well, does your solution work?
4*4-9 = 16-9 = 7
Nope
4x-y-9=0
4x-(x-3)-9=0
You messed up right here
It would be easier just to use the two values of y to find x:
4x-9 = x-3
3x = 6
x = 2
y = -1
4*2-9 = -1
2-3 = -1
That's better...
Solve each system algebraically
Y=4x-9 and y=x-3
This is what I have done
4x-y-9=0
x-3+4x-9=0
3x=9+3
3x=12
X=4
Than I substituted x in one equation
y=4-3
y=1
Please let me know if I am doing this correctly.
Thx
1 answer
1 answer