Question

question
find all solutions of the system of equations algebraically. write your solutions as coordinate points.
$y = -x^2 + 2x - 11$
$15 = 2x - y$
answer attempt 1 out of 2
two solutions
and

Explanation:

Step1: Solve the second equation for y

From \(15 = 2x - y\), we can rearrange it to get \(y = 2x - 15\).

Step2: Substitute y into the first equation

Substitute \(y = 2x - 15\) into \(y = -x^2 + 2x - 11\). So we have:
\(2x - 15 = -x^2 + 2x - 11\)

Step3: Simplify the equation

Subtract \(2x\) from both sides:
\(-15 = -x^2 - 11\)
Add \(x^2\) and 11 to both sides:
\(x^2 - 4 = 0\)
Factor the left side (difference of squares):
\((x - 2)(x + 2)=0\)

Step4: Solve for x

Set each factor equal to zero:
\(x - 2 = 0\) gives \(x = 2\)
\(x + 2 = 0\) gives \(x = -2\)

Step5: Find the corresponding y values

For \(x = 2\), substitute into \(y = 2x - 15\): \(y = 2(2)-15 = 4 - 15 = -11\)
For \(x = -2\), substitute into \(y = 2x - 15\): \(y = 2(-2)-15 = -4 - 15 = -19\)

Answer:

\((2, -11)\) and \((-2, -19)\)