Let's solve the given system of equations:

1. $y = -30x + 9$
2. $y = x^2 - 29x - 33$

We will solve it by substitution.

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Step 1: Set the two equations equal

Since both equations are equal to $y$, we equate them:

$-30x + 9 = x^2 - 29x - 33$

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Step 2: Rearrange into standard quadratic form

Move all terms to one side:

$x^2 - 29x + 30x - 33 - 9 = 0$

Simplify:

$x^2 + x - 42 = 0$

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Step 3: Solve the quadratic equation

We use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 1$, and $c = -42$.

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-42)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 168}}{2}$ $x = \frac{-1 \pm \sqrt{169}}{2}$ $x = \frac{-1 \pm 13}{2}$

Thus, the two solutions are:
$x = \frac{-1 + 13}{2} = 6 \quad \text{and} \quad x = \frac{-1 - 13}{2} = -7$

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Step 4: Find the corresponding $y$-values

We now substitute $x = 6$ and $x = -7$ into the first equation $y = -30x + 9$.

• For $x = 6$: $y = -30(6) + 9 = -180 + 9 = -171$

• For $x = -7$: $y = -30(-7) + 9 = 210 + 9 = 219$

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Step 5: Write the solutions

The solutions to the system are:

$(6, -171) \quad \text{and} \quad (-7, 219)$

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Would you like further details or have any questions?

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5 Related Questions

1. How do you graphically solve systems of nonlinear equations?
2. What other methods exist for solving systems of equations besides substitution?
3. How does the quadratic formula work for non-real solutions?
4. What is the discriminant in a quadratic equation, and how does it affect the solutions?
5. How can systems of equations be solved using matrices?

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Tip

When equating two functions, always rearrange to form a standard equation (like a quadratic) for easier solving.