Solve the following system of equations:?

Question

9x² + y² = 9
3x - y = 3

Answer 1

OK, use substitution for this one.
Isolate y on the second equation so that you get y = 3x - 3.
Now substitute that in for the y term in the first equation.
9x² + (3x-3)² = 9
9x² + 9x² - 18x + 9 = 9
18x² - 18x = 0
18x(x - 1) = 0
So 18x = 0
or
x - 1 = 0
That means that there are two solutions for x: x = 0 and x = 1.
Now put each of those back into the originals to get the corresponding y coordinates.
Your solutions: (0,-3) and (1,0)

That makes sense since the first equation represents a circle and the second a line. These can intersect at two points just like a line and a parabola can.

Answer 2

9x² + y² = 9
3x - y = 3 --> y = 3x - 3 replace to the top equation

9x^2 + 9x^2 - 18x + 9 = 9
--> x^2 - x = 0 --> x = 0 or x = 1
--> y = -3 or y = 0

Answer 3

9x^2 + y^2 = 9 -----------eqn(1)

3x - y = 3 ----------------eqn(2)

from eqn (2) y = 3x - 3 = 3(x-1)

substitute y value in eqn(1)

9x^2 + [3(x-1)]^2 = 9

9x^2 + 9(x^2 + 1 -2x) = 9

9x^2 + 9x^2 + 9 - 18x = 9

18x^2 - 18x = 0

x^2 - x = 0

x(x-1) = 0

x = 0 or 1

substituting in eqn (2)

case1: 0 - y = 3 : y = -3

case 2: 3 - y = 3 : y = 0

so (x,y) = (0 ,-3 ) or (1, 0)

Answer 4

y = 3x-3 --> y^2 = 9x^2 -18x + 9
9x^2 +9x^2 -18x + 9 = 9
2x^2 -2x +1 = 1
2x^2-2x = 0
2x(x -1) = 0
x = 0 or 1
y = -3 or 0
(0,-3) and (1,0) are solutions

Answer 5

3x-y=3 <=> y=3x-3

So, then,

9x^2 + (3x-3)^2 = 9
9x^2 + 9x^2 - 18x + 9 =9
18x^2 - 18x =0
x(18x-18)=0
x=0 or 18x-18=0
x=0 or x = 1

And, now, back to the first one:
If x=0, then y=-3
If x=1, then y= 0

The solutions are (0,-3) and (1,0)