Question 1127607: At one store, 5 pairs of jeans and 2 sweatshirts costs $213, while 3 pairs of jeans and 4 sweatshirts costs $195. Find the cost one sweatshirt.
A. $21
B. $35
C. $33
D. $24
Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20850)   (Show Source):
Answer by MathTherapy(10553)  (Show Source):
You can put this solution on YOUR website! At one store, 5 pairs of jeans and 2 sweatshirts costs $213, while 3 pairs of jeans and 4 sweatshirts costs $195. Find the cost one sweatshirt.
A. $21
B. $35
C. $33
D. $24
Never do a problem like this the way the other person did. It's an extremely inefficient, very time-consuming, and error-prone way to solve a SIMPLE system-in-2-variable equation.
Let cost of a pair of jeans, and a sweatshirt be J, and S, respectively
Then we get: 5J + 2S = 213 --- eq (i)
Also, 3J + 4S = 195 ---------- eq (ii)
- 10J - 4S = - 426 -------- Multiplying eq (i) by - 2 ------ eq (iii)
- 7J = - 231 -------- Adding eqs (iii) and (ii)
J, or cost of a pair of jeans = 
5(33) + 2S = 213 ------ Substituting 33 for J in eq (i)
165 + 2S = 213
2S = 213 - 165
2S = 48
S, or cost of a sweatshirt = 
That's all!! Again, never follow anyone who advises you to do it the way she did, with those MONSTROUS fractions. Who wants to deal with those???
I can't believe these people are claiming to be helping others. Is this really help?
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Just posting this to emphasize the point made by tutor @MathTherapy....
When the given problem produces a system of two equations both of the form Ax+By=C, then a solution by elimination is FAR easier than a solution by substitution.
That should be clear by comparing his solution to the one by the other tutor....
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