Questions: Solve each system algebraically. 8. y=x^2-6x+2 2x-y=5

Solution Steps

To solve the given system of equations algebraically, we can use substitution. First, solve the second equation for \( y \) in terms of \( x \). Then, substitute this expression into the first equation to find the values of \( x \). Finally, use these \( x \) values to find the corresponding \( y \) values.

Given the system of equations: \[ \begin{cases} y = x^2 - 6x + 2 \\ 2x - y = 5 \end{cases} \]

First, solve the second equation for \( y \): \[ 2x - y = 5 \implies y = 2x - 5 \]

Substitute \( y = 2x - 5 \) into the first equation: \[ 2x - 5 = x^2 - 6x + 2 \]

Rearrange the equation to form a quadratic equation: \[ x^2 - 6x + 2 - 2x + 5 = 0 \implies x^2 - 8x + 7 = 0 \]

Solve the quadratic equation: \[ x = \frac{8 \pm \sqrt{64 - 28}}{2} = \frac{8 \pm \sqrt{36}}{2} = \frac{8 \pm 6}{2} \]

Thus, the solutions for \( x \) are: \[ x = \frac{8 + 6}{2} = 7 \quad \text{and} \quad x = \frac{8 - 6}{2} = 1 \]

Substitute \( x = 7 \) and \( x = 1 \) back into \( y = 2x - 5 \) to find the corresponding \( y \) values: \[ \text{For } x = 7: \quad y = 2(7) - 5 = 14 - 5 = 9 \] \[ \text{For } x = 1: \quad y = 2(1) - 5 = 2 - 5 = -3 \]

Final Answer