SOLUTION: This is the addition method, you have to multiply before you add to get the solution set. I thought that the first problem was set up as 1(x+y)=(48)1 then I got lost??? x+y

Question 155406This question is from textbook Elementary and Intermediate Algebra
: This is the addition method, you have to multiply before you add to get the solution set.
I thought that the first problem was set up as 1(x+y)=(48)1 then I got lost???
x+y=48
12x+14y=628 This question is from textbook Elementary and Intermediate Algebra

Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
If you started with:
x + y = 48
12x + 14y = 628
.
Start by reordering:
12x + 14y = 628
x + y = 48
.
Now, multiply both sides of the second equation by -12:
12x + 14y = 628
-12(x + y) = -12(48)
.
12x + 14y = 628
-12x + (-12y) = -576
.
Now, we're ready to combine equation 1 with equation 2:
12x + 14y = 628
-12x + (-12y) = -576
-----------------------
2y = 52
y = 26 (solution for y)
.
Use the above definition of y and plug it back into:
x + y = 48
to solve for 'x'.
.
x + y = 48
x + 26 = 48
x = 48-26
x = 22 (solution for x)
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