins: 16
How many nickels and quarters did Joey put in the bank?

How to solve a word problem using a system of 3 equations with 3 variable?

Example:
At a store, Mary pays $34 for 2 pounds of apples, 1 pound of berries and 4 pounds of cherries. Tom Pays $35 for 3 pounds of apples, 2 pounds of berries, and 2 pounds of cherries. Lee Pays $49 for 5 pounds of apples, 3 pounds of berries, and 2 pounds of cherries. What is the price per pound for apples, for berries, and for cherries?

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Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
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Introduction to Systems	Distance Word Problem
Solving Systems by Graphing	Which Plumber Problem
Solving Systems with Substitution	Geometry Word Problem
Solving Systems with Linear Combination or Elimination	Work Problem
Types of Equations	Three Variable Word Problem
Systems with Three Equations	The “Candy” Problem
Algebra Word Problems with Systems:	Right Triangle Trigonometry Systems Problem
Investment Word Problem	Inequality Word Problem (in Linear Programming section)
Mixture Word Problems	More Practice

Note that we solve Algebra Word Problems without Systems here, and we solve systems using matrices in the Matrices and Solving Systems with Matrices section here.

Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is (y=mx+b). Let’s say we have the following situation:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50.  You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?  

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “(j)”) and how many dresses we want to buy (let’s say “(d)”). Always write down what your variables will be:

Let (j=) the number of jeans you will buy
Let (d=) the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

English	Math	Explanation
“You really, really want to take home 6 items of clothing because you need that many.”	(j+d=6) (Number of Items)	If you add up the pairs of jeans and dresses, you want to come up with 6 items.
“… you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50.”	(25j+50d=200) (Money)	This one’s a little trickier. Use easier numbers if you need to: if you buy 2 pairs of jeans and 1 dress, you spend (left( {2times $25} right)+left( {1times $50} right)). Now you can put the variables in with their prices, and they have to add up to $200.

Now we have the 2 equations as shown below. Notice that the (j) variable is just like the (x) variable and the (d) variable is just like the (y). It’s easier to put in (j) and (d) so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for 2 here. The cool thing is to solve for 2 variables, you typically need 2 equations, to solve for 3 variables, you need 3 equations, and so on. That’s easy to remember, right?

We need to get an answer that works in both equations; this is what we’re doing when we’re solving; this is called solving simultaneous systems, or solving system simultaneously. There are several ways to solve systems; we’ll talk about graphing first.

Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or (x/y) combinations) that make the equation work. In systems, you have to make both equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later). The points of intersections satisfy both equations simultaneously. 

Put these equations into the (y=mx+b) ((d=mj+b)) format, by solving for the (d) (which is like the (y)):

(displaystyle j+d=6;text{ },text{ }text{solve for }d:text{ }d=-j+6text{ })

(displaystyle 25j+50d=200;text{ },,text{solve for }d:text{ }d=frac{{200-25j}}{{50}}=-frac{1}{2}j+4)

Now graph both lines:

Solving Systems using Graph

Explanation

To graph, solve for the “(y)” value (“(d)” in our case) to use the slope-intercept method, or keep the equations as is and use the cover-up, or intercept method. The easiest way to graph the second equation is the intercept method; when we put 0 in for “(d)”, we get 8 for the “(j)” intercept; when we put 0 in for “(j)”, we get 4 for the “(d)” intercept. We can do this for the first equation too, or just solve for “(d)” to see that the slope is (-1) and the (y)-intercept is (6). The two graphs intercept at the point ((4,2)). This means that the numbers that work for both equations are 4 pairs of jeans and 2 dresses!

We can also use our graphing calculator to solve the systems of equations:

Graphing Calculator Instructions	Screens
(displaystyle begin{array}{c}j+d=6text{ }\25j+50d=200end{array}) Solve for (y,left( d right)) in both equations. Push (Y=) and enter the two equations in ({{Y}_{1}}=) and ({{Y}_{2}}=), respectively. Note that we don’t have to simplify the equations before we have to put them in the calculator. Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and/or “ZOOM 0” (ZoomFit) to make sure you see the lines crossing in the graph. (You can also use the WINDOW button to change the minimum and maximum values of your (x)- and (y)-values.)
To get the point of intersection, push “2nd TRACE” (CALC), and then either push 5, or move cursor down to intersect. You should see “First curve?” at the bottom. Then push ENTER. Now you should see “Second curve?” and then press ENTER again. Now you should see “Guess?”. Push ENTER one more time, and you will get the point of intersection on the bottom! Pretty cool!

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the Exponents and Radicals in Algebra section. Also, there are some examples of systems of inequality here in the Linear Inequalities section.

Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. You discover a store that has all jeans for $25 and all dresses for $50.  You really, really want to take home 6 items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole $200 (tax not included)?

Below are our two equations, and let’s solve for “(d)” in terms of “(j)” in the first equation. Then, let’s substitute what we got for “(d)” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

Steps Using Substitution	Notes
(displaystyle begin{array}{c}j+d=text{ }6;,,,,d=6-j\25j+50d=200end{array}) (displaystyle begin{array}{c}25j+50(6-j)=200\25j+300-50j=200\-25j=-100,,\j=4,\d=6-j=6-4=2end{array})	Solve for (d): (displaystyle d=6-j). Plug this in for (d) in the second equation and solve for (j). When you get the answer for (j), plug this back in the easier equation to get (d): (displaystyle d=6-(4)=2). The solution is ((4,2)).

We could buy 4 pairs of jeans and 2 dresses. Note that we could have also solved for “(j)” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

Steps Using Substitution	Notes
(displaystyle begin{array}{c}color{#800000}{begin{array}{c}37x+4y=124,\x=4,end{array}}\\37(4)+4y=124\4y=124-148\4y=-24\y=-6end{array})	This one is actually easier: we already know that (x=4). Now plug in 4 for the second equation and solve for (y). The solution is ((4,-6)).

Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “(y=)” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and/or Division Property of Equality that we saw here in the Types of Numbers and Algebraic Properties section:

(displaystyle begin{array}{c},,,3,,=,,3\underline{{+4,,=,,4}}\,,,7,,=,,7end{array})

(displaystyle begin{array}{l},,,12,=,12\,underline{{-8,,=,,,8}}\,,,,,4,,=,,4end{array})

(displaystyle begin{array}{c}3,,=,,3\4times 3,,=,,4times 3\12,,=,,12end{array})

(displaystyle begin{array}{c}12,,=,,12\frac{{12}}{3},,=,,frac{{12}}{3}\4,,=,,4end{array})

If we have a set of 2 equations with 2 unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

Linear Elimination Steps

Notes

(displaystyle begin{array}{c}color{#800000}{begin{array}{c}j+d=6text{ }\25j+50d=200end{array}}\\,left( {-25} right)left( {j+d} right)=left( {-25} right)6text{ }\,,,,-25j-25d,=-150,\,,,,,underline{{25j+50d,=,200}}text{ }\,,,0j+25d=,50\\25d,=,50\d=2\\d+j,,=,,6\,2+j=6\j=4end{array})

Since we need to eliminate a variable, we can multiply the first equation by –25. Remember that we need to multiply every term (anything separated by a plus, minus, or (=)  sign) by the –25. Then we add the two equations to get “(0j)” and eliminate the “(j)” variable (thus, the name “linear elimination”). We then solve for “(d)”. Now that we get (d=2), we can plug in that value in the either original equation (use the easiest!) to get the other variable. The solution is ((4,2)):  (j=4) and (d=2).

We could buy 4 pairs of jeans and 2 dresses.

Here’s another example:

Linear Elimination Steps

Notes

(displaystyle begin{array}{l}color{#800000}{{2x+5y=-1}},,,,,,,text{multiply by –}3\color{#800000}{{7x+3y=11}}text{ },,,,,,,text{multiply by }5end{array}) (displaystyle begin{array}{l}-6x-15y=3,\,underline{{35x+15y=55}}text{ }\,29x,,,,,,,,,,,,,,,=58\,,,,,,,,,,,,,x=2\,,,,,,,,,,,,,,,\2(2)+5y=-1\,,,,,,4+5y=-1\,,,,,,,,,,,,,,,5y=-5\,,,,,,,,,,,,,,,,,y=-1end{array})

Since we need to eliminate a variable, we can multiply the first equation by –3 and the second one by 5. There are many ways to do this, but we want to make sure that either the (x) or (y) will be eliminated when adding the two equations.  (We could have also picked multiplying the first by –7 and the second by 2). We then get the second set of equations to add, and the (y)’s are eliminated.  Solving for (x), we get (x=2). Now we can plug in that value in either original equation (use the easiest!) to get the other variable. The solution is ((2,-1)).

Types of equations

In the example above, we found one unique solution to the set of equations. Sometimes, however, for a set of equations, there are no solutions (when lines are parallel) or an infinite number of solutions or infinitely many solutions (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is at least one solution, the equations are consistent equations, since they have a solution. When there is only one solution, the system is called independent, since they cross at only one point. When equations have infinite solutions, they are the same equation, are consistent, and are called dependent or coincident (think of one just sitting on top of the other).

When equations have no solutions, they are called inconsistent equations, since we can never get a solution. 

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

Systems of Equations Calculator Screens	Notes
	(displaystyle begin{array}{l}y=-x+4\y=-x-2end{array}) Notice that the slope of these two equations is the same, but the (y)-intercepts are different. In this situation, the lines are parallel, as we can see from the graph. These types of equations are called inconsistent, since there are no solutions. If we were to “solve” the two equations, we’d end up with “(4=-2)”; no matter what values we give to (x) or (y), (4) can never equal (-2). Thus, there are no solutions. The symbol (emptyset ) is sometimes used for no solutions; it is called the “empty set”.
	(displaystyle x+y=6,,,,,,,text{or},,,,,,,y=-x+6) (displaystyle 2x+2y=12,,,,,,,text{or},,,,,,,y=frac{{-2x+12}}{2}=-x+6) Sometimes we have a situation where the system contains the same equations even though it may not be obvious. See how we may not know unless we actually graph, or simplify them? These types of equations are called dependent or coincident since they are one and the same equation and they have an infinite number of solutions, since one “sits on top of” the other. Since they have at least one solution, they are also consistent. If we were to “solve” the two equations, we’d end up with “(6=6)”, and no matter what values we give to (x) or (y), (6) always equals (6). Thus, there are an infinite number of solutions (infinitely many), but (y) always has to be equal to (-x+6). We can also write the solution as ((x,-x+6)).

Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for $20 and we managed to get $60 more to spend! Now we have a new problem. To spend the even $260, how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly 10 total items (Remember that jeans cost $25 each and dresses cost $50 each).

Let’s let (j=) the number of pair of jeans, (d=) the number of dresses, and (s=) the number of pairs of shoes we should buy. So far, we’ll have the following equations:

(displaystyle begin{array}{c}j+d+s=10text{ }\25j+text{ }50d+,20s=260end{array})

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: We want twice as many pairs of jeans as pairs of shoes. Now, since we have the same number of equations as variables, we can potentially get one solution for the system of equations. Here are the three equations:

(displaystyle begin{array}{c}j+d+s=10text{ }\25j+50d+,20s=260\j=2send{array})

Note that when we say “we have twice as many pairs of jeans as pair of shoes”, it doesn’t translate that well into math. We can think in terms of real numbers, such as if we had 8 pairs of jeans, we’d have 4 pairs of shoes. Then it’s easier to put it in terms of the variables.

We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an infinite number of solutions: (4=4)  (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are no solutions: (5=2) (variables are gone and two numbers are left and they don’t equal each other).

And another note: equations with three variables are represented by planes, not lines (you’ll learn about this in Geometry). If all the planes crossed in only one point, there is one solution, and if, for example, any two were parallel, we’d have no solution, and if, for example, two or three of them crossed in a line, we’d have an infinite number of solutions.

Let’s solve our system:     (displaystyle begin{array}{c}j+d+s=10text{ }\25j+text{ }50d+20s=260\j=2send{array}):

Solving Systems Steps

Notes

(displaystyle begin{array}{c}j+d+s=10text{ }\25j+50d+20s=260\j=2send{array}) (displaystyle begin{array}{c}2s+d+s=10,,,,,,,,,,Rightarrow ,,,,,,,,,,,,,,3s+d=10\25(2s)+50d+,20s=260,,,,,,Rightarrow ,,,,70s+50d=260end{array}) (displaystyle begin{array}{l}-150s-50d=-500\,,,,,underline{{,,70s+50d=,,,,260}}\,,-80s,,,,,,,,,,,,,,,,=-240\,,,,,,,,,,,,,,,,,,,s=3\\3(3)+d=10;,,,,,d=1,\j=2s=2(3);,,,,,,j=6end{array})

Use substitution since the last equation makes that easier. We’ll substitute (2s) for (j) in the other two equations and then we’ll have 2 equations and 2 unknowns. We then multiply the first equation by –50 so we can add the two equations to get rid of the (d). We could have also used substitution again. First, we get that (s=3), so then we can substitute this in one of the 2 equations we’re working with. Now we know that (d=1), so we can plug in (d) and (s) in the original first equation to get (j=6). The solution is ((6,1,3)).

We could buy 6 pairs of jeans, 1 dress, and 3 pairs of shoes.

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

(displaystyle begin{align}5x-6y-,7z,&=,7\6x-4y+10z&=,-34\2x+4y-,3z,&=,29end{align})

Solving Systems Steps

Notes

(displaystyle begin{array}{l}5x-6y-,7z,=,,7\6x-4y+10z=,-34\2x+4y-,3z,=,29,end{array})    (displaystyle begin{array}{l}6x-4y+10z=-34\underline{{2x+4y-,3z,=,29}}\8x,,,,,,,,,,,,,+7z=-5end{array}) (require{cancel} displaystyle begin{array}{l}cancel{{5x-6y-7z=7}},,,,,,,,,,,,,,20x-24y-28z,=,28,\cancel{{2x+4y-,3z,=29,,}},,,,,,,,underline{{12x+24y-18z=174}}\,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,32x,,,,,,,,,,,,,,-46z=202end{array}) (displaystyle begin{array}{l},,,cancel{{8x,,,+7z=,-5}},,,,,-32x,-28z=,20\32x,-46z=202,,,,,,,,,,,,underline{{,,32x,-46z=202}}\,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-74z=222\,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,z=-3end{array}) (displaystyle begin{array}{l}32x-46(-3)=202,,,,,,,,,,,,,x=frac{{202-138}}{{32}}=frac{{64}}{{32}}=2\\5(2)-6y-7(-3)=7,,,,,,,,y=frac{{-10+-21+7}}{{-6}}=4end{array})

We first pick any 2 equations and eliminate a variable; we’ll use equations 2 and 3 since we can add them to eliminate the (y). We then use 2 different equations (one will be the same!) to also eliminate the (y); we’ll use equations 1 and 3. To eliminate the (y), we can multiply the first by 4, and the second by 6. Now we use the 2 equations we’ve just created without the (y)’s and solve them just like a normal set of systems. We can multiply the first by –4 to eliminate the (x)’s to get the (z), which is –3. We can then get the (x) from either of the equations we just worked with. Since we have the (x) and the (z), we can use any of the original equations to get the (y). The solution is ((2,4,-3)).

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the Algebra Word Problems section, but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

• If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!
• If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!

Here are some problems:

Investment Word Problem

Investment Word Problem

Solution

Suppose Lindsay’s mom invests $10,000, part at 3%, and the rest at 2.5%, in interest bearing accounts. The totally yearly investment income (interest) is $283. How much did Lindsay’s mom invest at each rate?

Define a variable, and look at what the problem is asking. Use two variables: let (x=) the amount of money invested at 3%, and (y=) the amount of money invested at 2.5%. The yearly investment income or interest is the amount that we get from the yearly percentages. (This is the amount of money that the bank gives us for keeping our money there.) To get the interest, multiply each percentage by the amount invested at that rate. Add these amounts up to get the total interest. We have two equations and two unknowns. The total amount ((x+y)) must equal $10000, and the interest ((.03x+.025y)) must equal $283: (displaystyle begin{array}{c}x,+,y=10000\.03x+.025y=283end{array})          (displaystyle begin{array}{c}y=10000-x\.03x+.025(10000-x)=283\,,,.03x,+,250,-.025x=283\,.005x=33;,,,,x=6600,,\,,y=10000-6600=3400end{array}) Turn the percentages into decimals: move the decimal point two places to the left. Substitution is the easiest way to solve. Lindsay’s mom invested $6600 at 3% and $3400 at 2.5%.

We also could have set up this problem with a table:

	Amount	Turn % to decimal	Total
Amount at 3%	(x)	(.03)	(.03x)	Multiply across
Amount at 2.5%	(y)	(.025)	(.025y)	Multiply across
Total	(10000)		(283)	Do Nothing Here
	Add Down: (x+y=10000)	Do Nothing Here	Add Down: (.03x+.025y = 283) and solve the system

Mixture Word Problems

Here’s a mixture word problem. With mixture problems, remember if the problem calls for a pure solution or concentrate, use 100% (if the percentage is that solution) or 0% (if the percentage is another solution).

Mixture Word Problem

Solution

Two types of milk, one that has 1% butterfat, and the other that has 3.5% of butterfat, are mixed.  How many liters of these two different kinds of milk are to be mixed together to produce 10 liters of low-fat milk, which has 2% butterfat?

(Note that we did a similar mixture problem using only one variable here in the Algebra Word Problems section.) First define variables for the number of liters of each type of milk. Let (x=) the number of liters of the 1% milk, and (y=) the number of liters of the 3.5% milk. Use a table again: Amount Turn % to decimal Total 1% Milk (x) (.01) (.01x) Multiply across 3.5% Milk (y) (.035) (.035y) Multiply across Total (10) (.02) (10(.02)=.2) Multiply across Add Down: (x+y=10) Do Nothing Here Add Down: (.01x+.035y=.2) and solve the system We can also set up mixture problems with the type of figure below. We add up the terms inside the box, and then multiply the amounts in the boxes by the percentages above the boxes, and then add across. This will give us the two equations. Let’s do the math (use substitution)! (displaystyle begin{array}{c}x,,+,,y=10\.01x+.035y=10(.02)end{array})          (displaystyle begin{array}{c},y=10-x\.01x+.035(10-x)=.2\.01x,+,.35,,-,.035x=.2\,-.025x=-.15;,,,,,x=6\,y=10-6=4end{array}) We would need 6 liters of the 1% milk, and 4 liters of the 3.5% milk.

Here’s another mixture problem:

Mixture Word Problem with Money

Solution

A store sells two different types of coffee beans; the more expensive one sells for $8 per pound, and the cheaper one sells for $4 per pound. The beans are mixed to provide a mixture of 50 pounds that sells for $6.40 per pound. How much of each type of coffee bean should be used to create 50 pounds of the mixture?

First define variables for the number of pounds of each type of coffee bean. Let (x=) the number of pounds of the $8 coffee, and (y=) the number of pounds of the $4 coffee. Use a table again: Amount Cost Total $8 Coffee Beans (lbs) (x) (8) (8x) Multiply across $4 Coffee Beans (lbs) (y) (4) (4y) Multiply across Total (50) (6.40) (50(6.4)=320) Multiply across Add Down: (x+y=50) Do Nothing Here Add Down: (8x+4y=320) and solve the system Let’s do the math (use substitution)! (displaystyle begin{array}{c}x+y=50\8x+4y=50left( {6.4} right)end{array})                   (displaystyle begin{array}{c}y=50-x\8x+4left( {50-x} right)=320\8x+200-4x=320\4x=120,;,,,,x=30\y=50-30=20\8x+4y=50(6.4)end{array}) We would need 30 pounds of the $8 coffee bean, and 20 pounds of the $4 coffee bean. See how similar this problem is to the one where we use percentages?

Distance Word Problem:

Here’s a distance word problem using systems; distance problems have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the Algebra Word Problems section, there’s an example of a Parametric Distance Problem here in the Parametric Equations section.

Distance Word Problem

Solution

Lia walks to the mall from her house at 5 mph. 10 minutes later, Lia’s sister Megan starts riding her bike at 15 mph (from the same house) to the mall to meet Lia. They arrive at the mall the same time. How far is the mall from the sisters’ house? How long did it take Megan to get there?

Remember always that (text{distance}=text{rate}times text{time}). It’s difficult to know how to define the variables, but usually in these types of distance problems, we want to set the variables to time, since we have rates, and we’ll want to set distances equal to each other in this case (the house is always the same distance from the mall). (Sometimes we’ll need to add the distances together instead of setting them equal to each other.) Let (L) equal the how long (in hours) it will take Lia to get to the mall, and (M) equal to how long (in hours) it will take Megan to get to the mall. The rates of the Lia and Megan are 5 mph and 15 mph respectively. (Usually a rate is “something per something”). Lia’s time is Megan’s time plus (displaystyle frac{{10}}{{60}}=frac{1}{6},,,,(L=M+frac{1}{6})), since Lia left 10 minutes earlier than Megan (convert minutes to hours by dividing by 60 – try real numbers to see this). Use the distance formula for each of them separately, and then set their distances equal, since they are both traveling the same distance (house to mall). Then use substitution to solve the system for Megan’s time: after dividing both sides by 5, multiply both sides by 6 to get rid of the fractions.       (begin{array}{c}L=M+frac{1}{6};,,,,,5L=15M\5left( {M+frac{1}{6}} right)=15M;,,,M+frac{1}{6}=3M,,\6M+1=18M;,,,12M=1;,,M,,=frac{1}{{12}},,text{hr}text{.}\D=15left( {frac{1}{{12}}} right)=1.25,,text{miles}end{array}) Megan’s time is (displaystyle frac{1}{{12}}) of any hour, which is 5 minutes. The distance to the mall is rate times time, which is 1.25 miles.

Which Plumber Problem

Many word problems you’ll have to solve have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the initial charge will be the (boldsymbol {y})-intercept, and the rate will be the slope. Here is an example:

“Which Plumber” Systems Word Problem

Solution

Michaela’s mom is trying to decide between two plumber companies to fix her sink. The first company charges $50 for a service call, plus an additional $36 per hour for labor. The second company charges $35 for a service call, plus an additional $39 per hour of labor. At how many hours will the two companies charge the same amount of money?

The money spent depends on the plumber’s set up charge and number of hours, so let (y=) the total cost of the plumber, and (x=) the number of hours of labor. Again, set up charges are typical (boldsymbol {y})-intercepts, and rates per hour are slopes. The total price of the plumber’s house call will be the initial or setup charge, plus the number of hours ((x)) at the house times the price per hour for labor. To get the number of hours when the two companies charge the same amount of money, we just put the two (y)’s together and solve for (x) (substitution, right?): First plumber’s total price:  (displaystyle y=50+36x) Second plumber’s total price:  (displaystyle y=35+39x) (displaystyle 50+36x=35+39x;,,,,,,x=5) Here’s what a graph would look like: At 5 hours, the two plumbers will charge the same. At this time, the (y)-value is 230, so the total cost is $230. Note that, in the graph, before 5 hours, the first plumber will be more expensive (because of the higher setup charge), but after the first 5 hours, the second plumber will be more expensive. Thus, the plumber would be chosen based on how many hours Michaela’s mom thinks the plumber will be there.

Geometry Word Problem:

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

Geometry Systems Word Problem

Solution

Two angles are supplementary. The measure of one angle is 30 degrees smaller than twice the other.  Find the measure of each angle.

From Geometry, we know that two angles are supplementary if their angle measurements add up to 180 degrees (and remember also that two angles are complementary if their angle measurements add up to 90 degrees). Define the variables and turn English into Math. Let (x=) the first angle, and (y=) the second angle. We really don’t need to worry at this point about which angle is bigger; the math will take care of itself. (x) plus (y) must equal 180 degrees by definition, and also (x=2y-30) (Remember the English-to-Math chart?) Solve, using substitution: (displaystyle begin{array}{c}x+y=180\x=2y-30end{array}) (displaystyle begin{array}{c}2y-30+y=180\3y=210;,,,,,,,,y=70\x=2left( {70} right)-30=110end{array}) The larger angle is 110°, and the smaller is 70°. Let’s check our work: The two angles do in fact add up to 180°, and the larger angle (110°) is 30° less than twice the smaller (70°).

See – these are getting easier! Here’s one that’s a little tricky though:

Work Problem: 

Let’s do a “work problem” that is typically seen when studying Rational Equations (fraction with variables in them) and can be found here in the Rational Functions, Equations and Inequalities section.

Note that there’s also a simpler version of this problem here in the Direct, Inverse, Joint and Combined Variation section.

Work Word Problem (Systems)

Solution

8 women and 12 girls can paint a large mural in 10 hours. 6 women and 8 girls can paint it in 14 hours.  Find the time to paint the mural, by 1 woman alone, and 1 girl alone.

Let’s let (w=) the part of the job by 1 woman in 1 hour, and (g=) the part of the job by 1 girl in 1 hour. We have 10 hours with 8 women and 12 girls that paint the mural (do 1 job), and 14 hours with 6 women and 8 girls that paint the mural (do 1 job). Since (w=) the part of the job that is completed by 1 woman in 1 hour, then (8w=) the amount of the job that is completed by 8 women in 1 hour. Also, if (8w=) the amount of the job that is completed by 8 women in 1 hour, (10times 8w) is the amount of the job that is completed by 8 women in 10 hours. Similarly, (10times 12g) is the amount of the job that is completed by 12 girls in 10 hours. Add these two amounts and we get (displaystyle 10left( {8w+12g} right)), which will be the whole job. Use the same logic for the 6 women and 8 girls to paint the mural in 14 hours. The whole job is 1 (this is typical in work problems), and we can set up two equations that equal 1 to solve the system. Use linear elimination to solve the equations; it gets a little messy with the fractions, but we can get it! (displaystyle begin{array}{c}10left( {8w+12g} right)=1,text{ or }8w+12g=frac{1}{{10}}\,14left( {6w+8g} right)=1,text{ or },6w+8g=frac{1}{{14}}end{array}) (displaystyle begin{array}{c}text{Use elimination:}\left( {-6} right)left( {8w+12g} right)=frac{1}{{10}}left( {-6} right)\left( 8 right)left( {6w+8g} right)=frac{1}{{14}}left( 8 right)\cancel{{-48w}}-72g=-frac{3}{5}\cancel{{48w}}+64g=frac{4}{7},\,-8g=-frac{1}{{35}};,,,,,g=frac{1}{{280}}end{array})             (begin{array}{c}text{Substitute in first equation to get }w:\,10left( {8w+12cdot frac{1}{{280}}} right)=1\,80w+frac{{120}}{{280}}=1;,,,,,,w=frac{1}{{140}}\g=frac{1}{{280}};,,,,,,,,,,,w=frac{1}{{140}}end{array}) The answers we get is the part of the job that is completed by 1 woman or girl in 1 hour, so to get how long it would take them to do a whole job, we have to take the reciprocal. (Think about it; if we could complete (displaystyle frac{1}{3}) of a job in an hour, we could complete the whole job in 3 hours). Thus, it would take one of the women 140 hours to paint the mural by herself, and one of the girls 280 hours to paint the mural by herself.

Three Variable Word Problem:

Let’s do one more with three equations and three unknowns:

Three Variable Word Problem

Solution

A florist is making 5 identical bridesmaid bouquets for a wedding. She has $610 to spend (including tax) and wants 24 flowers for each bouquet. Roses cost $6 each, tulips cost $4 each, and lilies cost $3 each. She wants to have twice as many roses as the other 2 flowers combined in each bouquet. How many roses, tulips, and lilies are in each bouquet? The trick is to put real numbers in to make sure you’re doing the problem correctly, and also make sure you’re answering what the question is asking!

Look at the question being asked to define our variables: Let (r=) the number of roses, (t=) the number of tulips, and (l=) the number of lilies in each bouquet. Put the money terms together, and also the counting terms together: (begin{array}{l}5left( {6r+4t+3l} right)=610,,,text{(price of each flower times number of flowers x }5text{ bouquets= total price)}\,,,,,,,,,r=2(t+l)text{ },,,,,,,,,,text{ (two times the sum of the other two flowers = number of roses)}\,,,,,,r+t+l=24text{ },,,,,,,,,text{(total number of flowers in each bouquet is }24text{)}end{array}) Use substitution and put (r) from the middle equation in the other equations. Then, use linear elimination to put those two equations together: multiply the second by –5 to eliminate the (l). We typically have to use two separate pairs of equations to get the three variables down to two! (begin{array}{c}6r+4t+3l=122\r=2left( {t+l} right)\,r+t+l=24\\6left( {2t+2l} right)+4t+3l=122\,12t+12l+4t+3l=122\16t+15l=122\\left( {2t+2l} right)+t+l=24\3t+3l=24end{array})           (displaystyle begin{array}{c},16t+15l=122\,,,,,,,,,cancel{{3t+3l=24}}\,,,,underline{{-15t-15l=-120}}\,,,,,t,,,,,,,,,,,,,,,,,=2\16left( 2 right)+15l=122;,,,,l=6\\r=2left( {2+6} right)=16\,,,,,,,,,,r=16,,,,t=2,,,,l=6end{array}) We get (t=2). Solve for (l) in this same system, and (r) by using the value we got for (t) and (l) (most easily in the second equation at the top). Thus, for one bouquet, we’ll have 16 roses, 2 tulips, and 6 lilies. If we had solved for the total number of flowers, we would have had to divide each number by 5.

The “Candy” Problem

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

More Unknowns Than Variables Problem

Solution

Sarah buys 2 pounds of jelly beans and 4 pounds of chocolates for $4.00. She then buys 1 pound of jelly beans and 4 pounds of caramels for $3.00. She also buys 1 pound of jelly beans, 3 pounds of licorice and 1 pound of caramels for $1.50. How much will it cost to buy 1 pound of each of the four candies?

Look at the question being asked to define our variables: Let (j=) the cost of 1 pound of jelly beans, (o=) the cost of 1 pound of chocolates, (c=) the cost of 1 pound of caramels, and (l=) the cost of 1 pound of licorice. Here is the system of equations: (begin{array}{c}2j+4o=4\j+4c=3\j+3l+1c=1.5\text{Want: }j+o+c+lend{array}) Wait! Something’s not right since we have 4 variables and 3 equations. But note that they are not asking for the cost of each candy, but the cost to buy all 4! Maybe the problem will just “work out” so we can solve it; let’s try and see. From our three equations above (using substitution), we get values for (o), (c) and (l) in terms of (j). (displaystyle begin{align}o=frac{{4-2j}}{4}=frac{{2-j}}{2},,,,,,,,,c=frac{{3-j}}{4},\j+3l+1left( {frac{{3-j}}{4}} right)=1.5\4j+12l+3-j=6\,l=frac{{6-3-3j}}{{12}}=frac{{3-3j}}{{12}}=frac{{1-j}}{4}end{align})               (require{cancel} displaystyle begin{align}j+o+c+l&=j+frac{{2-j}}{2}+frac{{3-j}}{4}+frac{{1-j}}{4}\&=cancel{j}+1-cancel{{frac{1}{2}j}}+frac{3}{4}cancel{{-frac{j}{4}}}+frac{1}{4}cancel{{-frac{j}{4}}}=2end{align}) When we substitute back in the sum (text{ }j+o+c+l), all in terms of (j), our (j)’s actually cancel out, which is very unusual! We can’t really solve for all the variables, since we don’t know what (j) is. But we can see that the total cost to buy 1 pound of each of the candies is $2. Pretty cool!

There are more Systems Word Problems in the Matrices and Solving Systems with Matrices section, Linear Programming section, and Right Triangle Trigonometry section.

Understand these problems, and practice, practice, practice!

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For Practice: Use the Mathway widget below to try a Systems of Equations problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve by Substitution or Solve by Addition/Elimination to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Algebraic Functions, including Domain and Range – you’re ready! 

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• mathrm{Lauren’s:age:is:half:of:Joe’s:age.:Emma:is:four:years:older:than:Joe.:The:sum:of:Lauren,:Emma,:and:Joe’s:age:is:54.:How:old:is:Joe?}

• mathrm{Kira:went:for:a:drive:in:her:new:car.:She:drove:for:142.5:miles:at:a:speed:of:57:mph.:For:how:many:hours:did:she:drive?}

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• mathrm{Two:men:who:are:traveling:in:opposite:directions:at:the:rate:of:18:and:22:mph:respectively:started:at:the:same:time:at:the:same:place.:In:how:many:hours:will:they:be:250:apart?}

• mathrm{If:2:tacos:and:3:drinks:cost:12:and:3:tacos:and:2:drinks:cost:13:how:much:does:a:taco:cost?}

Frequently Asked Questions (FAQ)

• How do you solve word problems?

• To solve word problems start by reading the problem carefully and understanding what it’s asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.

• How do you identify word problems in math?

• Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.

• Is there a calculator that can solve word problems?

• Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.

• What is an age problem?

• An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as «x years ago,» «in y years,» or «y years later,» which indicate that the problem is related to time and age.

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After you finish this lesson, view all of our Pre-Algebra lessons and practice problems. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

Solving Word Problems

To solve a word problem using a system of equations, it is important to;
– Identify what we don’t know
– Declare variables
– Use sentences to create equations

An example on how to do this:

Mary and Jose each bought plants from the same store. Mary spent $188 on 7 cherry trees and 11 rose bushes. Jose spent $236 on 13 cherry trees and 11 rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

Cost of a cherry tree:
Cost of a rose bush:
7 cherry trees and 11 rose bushes = $188

The y-values cancel each other out, so now you are left with only x-values and real numbers.

Then, you plug in your x-value into an original equation in order to find the y-value.

Cost of a cherry tree: $8
Cost of a rose bush: $12

Example 1

Three coffees and a cupcake cost a total of dollars. Two coffees and four cupcake cost a total of dollars. What is the individual price for a single coffee and a single cupcake?

Let’s solve this by following steps.

1. What we don’t know:

Cost of a single coffee

Cost of single cupcake

2. Declare variables:

Cost of a single coffee=

Cost of single cupcake=

3. Use sentences to create equations.

Three coffees and a cupcake cost a total of dollars.

Two coffees and four cupcake cost a total of dollars.

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

Now, solve for the value of using the first equation.

Let’s solve the value of by substituting the value of to the bottom equation.

Distribute to each terms inside the parenthesis

Combine like terms

Now, let’s isolate the by subtracting on both sides.

Then divide both sides by ,

And we’ll have

Then, let’s plug the value of into one equation to get the value of .

Cost of a single coffee=

Cost of single cupcake=

Example 2

The senior class at High School A rented and filled vans and buses with students. High School B rented and filled vans and bus with students. Every van had the same number of students in it as did the buses. Find the number of students in each van and in each bus.

Let’s solve this by following steps.

1. What we don’t know:

Students in each van

Students in each bus

2. Declare variables:

Students in each van=

Students in each bus=

3. Use sentences to create equations.

High School A rented and filled 8 vans and 8 buses with 240 students.

High School B rented and filled 4 vans and 1 bus with 54 students.

Now, we have a system of equations:

Let’s solve for one of the variables in one of the equations and then use that to substitute into the other.

Now, solve for the value of using the second equation.

Let’s solve the value of by substituting the value of to the bottom equation.

Distribute to each terms inside the parenthesis

Combine like terms

Now, let’s isolate the by subtracting on both sides.

Then divide both sides by ,

And we’ll have

Then, let’s plug the value of into one equation to get the value of .

Students in each van=

Students in each bus=

Video-Lesson Transcript

To solve a word problem using system of equations, it is important to:
1. Identify what we don’t know
2. Declare variables.
3. Use sentences to create equations.

Let’s have an example:

Mary and Jose each bought plants from the same store. Mary spent on cherry trees and rose bushes. Jose spent on cherry trees and rose bushes. Find the cost of one cherry tree and the cost of one rose bush.

Let’s solve this by following steps above.

1. What we don’t know:
cost of a cherry tree
cost of a rose bush

2. Declare variables:
cost of a cherry tree :
cost of a rose bush :

3. Use sentences to create equations.

For Mary:
cherry trees and rose bushes

For Jose:
cherry trees and rose bushes

Now, we have a system of equations

We can solve this by process of substitution, elimination or fraction.

Since the value of is the same for both equations, let’s do the process of elimination.

First, let’s multiply the first equation by

Here we’ll have a negated equation

Let’s do the process of elimination now

We’ll have

Then, let’s isolate by dividing both sides by

Now, we have

Remember, our declared variable?

cost of a cherry tree :

Since

Now we can say that

cost of a cherry tree :

Now, let’s solve for the value of by getting one equation and plugging the value of .

Let’s use the first equation to plug in

Let’s isolate by subtracting on both sides of the equation

Then divide by

And we get

Now, we know that cost of a rose bush is .

This problem has to do with mowing lawns, someone who’s a contractor of a landscaping business works in a way where they’re ordering different people to do different jobs. In this situation Brooks and Jeremy are asked to mow a lawn together. Here’s what we’re going to do to solve it. Brooks can mow a lawn in 4 hours, Jeremy could mow the same lawn twice as fast. How long would it take them working together? Before we do that let’s figure out how fast Jeremy mows, he goes twice as fast as Brooks so if Brooks takes 4 hours then Jeremy is going to take 2 hours right? Does that make sense? Twice as fast means half as much time so Jeremy is going to be 2 hours to do the job by himself. 2 hours alone, okay so what I’m going to be doing is writing fractions for each of these guys and then adding those fractions together to see how quickly they could do this job working together.
Brooks could mow the lawn in 4 hours that means every hour he does 1 fourth of the job does that make sense think about it I’m going to say it again. Brooks could mow the lawn in 4 hours that means he does 1 fourth of the job in every hour. Similarly Jeremy could mow the lawn in 2 hours. So every hour he’s going to do half of the job. This represents 1 hour together one half plus one fourth when you find a common denominator instead of one half I’m going to write that as two fourths, one fourth plus 2 fourths equals 3 fourths. What that means is that in 1 hour working together they could do 3 fourths of the jobs. Does that make sense we say it again 1 hour working together, they could do 3 fourths of the job, they’re almost done not all the way done.
The way to figure out how long it would take them and to complete the job all the way would be to use the formula fraction they could do together times sum amount of time that they’re working equals 1 total job, 1 complete job. All we need to do here is solve for x. Well the way to undo multiplying by a fraction is to either divide by the fraction, or multiply by the reciprocal. I’m going to multiply both sides here by 4 thirds, 4 thirds these cancel out so I’m left with x equals 4 thirds. What that tells me is it would take them 4 thirds of an hour working together in order to mow this lawn.
Before you start doing your homework problems make sure you’re not doing the meatball answer, the meatball answer for this kind of a problem would be well if Brooks mows it in 4 hours, Jeremy mows it in 2 hours the meatball would say it takes them 6 hours working together what it does it you guys when they work together it goes faster, that’s why again the answer that’s faster than either guy working alone. Keep that straight in your head and don’t pick the silly multiple choice question. Answer.

Presentation on theme: «6.3 Systems Word Problems A2 U1 §3.5 Systems Word Problems.»— Presentation transcript:

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6.3 Systems Word Problems A2 U1 §3.5 Systems Word Problems

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Kristin spent $131 on shirts
Kristin spent $131 on shirts. Fancy shirts cost $28 and plain shirts cost $15. If she bought a total of 7 then how many of each kind did she buy?

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There are 13 animals in the barn. Some are chickens and some are pigs
There are 13 animals in the barn. Some are chickens and some are pigs. There are 40 legs in all. How many of each animal is there?

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A farmhouse shelters 10 animals. Some are pigs and some are ducks
A farmhouse shelters 10 animals. Some are pigs and some are ducks. Altogether there are 36 legs. How many of each animal is there?

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A class of 195 students went on a field trip
A class of 195 students went on a field trip. They took 7 vehicles, some cars and some buses. Find the number of cars and the number of buses they took if each car holds 5 students and each bus hold 45 students.

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At Elisa’s Printing Company LLC there are two kinds of printing presses: Model A which can print 70 books per day and Model B which can print 55 books per day. The company owns 14 total printing presses and this allows them to print 905 books per day. How many of each type of press do they have?

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Molly’s Custom Kitchen Supplies sells handmade forks and spoons
Molly’s Custom Kitchen Supplies sells handmade forks and spoons. It costs the store $1.70 to buy the supplies to make a fork and $1.30 to buy the supplies to make a spoon. The store sells forks for $5.60 and spoons for $5.40. Last April Molly’s Custom Kitchen Supplies spent $37.90 on materials for forks and spoons. They sold the finished products for a total of $ How many forks and how many spoons did they make last April?

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A company borrowed 600,000 for one year
A company borrowed 600,000 for one year. Part was at 8% interest; part was at 10% interest rate. How much at each rate if annual interest is $53,000?

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A community center sells a total of 301 tickets for a basketball game
A community center sells a total of 301 tickets for a basketball game. An adult ticket costs $3. A student ticket costs $1. The sponsors collect $487 in ticket sales. Find the number of each type of ticket sold.

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6.4 Systems Word Problems

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A farmer raises wheat and soybeans on 215 acres
A farmer raises wheat and soybeans on 215 acres. If he wants to plant 31 more acres in wheat than soybeans, how many acres of each should be planted?

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A grocer plans to mix candy that sells for $2 a pound with candy that sells for $4 a pound to get a mixture that sells for $2.75 a pound.  How much of each type of candy should she use if she wants a 40-pound mixture?

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A total of 512 tickets were sold for the school play
A total of 512 tickets were sold for the school play. The number of student tickets sold was three times the number of adult tickets sold. How many adult tickets were sold?

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Amy took a test that had a math section and an English section
Amy took a test that had a math section and an English section. She earned a total score of 88. Her math score was 6 points higher than her English score. What were her scores on each section?

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April buys eight books for $44
April buys eight books for $44. Paperback books cost $4 and hardback cost $8. How much of each type did she buy?

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At the afternoon matinee movie 3 adult tickets and 4 child tickets cost $41, and 4 adult tickets and 3 child tickets cost $43. Find the cost of an adult ticket and the cost of a child ticket.

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Copper that was 63% pure was melted together with copper that was 90% pure to make 18 kilograms of an alloy that was 75% pure. How many kilograms of each kind were used?

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Tickets to a movie cost $7. 25 for adults and $5. 50 for students
Tickets to a movie cost $7.25 for adults and $5.50 for students. A group of friends purchased 8 tickets for $52.75.  Write a system of equations to represent this situation.

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A farmhouse shelters 11 animals. Some are goats and some are ducks
A farmhouse shelters 11 animals. Some are goats and some are ducks. Altogether there are 34 legs. How many of each animal is there?

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All 231 students in the Math Club went on a field trip
All 231 students in the Math Club went on a field trip. Some students’ rode in vans that hold 7 students each and some students rode in buses that hold 25 students each. How many of each type of vehicle did they use if there were 15 vehicles total?

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A chemist wants to make 200mL of acid solution with a concentration of 48%. He wants to make this from two solutions with 60% and 40% concentrations respectively. How much of each solution should he use?

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To make 10 kg of aluminum alloy with 60% aluminum, a scientist wants to use two metals with 45% and 70% aluminum content respectively. How much of each metal should she use?

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