Solve the System of Equations y+5=x y=x^2-3x-5

$y + 5 = x$ $y = x^{2} - 3 x - 5$

Replace all occurrences of $y$ with $x^{2} - 3 x - 5$ in each equation.

Replace all occurrences of $y$ in $y + 5 = x$ with $x^{2} - 3 x - 5$ .

$(x^{2} - 3 x - 5) + 5 = x$

$y = x^{2} - 3 x - 5$

Simplify the left side.

Combine the opposite terms in $(x^{2} - 3 x - 5) + 5$ .

Add $- 5$ and $5$ .

$x^{2} - 3 x + 0 = x$

$y = x^{2} - 3 x - 5$

Add $x^{2} - 3 x$ and $0$ .

$x^{2} - 3 x = x$

$y = x^{2} - 3 x - 5$

$x^{2} - 3 x = x$

$y = x^{2} - 3 x - 5$

$x^{2} - 3 x = x$

$y = x^{2} - 3 x - 5$

$x^{2} - 3 x = x$

$y = x^{2} - 3 x - 5$

Solve for $x$ in $x^{2} - 3 x = x$ .

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

Subtract $x$ from both sides of the equation.

$x^{2} - 3 x - x = 0$

$y = x^{2} - 3 x - 5$

Subtract $x$ from $- 3 x$ .

$x^{2} - 4 x = 0$

$y = x^{2} - 3 x - 5$

$x^{2} - 4 x = 0$

$y = x^{2} - 3 x - 5$

Factor $x$ out of $x^{2} - 4 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 4 x = 0$

$y = x^{2} - 3 x - 5$

Factor $x$ out of $- 4 x$ .

$x \cdot x + x \cdot - 4 = 0$

$y = x^{2} - 3 x - 5$

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

$x (x - 4) = 0$

$y = x^{2} - 3 x - 5$

$x (x - 4) = 0$

$y = x^{2} - 3 x - 5$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 4 = 0$

$y = x^{2} - 3 x - 5$

Set $x$ equal to $0$ .

$x = 0$

$y = x^{2} - 3 x - 5$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

$y = x^{2} - 3 x - 5$

Add $4$ to both sides of the equation.

$x = 4$

$y = x^{2} - 3 x - 5$

$x = 4$

$y = x^{2} - 3 x - 5$

The final solution is all the values that make $x (x - 4) = 0$ true.

$x = 0, 4$

$y = x^{2} - 3 x - 5$

$x = 0, 4$

$y = x^{2} - 3 x - 5$

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 3 x - 5$ with $0$ .

$y = {(0)}^{2} - 3 \cdot 0 - 5$

$x = 0$

Simplify the right side.

Simplify ${(0)}^{2} - 3 \cdot 0 - 5$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 - 3 \cdot 0 - 5$

$x = 0$

Multiply $- 3$ by $0$ .

$y = 0 + 0 - 5$

$x = 0$

$y = 0 + 0 - 5$

$x = 0$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 5$

$x = 0$

Subtract $5$ from $0$ .

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 3 x - 5$ with $4$ .

$y = {(4)}^{2} - 3 \cdot 4 - 5$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} - 3 \cdot 4 - 5$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$y = 16 - 3 \cdot 4 - 5$

$x = 4$

Multiply $- 3$ by $4$ .

$y = 16 - 12 - 5$

$x = 4$

$y = 16 - 12 - 5$

$x = 4$

Simplify by subtracting numbers.

Subtract $12$ from $16$ .

$y = 4 - 5$

$x = 4$

Subtract $5$ from $4$ .

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

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

$(0, - 5)$

$(4, - 1)$

The result can be shown in multiple forms.

Point Form:

$(0, - 5), (4, - 1)$

Equation Form:

$x = 0, y = - 5$

$x = 4, y = - 1$

Solve the System of Equations y+5=x y=x^2-3x-5

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