Asked by Dennis
how do i solve 3x+2y=40,x+y=18 i would like the work shown.. thanks.
Answers
Answered by
Reiny
I would use substitution
from the 2nd equation....
x + y = 18 ----> y = 18-x
sub into the 1st
3x + 2(18-x) = 40
3x + 36 - 2x = 40
x = 40-36 = 4
sub that into y = 18-x = 18-4 = 14
x=4 , y = 14
or.....
double the 2nd
2x + 2y = 36
3x + 2y = 40 , now subtract them
-x + 0 = -4
x = 4
into the 2nd:
x+y=18
4+y=18
y = 18-4 = 14
from the 2nd equation....
x + y = 18 ----> y = 18-x
sub into the 1st
3x + 2(18-x) = 40
3x + 36 - 2x = 40
x = 40-36 = 4
sub that into y = 18-x = 18-4 = 14
x=4 , y = 14
or.....
double the 2nd
2x + 2y = 36
3x + 2y = 40 , now subtract them
-x + 0 = -4
x = 4
into the 2nd:
x+y=18
4+y=18
y = 18-4 = 14
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