To solve the system of equations by substitution, we follow these steps:

Step 1: Write down the given equations.

$y = 4x^2 - 15x - 1 \quad \text{(Equation 1)}$ $y = x - 17 \quad \text{(Equation 2)}$

Step 2: Substitute the expression for $y$ from Equation 2 into Equation 1.

Since $y = x - 17$, substitute this into Equation 1: $x - 17 = 4x^2 - 15x - 1$

Step 3: Rearrange the equation to set it equal to zero.

$4x^2 - 15x - 1 - x + 17 = 0$ Simplify by combining like terms: $4x^2 - 16x + 16 = 0$

Step 4: Solve the quadratic equation.

Divide the entire equation by 4 to simplify: $x^2 - 4x + 4 = 0$ This is a perfect square trinomial: $(x - 2)^2 = 0$

Step 5: Solve for $x$.

$x - 2 = 0$ $x = 2$

Step 6: Find the corresponding $y$ value.

Substitute $x = 2$ back into Equation 2 to find $y$: $y = 2 - 17 = -15$

Step 7: Write the solution as an ordered pair.

The solution is $(2, -15)$.

Step 8: Verify the solution.

Substitute $x = 2$ into Equation 1 to verify: $y = 4(2)^2 - 15(2) - 1 = 16 - 30 - 1 = -15$ Since both equations give $y = -15$, the solution is correct.

Final Answer:

A. The solution is $(2, -15)$.

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. How would the solution change if the second equation was $y = 2x - 17$?
2. Can you factorize the quadratic expression $4x^2 - 16x + 16$ directly?
3. What happens if a system of equations leads to a quadratic that has no real roots?
4. How can you graphically interpret the solution to this system of equations?
5. If the equation was $y = 4x^2 - 15x + 3$, how many solutions would there be?

Tip: Always check your solutions by substituting back into the original equations. This helps verify that you didn’t make any algebraic errors.