You buy 3 shirts and 1 pair of pants for $32. Your friend buys 2 shirts and 3 pairs of pants for $40.

Answer 1

I like the substitution method to solve 2 variable equations. You move the variables to separate sides and use substitution to solve.

3s+p=32

2s+3p=40

First, isolate p in the first equation. You do this by subtracting 3s from both sides. That leaves you with p=32-3s

Next, substitute 32-3s for p in the second equation to solve. 2s+3(32-3s)=40

1. Following the PEMDAS model, start by multiplying each term in the parenthesis by 3
2. 2s+96-9s=40. Now, combine like terms. 2s and -9s become -7s
3. -7s+96=40. Further isolate by subtracting 96 from both sides
4. -7s=-56. Finally, divide both sides by -7
5. s=8

Now, substitute 8 for s in the first equation to solve for p.

1. 3(8)+p=32. Multiply 3 and 8
2. 24+p=32. Finally, subtract 24 from both sides
3. p=8

Check your work! We have concluded that both the shirts and the pants are $8. Plug those into the original equations to see if this is correct.

Does 3(8)+8 = 32?

Does 2(8)+3(8)=40?

Answer 2

Hi Kira G

First we need to know what your question is?

If your question is to find out the cost of each item, then you can use a system of equations with either substitution or elimination to solve your problem

Let s = shirt

Let p = pair of pants

You are given two situations you buy 3 shirts and a pair of pants for $32

3s + p = 32

Your friend buys 2 shirts and 3 pairs of pants for $40

2s + 3p = 40

You have to use the two equations to solve for the cost of a shirt and the cost of a pair of pants. I’d suggest elimination of p to get things started

3s + p = $32

2s + 3p = $40

If the cost of each item is what you need you can continue from here?

I hope this helps and if you have any questions just write back.