To solve the system of equations using the substitution method, we have:
- \( y = x^2 - x - 40 \) (Equation 1)
- \( y = -3x - 5 \) (Equation 2)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 - x - 40 = -3x - 5 \]
Now, we'll rearrange the equation:
- Move all terms to one side of the equation:
\[ x^2 - x + 3x - 40 + 5 = 0 \]
This simplifies to:
\[ x^2 + 2x - 35 = 0 \]
- Next, we factor the quadratic equation:
\[ (x + 7)(x - 5) = 0 \]
This gives us the solutions for \( x \):
\[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \]
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]
- Now we can find the corresponding values of \( y \) for both values of \( x \):
a. For \( x = 5 \):
Using Equation 2:
\[ y = -3(5) - 5 = -15 - 5 = -20 \]
So one solution is \( (5, -20) \).
b. For \( x = -7 \):
Again using Equation 2:
\[ y = -3(-7) - 5 = 21 - 5 = 16 \]
So the other solution is \( (-7, 16) \).
- The solutions to the system are:
- \( (5, -20) \)
- \( (-7, 16) \)
Therefore, the correct answer is C. (5, -20) and (-7, 16).
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