To solve the system of equations given by
[tex]\[
\begin{cases}
x - y = 7 \\
x^2 + y = 149
\end{cases}
\][/tex]
we will use substitution and solve these step-by-step.
1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from the first equation:
[tex]\[
x - y = 7 \implies y = x - 7
\][/tex]
2. Substitute this expression for [tex]\( y \)[/tex] into the second equation:
[tex]\[
x^2 + (x - 7) = 149
\][/tex]
Simplify the equation:
[tex]\[
x^2 + x - 7 = 149
\][/tex]
Combine like terms:
[tex]\[
x^2 + x - 156 = 0
\][/tex]
3. Solve the quadratic equation:
[tex]\[
x^2 + x - 156 = 0
\][/tex]
To solve this quadratic equation, we can factorize it. We need to find two numbers that multiply to [tex]\(-156\)[/tex] and add up to [tex]\(1\)[/tex].
These numbers are [tex]\(13\)[/tex] and [tex]\(-12\)[/tex]. Hence, we factorize:
[tex]\[
x^2 + 13x - 12x - 156 = 0
\][/tex]
Group the terms:
[tex]\[
x(x + 13) - 12(x + 13) = 0
\][/tex]
Factor by grouping:
[tex]\[
(x - 12)(x + 13) = 0
\][/tex]
4. Find the zeroes of the factored equation:
[tex]\[
x - 12 = 0 \implies x = 12
\][/tex]
[tex]\[
x + 13 = 0 \implies x = -13
\][/tex]
5. Substitute these [tex]\( x \)[/tex]-values back into the expression [tex]\( y = x - 7 \)[/tex] to find the corresponding [tex]\( y \)[/tex]-values:
For [tex]\( x = 12 \)[/tex]:
[tex]\[
y = 12 - 7 = 5
\][/tex]
For [tex]\( x = -13 \)[/tex]:
[tex]\[
y = -13 - 7 = -20
\][/tex]
6. Write down the solutions:
The solutions to the system are:
[tex]\[
(x, y) = (12, 5) \quad \text{and} \quad (x, y) = (-13, -20)
\][/tex]
So, the system of equations has two solutions:
[tex]\[
(x, y) = (12, 5) \quad \text{and} \quad (x, y) = (-13, -20)
\][/tex]
