Question 66341This question is from textbook
: This is also supposed to be solved using the substitution method.
r/2 + s/3 = 1
r/4 + 2s/3 = -1
This question is from textbook
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! r/2 + s/3 = 1
r/4 + 2s/3 = -1
:
First, get rid of those denominators, mult the 1st eq by 6, the 2nd eq by 12
3r + 2s = 6
3r + 8s = -12
:
Use the 1st eq for substitution for s
3r + 2s = 6
2s = 6 - 3r
divide by 2
s = 3 - 1.5r
:
Substitute (3-1.5r) of s in the 2nd equation:
3r + 8s = -12
3r + 8(3-1.5r) = -12
3r + 24 - 12r = - 12
3r - 12r = -12 - 24
Get rid of all those negatives, mult eq by -1
-3r + 12r = 12 + 24
9r = 36
r = 36/9
r = 4
:
Use 3r + 2s = 6 to find s
3(4) + 2s = 6
12 + 2s = 6
2s = 6 - 12
2s = -6
s = -3
:
Check solutions in the original eq:
r/2 + s/3 = 1
4/2 + (-3/3) = 1
2 - 1 = 1
:
You can check our solutions in the original 2nd equation, to be sure
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