QUESTION IMAGE
Question
solve the following system of equations algebraically:
$y = x^2 - 10x + 9$
$y = -x - 9$
answer
two solutions
and
Step1: Set equations equal
Since $y = x^2 - 10x + 9$ and $y = -x - 9$, substitute to get:
$$x^2 - 10x + 9 = -x - 9$$
Step2: Rearrange to standard quadratic form
Bring all terms to one side:
$$x^2 - 10x + x + 9 + 9 = 0$$
$$x^2 - 9x + 18 = 0$$
Step3: Factor the quadratic
Find two numbers that multiply to 18 and add to -9:
$$(x - 3)(x - 6) = 0$$
Step4: Solve for $x$
Set each factor equal to 0:
$x - 3 = 0 \implies x = 3$
$x - 6 = 0 \implies x = 6$
Step5: Find corresponding $y$ values
For $x=3$: $y = -(3) - 9 = -12$
For $x=6$: $y = -(6) - 9 = -15$
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$(3, -12)$ and $(6, -15)$