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

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...
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