You have a system of two equations:

1. $y - 3 = 3x$
2. $y + 11 = 50 - 10x + x^2$

We will solve this system step by step.

Step 1: Express $y$ in terms of $x$ from the first equation.

From the first equation:

$y - 3 = 3x$ $y = 3x + 3$

Step 2: Substitute $y = 3x + 3$ into the second equation.

Now, substitute this expression for $y$ into the second equation:

$y + 11 = 50 - 10x + x^2$

Substituting $y = 3x + 3$:

$(3x + 3) + 11 = 50 - 10x + x^2$ Simplifying:

$3x + 14 = 50 - 10x + x^2$

Step 3: Rearrange the equation.

Move all terms to one side:

$3x + 14 - 50 + 10x - x^2 = 0$

Simplifying further:

$13x - x^2 - 36 = 0$

Rearrange into standard quadratic form:

$-x^2 + 13x - 36 = 0$

Step 4: Solve the quadratic equation.

To make solving easier, multiply through by $-1$:

$x^2 - 13x + 36 = 0$

Now, solve the quadratic equation using the quadratic formula:

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

For the equation $x^2 - 13x + 36 = 0$, $a = 1$, $b = -13$, and $c = 36$:

$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}$ $x = \frac{13 \pm \sqrt{169 - 144}}{2}$ $x = \frac{13 \pm \sqrt{25}}{2}$ $x = \frac{13 \pm 5}{2}$

Thus, the two possible values for $x$ are:

$x = \frac{13 + 5}{2} = \frac{18}{2} = 9$ $x = \frac{13 - 5}{2} = \frac{8}{2} = 4$

Step 5: Find the corresponding $y$-values.

Now, substitute $x = 9$ and $x = 4$ into $y = 3x + 3$ to find the corresponding $y$-values.

• For $x = 9$:

$y = 3(9) + 3 = 27 + 3 = 30$

• For $x = 4$:

$y = 3(4) + 3 = 12 + 3 = 15$

Step 6: Conclusion.

The solutions to the system of equations are:

$(x, y) = (9, 30) \quad \text{or} \quad (x, y) = (4, 15)$

Let me know if you'd like further clarification or details on any part of the process!

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Here are 5 related questions to expand the concepts:

1. How would you solve a system of equations if both equations were quadratic?
2. What does the discriminant of a quadratic equation tell you about its solutions?
3. How do you solve a system where one equation is linear and the other is quadratic using graphing?
4. How can you verify the solutions algebraically after solving a system of equations?
5. What are some real-life situations that can be modeled using systems of equations?

Tip: When solving quadratic equations, always check for factorability before resorting to the quadratic formula—it can save time if the equation is easy to factor!