Precalculus Examples

$x - y^{2} = - 1$ , $x - 2 y = 2$

Step 1

Add $y^{2}$ to both sides of the equation.

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

$x - 2 y = 2$

Step 2

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

Tap for more steps...

Step 2.1

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

$(- 1 + y^{2}) - 2 y = 2$

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

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Remove parentheses.

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

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

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

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

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

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

Step 3

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

Tap for more steps...

Step 3.1

Subtract $2$ from both sides of the equation.

$- 1 + y^{2} - 2 y - 2 = 0$

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

Step 3.2

Subtract $2$ from $- 1$ .

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

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

Step 3.3

Factor $y^{2} - 2 y - 3$ using the AC method.

Tap for more steps...

Step 3.3.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

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

Step 3.3.2

Write the factored form using these integers.

$(y - 3) (y + 1) = 0$

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

$(y - 3) (y + 1) = 0$

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

Step 3.4

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

$y - 3 = 0$

$y + 1 = 0$

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

Step 3.5

Set $y - 3$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 3.5.1

Set $y - 3$ equal to $0$ .

$y - 3 = 0$

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

Step 3.5.2

Add $3$ to both sides of the equation.

$y = 3$

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

$y = 3$

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

Step 3.6

Set $y + 1$ equal to $0$ and solve for $y$ .

Tap for more steps...

Step 3.6.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

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

Step 3.6.2

Subtract $1$ from both sides of the equation.

$y = - 1$

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

$y = - 1$

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

Step 3.7

The final solution is all the values that make $(y - 3) (y + 1) = 0$ true.

$y = 3, - 1$

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

$y = 3, - 1$

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

Step 4

Replace all occurrences of $y$ with $3$ in each equation.

Tap for more steps...

Step 4.1

Replace all occurrences of $y$ in $x = - 1 + y^{2}$ with $3$ .

$x = - 1 + {(3)}^{2}$

$y = 3$

Step 4.2

Simplify the right side.

Tap for more steps...

Step 4.2.1

Simplify $- 1 + {(3)}^{2}$ .

Tap for more steps...

Step 4.2.1.1

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

$x = - 1 + 9$

$y = 3$

Step 4.2.1.2

Add $- 1$ and $9$ .

$x = 8$

$y = 3$

$x = 8$

$y = 3$

$x = 8$

$y = 3$

$x = 8$

$y = 3$

Step 5

Replace all occurrences of $y$ with $- 1$ in each equation.

Tap for more steps...

Step 5.1

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

$x = - 1 + {(- 1)}^{2}$

$y = - 1$

Step 5.2

Simplify the right side.

Tap for more steps...

Step 5.2.1

Simplify $- 1 + {(- 1)}^{2}$ .

Tap for more steps...

Step 5.2.1.1

Raise $- 1$ to the power of $2$ .

$x = - 1 + 1$

$y = - 1$

Step 5.2.1.2

Add $- 1$ and $1$ .

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

$x = 0$

$y = - 1$

Step 6

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

$(8, 3)$

$(0, - 1)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(8, 3), (0, - 1)$

Equation Form:

$x = 8, y = 3$

$x = 0, y = - 1$

Step 8