Question

Solve the system of equations.
y=x2+48x+31
y=40x+40
Write the coordinates in exact form. Simplify all fractions and radicals.
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Answer 1

To solve the system of equations given by

1. \( y = x^2 + 48x + 31 \)
2. \( y = 40x + 40 \)

we can set these two equations equal to each other:

\[
x^2 + 48x + 31 = 40x + 40
\]

Next, we can rearrange the equation to one side:

\[
x^2 + 48x + 31 - 40x - 40 = 0
\]

This simplifies to:

\[
x^2 + 8x - 9 = 0
\]

Now, we will factor the quadratic equation. We are looking for two numbers that multiply to \(-9\) and add to \(8\). The numbers are \(9\) and \(-1\). Therefore, we can factor the equation as:

\[
(x + 9)(x - 1) = 0
\]

Setting each factor to zero gives us:

\[
x + 9 = 0 \quad \Rightarrow \quad x = -9
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Now we have two values for \(x\). Next, we find the corresponding \(y\) values for each \(x\) using the linear equation \(y = 40x + 40\).

1. For \(x = -9\):

\[
y = 40(-9) + 40 = -360 + 40 = -320
\]

So one solution is \((-9, -320)\).

2. For \(x = 1\):

\[
y = 40(1) + 40 = 40 + 40 = 80
\]

So the other solution is \((1, 80)\).

Thus, the coordinates of the solutions are:

\[
\boxed{(-9, -320)} \quad \text{and} \quad \boxed{(1, 80)}
\]