SOLUTION: How do you solve this linear system with substitution? 8x+3y=0 -x-9y=92 All I have got to is... x=-9y-92 3y(-9y-92)=0 -6y-92=0 -6y=92 I dont think I did it right?

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: How do you solve this linear system with substitution? 8x+3y=0 -x-9y=92 All I have got to is... x=-9y-92 3y(-9y-92)=0 -6y-92=0 -6y=92 I dont think I did it right?      Log On


   

Question 186085This question is from textbook Algebra 1
: How do you solve this linear system with substitution?
8x+3y=0
-x-9y=92
All I have got to is...
x=-9y-92
3y(-9y-92)=0
-6y-92=0
-6y=92
I dont think I did it right?? This question is from textbook Algebra 1

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(1) 8x+%2B+3y+=+0
(2) -x+-+9y+=+92
Subtract 3y from both sides of (1)
(1) 8x+=+-3y
Divide both sides by -3
(1) y+=+-%288%2F3%29%2Ax
Multiply the right side by 3%2F3
(1) y+=+-%2824%2F9%29%2Ax
Substitute this in (2)
(2) -x+-+9y+=+92
(2) -x+-+9%2A%28-%2824%2F9%29%29%2Ax+=+92
(2) -x+%2B+24x+=+92
(2) 23x+=+92
(2) x+=+4
And, from
(1) 8x+%2B+3y+=+0
(1) 8%2A4+%2B+3y+=+0
(1) 3y+=+-32
(1) y+=+-%2832%2F3%29
check answers:
(1) 8%2A4+%2B+3%2A%28-%2832%2F3%29%29+=+0
(1) 32+-+32+=+0
OK
(2) -4+-+9%2A%28-%2832%2F3%29%29+=+92
(2) -4+%2B+96+=+92
(2) 92+=+92
OK