We are tasked with solving the given system of equations using the substitution method. The system is:

$5x - y = -28 \quad \text{(1)}$ $y = x^2 - 4 \quad \text{(2)}$

Step 1: Substitute $y$ from equation (2) into equation (1).

From equation (2), we know: $y = x^2 - 4.$

Substitute this expression for $y$ in equation (1): $5x - (x^2 - 4) = -28.$

Simplify the equation: $5x - x^2 + 4 = -28.$

Combine like terms: $-x^2 + 5x + 4 = -28.$

Add $28$ to both sides: $-x^2 + 5x + 32 = 0.$

Multiply through by $-1$ to simplify: $x^2 - 5x - 32 = 0.$

Step 2: Solve the quadratic equation.

The equation is: $x^2 - 5x - 32 = 0.$

We solve this using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = -5$, and $c = -32$.

Substitute these values: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-32)}}{2(1)}.$

Simplify: $x = \frac{5 \pm \sqrt{25 + 128}}{2}.$

$x = \frac{5 \pm \sqrt{153}}{2}.$

Thus: $x = \frac{5 + \sqrt{153}}{2} \quad \text{or} \quad x = \frac{5 - \sqrt{153}}{2}.$

Step 3: Find $y$ for each $x$.

From equation (2), $y = x^2 - 4$. Substitute each $x$ value:

1. For $x = \frac{5 + \sqrt{153}}{2}$:

   • Compute $x^2$ and subtract 4 to find $y$.

2. For $x = \frac{5 - \sqrt{153}}{2}$:

   • Similarly, compute $y$.

Final Answer:

The solutions are:

$x = \frac{5 + \sqrt{153}}{2}, \, y = \left( \frac{5 + \sqrt{153}}{2} \right)^2 - 4$ and $x = \frac{5 - \sqrt{153}}{2}, \, y = \left( \frac{5 - \sqrt{153}}{2} \right)^2 - 4.$

Would you like me to simplify these further or calculate the approximate decimal values? Let me know!

Related Questions:

1. How do you solve a system of equations using the elimination method?
2. How can you solve a quadratic equation using factoring instead of the quadratic formula?
3. What are the graphical interpretations of solutions to a system of nonlinear equations?
4. How do you determine if a system of equations has no solution or infinitely many solutions?
5. What is the role of the discriminant in solving quadratic equations?

Tip:

When solving systems with substitution, always isolate one variable first to simplify substitution into the other equation.