A certain circle can be represented by the following equation. X^2+y^2-6x-6y+2=0.
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So in this case, we have to replace the known value.
y=3
y=-2x+3
3=-2x+3
Then we leave our unknown value alone.
= x
In this case, our x value would be 0.
We check it...
3=-2(0)+3
3=0+3
3=3
So y=3 x=0
For the second one we have...
y=3x+2
y=-3x-4
For this we substitute the y in any of the equation...
3x+2=-3x-4
We move the unknown values to one side and the ones without unlown values to the other side...
3x+3x=-4-2
Then we solve
6x=-6
Then we leave the unknown value alone.
x=
Then solve for x.
x= -1
Then for our y value we return to one of the original equations and substitute the x value.
y=3x+2
y=3(-1)+2
y=-3+2
y=-1
y=-3x-4
y=-3(-1)-4
y=3-4
y=-1
So in this case we got that x= -1 and y= -1
y=3
y=-2x+3
3=-2x+3
Then we leave our unknown value alone.
In this case, our x value would be 0.
We check it...
3=-2(0)+3
3=0+3
3=3
So y=3 x=0
For the second one we have...
y=3x+2
y=-3x-4
For this we substitute the y in any of the equation...
3x+2=-3x-4
We move the unknown values to one side and the ones without unlown values to the other side...
3x+3x=-4-2
Then we solve
6x=-6
Then we leave the unknown value alone.
x=
Then solve for x.
x= -1
Then for our y value we return to one of the original equations and substitute the x value.
y=3x+2
y=3(-1)+2
y=-3+2
y=-1
y=-3x-4
y=-3(-1)-4
y=3-4
y=-1
So in this case we got that x= -1 and y= -1