Question 1179663: Sam bought 3 shirts and 2 pairs of pants for $85.50. Randy bought 4 shirts and 3 pairs of pants for $123. How much does one shirt cost? How much does one pair of pants cost?
Found 3 solutions by mananth, MathTherapy, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Sam bought 3 shirts and 2 pairs of pants for $85.50. Randy bought 4 shirts and 3 pairs of pants for $123. How much does one shirt cost? How much does one pair of pants cost?
x number of shirts
y number of pairs of pants
3.00 x + 2.00 y = 85.50
4.00 x + 3.00 y = 123.00 .............2
Eliminate y
multiply (1)by -3.00
Multiply (2) by 2.00
-9.00 x -6.00 y = -256.50
8.00 x 6.00 y = 246.00
Add the two equations
-1.00 x = -10.50
/ -1.00
x = 10.50
plug value of x in (1)
3.00 x + 2.00 y = 85.50
31.50 + 2.00 y = 85.50
2.00 y = 54.00
y = 27.00
Ans x = 11
y = 27
11 number of shirts
27 number of pairs of pants
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
Sam bought 3 shirts and 2 pairs of pants for $85.50. Randy bought 4 shirts and 3 pairs of pants for $123. How much does one shirt cost? How much does one pair of pants cost?
Let the cost of a shirt and pair of pants be S, and P, respectively
Then we get: 3S + 2P = 85.5 ------ eq (i)
Also, 4S + 3P = 123 ------ eq (ii)
Subtract eq (i) from eq (ii) to get: S + P = 37.5, or S = 37.5 - P
Substitute the value of S into eq (i) to get an equation in P.
From there, solve for P to get a pair of pants, or 
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
There are an endless number of ways to solve this problem using formal algebra; you have received two responses showing two of those ways.
I would certainly not use the method shown by tutor @Mananth. She has the strange habit of representing whole numbers using 2 decimal places.
The solution from tutor @MathTherapy is more straightforward.
I would start the way he did but then go a different direction with it.
Subtracting the two equations formed from the given information gives the result that the cost of one shirt and one pair of pants is $37.50:
Instead of solving that equation for one variable in terms of the other and finishing the problem using substitution, I would continue with elimination.
Informally, finishing the problem using elimination could go something like this:
one shirt and one pair of pants cost $37.50
so two shirts and two pairs of pants cost 2($37.50)=$75
but three shirts and two pairs of pants cost $85.50, so one shirt costs $85.50-$75=$10.50.
and then one pair of pants costs $37.50-$10.50=$27.
The formal algebra for finishing the problem by that path looks like this:
(1) x+y=$37.50 (from the two original equations)
(2) 2x+2y=2($37.50)=$75 (doubling (1))
(3) 3x+2y=$85.50 (one of the original equations)
(4) x=$85.50-$75=$10.50 (comparing (2) and (3))
(5) y=$37.50-$10.50=$27 (from (1) and (4))
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