Solve the System of Equations x+y=1 y=x^2-5

$x + y = 1$ $y = x^{2} - 5$

Step 1

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

Replace all occurrences of $y$ in $x + y = 1$ with $x^{2} - 5$ .

$x + x^{2} - 5 = 1$

$y = x^{2} - 5$

Simplify the left side.

Remove parentheses.

$x + x^{2} - 5 = 1$

$y = x^{2} - 5$

$x + x^{2} - 5 = 1$

$y = x^{2} - 5$

$x + x^{2} - 5 = 1$

$y = x^{2} - 5$

Step 2

Solve for $x$ in $x + x^{2} - 5 = 1$ .

Subtract $1$ from both sides of the equation.

$x + x^{2} - 5 - 1 = 0$

$y = x^{2} - 5$

Subtract $1$ from $- 5$ .

$x + x^{2} - 6 = 0$

$y = x^{2} - 5$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u + u^{2} - 6 = 0$

$y = x^{2} - 5$

Factor $u + u^{2} - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $1$ .

$- 2, 3$

$y = x^{2} - 5$

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

$y = x^{2} - 5$

$(u - 2) (u + 3) = 0$

$y = x^{2} - 5$

Replace all occurrences of $u$ with $x$ .

$(x - 2) (x + 3) = 0$

$y = x^{2} - 5$

$(x - 2) (x + 3) = 0$

$y = x^{2} - 5$

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

$x - 2 = 0$

$x + 3 = 0$

$y = x^{2} - 5$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = x^{2} - 5$

Add $2$ to both sides of the equation.

$x = 2$

$y = x^{2} - 5$

$x = 2$

$y = x^{2} - 5$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

$y = x^{2} - 5$

Subtract $3$ from both sides of the equation.

$x = - 3$

$y = x^{2} - 5$

$x = - 3$

$y = x^{2} - 5$

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

$y = x^{2} - 5$

$x = 2, - 3$

$y = x^{2} - 5$

Step 3

Replace all occurrences of $x$ with $2$ in each equation.

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

$y = {(2)}^{2} - 5$

$x = 2$

Simplify the right side.

Simplify ${(2)}^{2} - 5$ .

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

$y = 4 - 5$

$x = 2$

Subtract $5$ from $4$ .

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

$y = - 1$

$x = 2$

Step 4

Replace all occurrences of $x$ with $- 3$ in each equation.

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

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

$x = - 3$

Simplify the right side.

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

Raise $- 3$ to the power of $2$ .

$y = 9 - 5$

$x = - 3$

Subtract $5$ from $9$ .

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

Step 5

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

$(2, - 1)$

$(- 3, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, - 1), (- 3, 4)$

Equation Form:

$x = 2, y = - 1$

$x = - 3, y = 4$

Step 7

Solve the System of Equations x+y=1 y=x^2-5

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