Solve the System of Equations x=y^2-1 2x-y=4

$x = y^{2} - 1$ $2 x - y = 4$

Step 1

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

Replace all occurrences of $x$ in $2 x - y = 4$ with $y^{2} - 1$ .

$2 (y^{2} - 1) - y = 4$

$x = y^{2} - 1$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$2 y^{2} + 2 \cdot - 1 - y = 4$

$x = y^{2} - 1$

Multiply $2$ by $- 1$ .

$2 y^{2} - 2 - y = 4$

$x = y^{2} - 1$

$2 y^{2} - 2 - y = 4$

$x = y^{2} - 1$

$2 y^{2} - 2 - y = 4$

$x = y^{2} - 1$

$2 y^{2} - 2 - y = 4$

$x = y^{2} - 1$

Step 2

Solve for $y$ in $2 y^{2} - 2 - y = 4$ .

Subtract $4$ from both sides of the equation.

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

$x = y^{2} - 1$

Subtract $4$ from $- 2$ .

$2 y^{2} - y - 6 = 0$

$x = y^{2} - 1$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 6 = - 12$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- y$ .

$2 y^{2} - (y) - 6 = 0$

$x = y^{2} - 1$

Rewrite $- 1$ as $3$ plus $- 4$

$2 y^{2} + (3 - 4) y - 6 = 0$

$x = y^{2} - 1$

Apply the distributive property.

$2 y^{2} + 3 y - 4 y - 6 = 0$

$x = y^{2} - 1$

$2 y^{2} + 3 y - 4 y - 6 = 0$

$x = y^{2} - 1$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 y^{2} + 3 y) - 4 y - 6 = 0$

$x = y^{2} - 1$

Factor out the greatest common factor (GCF) from each group.

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

$x = y^{2} - 1$

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

$x = y^{2} - 1$

Factor the polynomial by factoring out the greatest common factor, $2 y + 3$ .

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

$x = y^{2} - 1$

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

$x = y^{2} - 1$

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

$2 y + 3 = 0$

$y - 2 = 0$

$x = y^{2} - 1$

Set $2 y + 3$ equal to $0$ and solve for $y$ .

Set $2 y + 3$ equal to $0$ .

$2 y + 3 = 0$

$x = y^{2} - 1$

Solve $2 y + 3 = 0$ for $y$ .

Subtract $3$ from both sides of the equation.

$2 y = - 3$

$x = y^{2} - 1$

Divide each term in $2 y = - 3$ by $2$ and simplify.

Divide each term in $2 y = - 3$ by $2$ .

$\frac{2 y}{2} = \frac{- 3}{2}$

$x = y^{2} - 1$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 3}{2}$

$x = y^{2} - 1$

Divide $y$ by $1$ .

$y = \frac{- 3}{2}$

$x = y^{2} - 1$

$y = \frac{- 3}{2}$

$x = y^{2} - 1$

$y = \frac{- 3}{2}$

$x = y^{2} - 1$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{3}{2}$

$x = y^{2} - 1$

$y = - \frac{3}{2}$

$x = y^{2} - 1$

$y = - \frac{3}{2}$

$x = y^{2} - 1$

$y = - \frac{3}{2}$

$x = y^{2} - 1$

$y = - \frac{3}{2}$

$x = y^{2} - 1$

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

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

$y - 2 = 0$

$x = y^{2} - 1$

Add $2$ to both sides of the equation.

$y = 2$

$x = y^{2} - 1$

$y = 2$

$x = y^{2} - 1$

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

$y = - \frac{3}{2}, 2$

$x = y^{2} - 1$

$y = - \frac{3}{2}, 2$

$x = y^{2} - 1$

Step 3

Replace all occurrences of $y$ with $- \frac{3}{2}$ in each equation.

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

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

$y = - \frac{3}{2}$

Simplify the right side.

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

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{3}{2}$ .

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

$y = - \frac{3}{2}$

Apply the product rule to $\frac{3}{2}$ .

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

$y = - \frac{3}{2}$

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

$y = - \frac{3}{2}$

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

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

$y = - \frac{3}{2}$

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

$x = \frac{3^{2}}{2^{2}} - 1$

$y = - \frac{3}{2}$

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

$x = \frac{9}{2^{2}} - 1$

$y = - \frac{3}{2}$

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

$x = \frac{9}{4} - 1$

$y = - \frac{3}{2}$

$x = \frac{9}{4} - 1$

$y = - \frac{3}{2}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = \frac{9}{4} - 1 \cdot \frac{4}{4}$

$y = - \frac{3}{2}$

Combine $- 1$ and $\frac{4}{4}$ .

$x = \frac{9}{4} + \frac{- 1 \cdot 4}{4}$

$y = - \frac{3}{2}$

Combine the numerators over the common denominator.

$x = \frac{9 - 1 \cdot 4}{4}$

$y = - \frac{3}{2}$

Simplify the numerator.

Multiply $- 1$ by $4$ .

$x = \frac{9 - 4}{4}$

$y = - \frac{3}{2}$

Subtract $4$ from $9$ .

$x = \frac{5}{4}$

$y = - \frac{3}{2}$

$x = \frac{5}{4}$

$y = - \frac{3}{2}$

$x = \frac{5}{4}$

$y = - \frac{3}{2}$

$x = \frac{5}{4}$

$y = - \frac{3}{2}$

$x = \frac{5}{4}$

$y = - \frac{3}{2}$

Step 4

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

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

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

$y = 2$

Simplify the right side.

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

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

$x = 4 - 1$

$y = 2$

Subtract $1$ from $4$ .

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

Step 5

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

$(\frac{5}{4}, - \frac{3}{2})$

$(3, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{5}{4}, - \frac{3}{2}), (3, 2)$

Equation Form:

$x = \frac{5}{4}, y = - \frac{3}{2}$

$x = 3, y = 2$

Step 7

Solve the System of Equations x=y^2-1 2x-y=4

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