To solve the given system of equations:

1. $y = x - 10$
2. $y = x^2 - 15x + 29$

We solve by substitution. Here's the step-by-step solution:

Step 1: Substitute $y = x - 10$ into $y = x^2 - 15x + 29$:

$x - 10 = x^2 - 15x + 29$

Step 2: Rearrange into standard quadratic form:

Move all terms to one side: $0 = x^2 - 15x + 29 - (x - 10)$

Simplify: $0 = x^2 - 15x - x + 29 + 10$ $0 = x^2 - 16x + 39$

Step 3: Solve the quadratic equation:

We solve $x^2 - 16x + 39 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -16$, $c = 39$.

Substitute into the formula: $x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(39)}}{2(1)}$ $x = \frac{16 \pm \sqrt{256 - 156}}{2}$ $x = \frac{16 \pm \sqrt{100}}{2}$ $x = \frac{16 \pm 10}{2}$

Step 4: Solve for $x$:

1. $x = \frac{16 + 10}{2} = \frac{26}{2} = 13$
2. $x = \frac{16 - 10}{2} = \frac{6}{2} = 3$

Thus, $x = 13$ and $x = 3$.

Step 5: Solve for $y$ using $y = x - 10$:

For $x = 13$: $y = 13 - 10 = 3$

For $x = 3$: $y = 3 - 10 = -7$

Step 6: Final Solution:

The solutions are: $(13, 3) \quad \text{and} \quad (3, -7)$

Verification:

1. For $(13, 3)$:
   • Substituting into $y = x^2 - 15x + 29$: $3 = 13^2 - 15(13) + 29 \quad \text{(True)}$
2. For $(3, -7)$:
   • Substituting into $y = x^2 - 15x + 29$: $-7 = 3^2 - 15(3) + 29 \quad \text{(True)}$

Both solutions are correct.

Final Answer:

The solutions are:
$(13, 3) \quad \text{and} \quad (3, -7)$

Do you need further clarification or verification? Let me know!

Related Questions:

1. How do you graph these two equations to visually verify the solutions?
2. What is the discriminant of the quadratic equation $x^2 - 16x + 39 = 0$, and what does it tell us?
3. How can you solve a system of equations using elimination instead of substitution?
4. What are the intersections of the line $y = x - 10$ with other parabolas?
5. How do you determine the vertex of the parabola $y = x^2 - 15x + 29$?

Tip:

For solving systems of equations, substitution works best when one equation is already solved for a variable.