How much did each item cost?

Question

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40. How much did each item cost?

Answer 1

You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40. How much did each item cost?

First let’s define a shirt as X and pants as Y!

In the first scenario, the equation will be 3X + Y =$32

In the second scenario, the equation will be 2X + 3Y = $40

Now because we have two different variables (X & Y) let’s solve for one of them (Ex: Y) to simplify the equations!

3X + Y= $32—> subtract 3X on both sides of the equation

-3X. - 3X

Net result: Y= 32- 3X

Now we can insert this into the second equation 2X + 3Y = $40 in order to solve for X variable

2X + 3(32-3X) = $40

2X + 96 - 9X = $40 —> simplify the equation

-7X + 96 = $40 —> substrate 96 from both sides

-7X = -56 —> divide both sides by -7

X= 8

Now reinsert 8= X into 3X + Y = $32

3(8) + Y = $32

24+ Y = $32 —> substract 24 from both sides

y = 8

ANSWER: X= 8, Y= 8

Answer 2

Let's use the variable, s, represent the price of one shirt and the variable, p, represent the price of one pair of pants.

We can write the following equations to represent the two purchase amounts:

3s + 1p = $32

2s + 3p = 40

Now we can solve these two equations with two unknowns to find the price of each item.

Multiply the first equation by -3 to find the price of the shirt first:

3s + 1p = $32 becomes -9s - 3p = -96

Now combine this "new" equation with the second one:

-9s - 3p = -96

2s + 3p = 40

-7s = -56

s = $8

The cost of one shirt is $8.

Now substitute $8 for s in either of the original equations to find the price p of one pair of pants (Let's use the second equation.)

2s + 3p = 40

2 • 8 + 3p = 40

16 + 3p = 40

16 + 3p - 16 = 40 - 16

3p = 24

p = $8

The price of a pair of pants is $8 also.

Answer 3

Hi Joe,

The way I would approach this is by first writing the sentences as equations. Let's let shirts be s and pants be p.

Our two equations are now:

3s+1p=32 and 2s+3p=40

So now we have to combine the equations in some way to get only 1 variable in an equation and solve for that variable. My approach to this is going to be by substituting the p's from one equation to another. To do this, we need the p's to have the same coefficient. We can do this by multiplying the first equation by 3.

9s+3p=96 and we can rearrange it to say 3p=96-9s

Now we can substitute this equation into 3p of the second equation.

2s+(96-9s)=40 By simplifying we get 56=7s and then s=8

Now we can plug this into either of the original equation to get the value of p.

3(8)+1p=32 -> 24+p=32 so p=8

We can verify that this solution works by plugging it into the second equation:

2(8)+3(8)=5(8)=40