To solve the system of equations given, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Here are the equations:
1. [tex]\( 13x - y = 90 \)[/tex]
2. [tex]\( y = x^2 - x - 42 \)[/tex]
Let's solve this system step by step:
### Step 1: Express [tex]\( y \)[/tex] from the first equation
From the first equation, we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
13x - y = 90 \implies y = 13x - 90
\][/tex]
### Step 2: Substitute [tex]\( y \)[/tex] into the second equation
Next, substitute this expression for [tex]\( y \)[/tex] into the second equation:
[tex]\[
y = x^2 - x - 42 \implies 13x - 90 = x^2 - x - 42
\][/tex]
### Step 3: Simplify the equation
Rearrange the equation to form a standard quadratic equation:
[tex]\[
13x - 90 = x^2 - x - 42 \implies x^2 - x - 42 - 13x + 90 = 0 \implies x^2 - 14x + 48 = 0
\][/tex]
### Step 4: Solve the quadratic equation
We solve the quadratic equation [tex]\( x^2 - 14x + 48 = 0 \)[/tex]. This can be done by factoring:
[tex]\[
x^2 - 14x + 48 = (x - 6)(x - 8) = 0
\][/tex]
Thus, we have two potential solutions for [tex]\( x \)[/tex]:
[tex]\[
x = 6 \quad \text{or} \quad x = 8
\][/tex]
### Step 5: Find corresponding [tex]\( y \)[/tex] values
For [tex]\( x = 6 \)[/tex]:
[tex]\[
y = 13(6) - 90 = 78 - 90 = -12
\][/tex]
For [tex]\( x = 8 \)[/tex]:
[tex]\[
y = 13(8) - 90 = 104 - 90 = 14
\][/tex]
### Conclusion
The solutions to the system of equations are:
[tex]\[
(6, -12) \quad \text{and} \quad (8, 14)
\][/tex]
So, the answers in the boxes should be:
(6, -12) and (8, 14)
