Question

ALG 2... PLEASE HELP! ILL GIVE A MEDAL! ^.^
What are the x-coordinates of the solutions to this system of equations?

x2 + y2 = 36
y = x - 6

2 and 0

6 and 2

6 and -2

6 and 0

Answer 1

Since the second equation is already solved for y, use the substitution method.

Answer 2

welp, i know its not 6 and 2...

Answer 3

@kmeis002
In this problem, you can't just plug in points to see if they work, because of the wording of the question.

Answer 4

To solve that original syatem of equations:

First note that we take y = x - 6 and substiutute that into y in the first equation,

So the FIRST equation becomes x^2 + (x-6)^2 = 36
Simplify the above, you get x^2 + x^2 - 12x + 36 = 36
2x^2 - 12 x = 0
2x(x - 6) = 0
x = 0 or x = 6
If x = 0, y = -6; if x = 6, y = 0

Final solutions: x = 0 and y = -6 AND x = 6 and y= 0

Answer 5

Ah I see, thank you @mathstudent55, I had misread.

Answer 6

\(x^2 + y^2 = 36\)
\(y = x - 6\)

\(x^2 + \color{red}{y}^2 = 36\)
\(y = \color{red}{x - 6}\)

\(x^2 + \color{red}{(x - 6)}^2 = 36\)

That is the substitution step.Now solve the bottom equation for x.

Answer 7

i still cant get it.. .-.

Answer 8

Now we solve that quadratic equation.
First, we square the binomial x - 6:

\(x^2 + (x - 6)^2 = 36\)

\(x^2 + x^2 - 12x + 36 = 36\)

\(2x^2 - 12x + 36 = 36\)

\(2x^2 - 12x = 0\)

\(x^2 - 6x = 0\)

\(x(x - 6) = 0\)

x = 0 or x - 6 = 0

x = 0 or x = 6

Answer 9

ohh, okay thank you!!

Answer 10

You're welcome.