SOLUTION: 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

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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): You can put this solution on YOUR website!

At one store, pairs of jeans and sweatshirts costs $,





.......eq.1


while pairs of jeans and sweatshirts costs $





..........eq.2

from eq.1 and eq.2, we have

........solve or













the cost one sweatshirt is $

so, answer is: D. $

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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