Solve by Substitution y=-x^2+x-5 y=x-5

$y = - x^{2} + x - 5$ $y = x - 5$

Eliminate the equal sides of each equation and combine.

$- x^{2} + x - 5 = x - 5$

Solve $- x^{2} + x - 5 = x - 5$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$- x^{2} + x - 5 - x = - 5$

Combine the opposite terms in $- x^{2} + x - 5 - x$ .

Subtract $x$ from $x$ .

$- x^{2} + 0 - 5 = - 5$

Add $- x^{2}$ and $0$ .

$- x^{2} - 5 = - 5$

Move all terms not containing $x$ to the right side of the equation.

Add $5$ to both sides of the equation.

$- x^{2} = - 5 + 5$

Add $- 5$ and $5$ .

$- x^{2} = 0$

Divide each term in $- x^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- x^{2} = 0$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{0}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x^{2} = 0$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Evaluate $y$ when $x = 0$ .

Substitute $0$ for $x$ .

$y = (0) - 5$

Substitute $0$ for $x$ in $y = (0) - 5$ and solve for $y$ .

Remove parentheses.

$y = 0 - 5$

Remove parentheses.

$y = (0) - 5$

Subtract $5$ from $0$ .

$y = - 5$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(0, - 5)$

The result can be shown in multiple forms.

Point Form:

$(0, - 5)$

Equation Form:

$x = 0, y = - 5$

Solve by Substitution y=-x^2+x-5 y=x-5

Related Questions

Support LCMGCF.com by disabling your adblocker.