https://r2.erweima.ai/mathbot/image/1724380372271m5167521.jpg

The problem in the image asks to solve the following system of equations algebraically:

1. \( y = x^2 + 10x - 5 \)
2. \( 2x = y - 5 \)

Let's solve the system step by step:

### Step 1: Express \( y \) in terms of \( x \) from the second equation.

From the second equation:

\[
2x = y - 5 \implies y = 2x + 5
\]

### Step 2: Substitute \( y \) from the second equation into the first equation.

Using the expression \( y = 2x + 5 \) in the first equation:

\[
2x + 5 = x^2 + 10x - 5
\]

### Step 3: Rearrange the equation to standard quadratic form.

\[
x^2 + 10x - 5 - 2x - 5 = 0 \implies x^2 + 8x - 10 = 0
\]

### Step 4: Solve the quadratic equation using the quadratic formula.

The quadratic formula is given by:

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

Here, \( a = 1 \), \( b = 8 \), and \( c = -10 \).

\[
x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-10)}}{2(1)}
\]
\[
x = \frac{-8 \pm \sqrt{64 + 40}}{2}
\]
\[
x = \frac{-8 \pm \sqrt{104}}{2}
\]
\[
x = \frac{-8 \pm 2\sqrt{26}}{2}
\]
\[
x = -4 \pm \sqrt{26}
\]

So the solutions for \( x \) are:

\[
x = -4 + \sqrt{26} \quad \text{and} \quad x = -4 - \sqrt{26}
\]

### Step 5: Find the corresponding values of \( y \).

Using \( y = 2x + 5 \):

1. For \( x = -4 + \sqrt{26} \):
   \[
   y = 2(-4 + \sqrt{26}) + 5 = -8 + 2\sqrt{26} + 5 = -3 + 2\sqrt{26}
   \]

2. For \( x = -4 - \sqrt{26} \):
   \[
   y = 2(-4 - \sqrt{26}) + 5 = -8 - 2\sqrt{26} + 5 = -3 - 2\sqrt{26}
   \]

### Solutions

The solutions to the system of equations are:

1. \( \left( -4 + \sqrt{26}, -3 + 2\sqrt{26} \right) \)
2. \( \left( -4 - \sqrt{26}, -3 - 2\sqrt{26} \right) \)

Would you like more details or have any questions?

Here are some additional related questions to explore:

1. How can you verify the solutions to the system of equations?
2. What does the graph of this system of equations look like?
3. How do you interpret the discriminant of the quadratic equation in this context?
4. Can this system of equations have a unique solution? Why or why not?
5. How would the solution change if the second equation was \( 2x = y + 5 \)?

**Tip:** When solving systems of equations, substituting one equation into another can simplify finding the solution.

Learn how to solve a system of equations algebraically involving a quadratic equation and a linear equation. Step-by-step solution provided for the system y = x^2 + 10x - 5 and 2x = y - 5. Explore quadratic equations and linear equations with detailed explanations.

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