We are tasked with solving the system of equations algebraically:

1. $y = -3x^2 - 18x - 47$
2. $17 = 3x - y$

Step 1: Substitute for $y$ from the first equation into the second equation

From equation (1), we already have $y = -3x^2 - 18x - 47$. Let's substitute this expression for $y$ into equation (2):

$17 = 3x - (-3x^2 - 18x - 47)$

Simplifying the right-hand side:

$17 = 3x + 3x^2 + 18x + 47$

Now combine like terms:

$17 = 3x^2 + (3x + 18x) + 47$

$17 = 3x^2 + 21x + 47$

Step 2: Move all terms to one side to set the equation equal to zero

$0 = 3x^2 + 21x + 47 - 17$

$0 = 3x^2 + 21x + 30$

Step 3: Simplify and solve the quadratic equation

Divide the entire equation by 3 to make the equation simpler:

$0 = x^2 + 7x + 10$

Step 4: Factor the quadratic equation

The quadratic equation $x^2 + 7x + 10$ can be factored as:

$(x + 5)(x + 2) = 0$

Step 5: Solve for $x$

Set each factor equal to zero:

1. $x + 5 = 0$ gives $x = -5$
2. $x + 2 = 0$ gives $x = -2$

Step 6: Find the corresponding $y$-values

Now, substitute these $x$-values back into the first equation $y = -3x^2 - 18x - 47$ to find the corresponding $y$-values.

For $x = -5$:

$y = -3(-5)^2 - 18(-5) - 47$ $y = -3(25) + 90 - 47$ $y = -75 + 90 - 47 = -32$

So, one solution is $(-5, -32)$.

For $x = -2$:

$y = -3(-2)^2 - 18(-2) - 47$ $y = -3(4) + 36 - 47$ $y = -12 + 36 - 47 = -23$

So, the other solution is $(-2, -23)$.

Final Answer:

The solutions to the system of equations are the points:

$(-5, -32) \quad \text{and} \quad (-2, -23)$

Would you like any further details or explanations?

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1. How do you solve quadratic equations using the quadratic formula?
2. What are other methods to solve systems of equations algebraically?
3. How does factoring work for solving quadratic equations?
4. Can every quadratic equation be factored easily?
5. How does substitution help in solving systems of equations?

Tip: Always check your solutions by plugging them back into the original system to verify correctness.