Algebra Examples

${(2 x)}^{2} - 3 y^{2} + 4 x = - 1$ $y = 2$

Step 1

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

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Step 1.1

Replace all occurrences of $y$ in ${(2 x)}^{2} - 3 y^{2} + 4 x = - 1$ with $2$ .

${(2 x)}^{2} - 3 {(2)}^{2} + 4 x = - 1$

$y = 2$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Apply the product rule to $2 x$ .

$2^{2} x^{2} - 3 {(2)}^{2} + 4 x = - 1$

$y = 2$

Step 1.2.1.2

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

$4 x^{2} - 3 {(2)}^{2} + 4 x = - 1$

$y = 2$

Step 1.2.1.3

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

$4 x^{2} - 3 \cdot 4 + 4 x = - 1$

$y = 2$

Step 1.2.1.4

Multiply $- 3$ by $4$ .

$4 x^{2} - 12 + 4 x = - 1$

$y = 2$

$4 x^{2} - 12 + 4 x = - 1$

$y = 2$

$4 x^{2} - 12 + 4 x = - 1$

$y = 2$

$4 x^{2} - 12 + 4 x = - 1$

$y = 2$

Step 2

Solve for $x$ in $4 x^{2} - 12 + 4 x = - 1$ .

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Step 2.1

Move all terms to the left side of the equation and simplify.

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Step 2.1.1

Add $1$ to both sides of the equation.

$4 x^{2} - 12 + 4 x + 1 = 0$

$y = 2$

Step 2.1.2

Add $- 12$ and $1$ .

$4 x^{2} + 4 x - 11 = 0$

$y = 2$

$4 x^{2} + 4 x - 11 = 0$

$y = 2$

Step 2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

$y = 2$

Step 2.3

Substitute the values $a = 4$ , $b = 4$ , and $c = - 11$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (4 \cdot - 11)}}{2 \cdot 4}$

$y = 2$

Step 2.4

Simplify.

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Step 2.4.1

Simplify the numerator.

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Step 2.4.1.1

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.2

Multiply $- 4 \cdot 4 \cdot - 11$ .

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Step 2.4.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{- 4 \pm \sqrt{16 - 16 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.2.2

Multiply $- 16$ by $- 11$ .

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.3

Add $16$ and $176$ .

$x = \frac{- 4 \pm \sqrt{192}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.4

Rewrite $192$ as $8^{2} \cdot 3$ .

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Step 2.4.1.4.1

Factor $64$ out of $192$ .

$x = \frac{- 4 \pm \sqrt{64 (3)}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.4.2

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

Step 2.4.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

Step 2.4.2

Multiply $2$ by $4$ .

$x = \frac{- 4 \pm 8 \sqrt{3}}{8}$

$y = 2$

Step 2.4.3

Simplify $\frac{- 4 \pm 8 \sqrt{3}}{8}$ .

$x = \frac{- 1 \pm 2 \sqrt{3}}{2}$

$y = 2$

$x = \frac{- 1 \pm 2 \sqrt{3}}{2}$

$y = 2$

Step 2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.2

Multiply $- 4 \cdot 4 \cdot - 11$ .

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Step 2.5.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{- 4 \pm \sqrt{16 - 16 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.2.2

Multiply $- 16$ by $- 11$ .

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.3

Add $16$ and $176$ .

$x = \frac{- 4 \pm \sqrt{192}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.4

Rewrite $192$ as $8^{2} \cdot 3$ .

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Step 2.5.1.4.1

Factor $64$ out of $192$ .

$x = \frac{- 4 \pm \sqrt{64 (3)}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.4.2

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

Step 2.5.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

Step 2.5.2

Multiply $2$ by $4$ .

$x = \frac{- 4 \pm 8 \sqrt{3}}{8}$

$y = 2$

Step 2.5.3

Simplify $\frac{- 4 \pm 8 \sqrt{3}}{8}$ .

$x = \frac{- 1 \pm 2 \sqrt{3}}{2}$

$y = 2$

Step 2.5.4

Change the $\pm$ to $+$ .

$x = \frac{- 1 + 2 \sqrt{3}}{2}$

$y = 2$

Step 2.5.5

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + 2 \sqrt{3}}{2}$

$y = 2$

Step 2.5.6

Factor $- 1$ out of $2 \sqrt{3}$ .

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

$y = 2$

Step 2.5.7

Factor $- 1$ out of $- 1 (1) - (- 2 \sqrt{3})$ .

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

$y = 2$

Step 2.5.8

Move the negative in front of the fraction.

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

$y = 2$

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

$y = 2$

Step 2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.2

Multiply $- 4 \cdot 4 \cdot - 11$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{- 4 \pm \sqrt{16 - 16 \cdot - 11}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.2.2

Multiply $- 16$ by $- 11$ .

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{16 + 176}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.3

Add $16$ and $176$ .

$x = \frac{- 4 \pm \sqrt{192}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.4

Rewrite $192$ as $8^{2} \cdot 3$ .

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Step 2.6.1.4.1

Factor $64$ out of $192$ .

$x = \frac{- 4 \pm \sqrt{64 (3)}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.4.2

Rewrite $64$ as $8^{2}$ .

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm \sqrt{8^{2} \cdot 3}}{2 \cdot 4}$

$y = 2$

Step 2.6.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

$x = \frac{- 4 \pm 8 \sqrt{3}}{2 \cdot 4}$

$y = 2$

Step 2.6.2

Multiply $2$ by $4$ .

$x = \frac{- 4 \pm 8 \sqrt{3}}{8}$

$y = 2$

Step 2.6.3

Simplify $\frac{- 4 \pm 8 \sqrt{3}}{8}$ .

$x = \frac{- 1 \pm 2 \sqrt{3}}{2}$

$y = 2$

Step 2.6.4

Change the $\pm$ to $-$ .

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

$y = 2$

Step 2.6.5

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - 2 \sqrt{3}}{2}$

$y = 2$

Step 2.6.6

Factor $- 1$ out of $- 2 \sqrt{3}$ .

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

$y = 2$

Step 2.6.7

Factor $- 1$ out of $- 1 (1) - (2 \sqrt{3})$ .

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

$y = 2$

Step 2.6.8

Move the negative in front of the fraction.

$x = - \frac{1 + 2 \sqrt{3}}{2}$

$y = 2$

$x = - \frac{1 + 2 \sqrt{3}}{2}$

$y = 2$

Step 2.7

The final answer is the combination of both solutions.

$x = - \frac{1 - 2 \sqrt{3}}{2}, - \frac{1 + 2 \sqrt{3}}{2}$

$y = 2$

$x = - \frac{1 - 2 \sqrt{3}}{2}, - \frac{1 + 2 \sqrt{3}}{2}$

$y = 2$

Step 3

Solve the system of equations.

$x = - \frac{1 - 2 \sqrt{3}}{2}, y = 2$

Step 4

Solve the system of equations.

$x = - \frac{1 + 2 \sqrt{3}}{2}, y = 2$

Step 5

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

$(- \frac{1 - 2 \sqrt{3}}{2}, 2)$

$(- \frac{1 + 2 \sqrt{3}}{2}, 2)$