Precalculus Examples

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

Step 1

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$(2 x - 1) (x - 2) = (x + 3) \cdot 4$

Step 2

Solve the equation for $x$ .

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Step 2.1

Simplify $(2 x - 1) (x - 2)$ .

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Step 2.1.1

Rewrite.

$0 + 0 + (2 x - 1) (x - 2) = (x + 3) \cdot 4$

Step 2.1.2

Simplify by adding zeros.

$(2 x - 1) (x - 2) = (x + 3) \cdot 4$

Step 2.1.3

Expand $(2 x - 1) (x - 2)$ using the FOIL Method.

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Step 2.1.3.1

Apply the distributive property.

$2 x (x - 2) - 1 (x - 2) = (x + 3) \cdot 4$

Step 2.1.3.2

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 2 - 1 (x - 2) = (x + 3) \cdot 4$

Step 2.1.3.3

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 2 - 1 x - 1 \cdot - 2 = (x + 3) \cdot 4$

Step 2.1.4

Simplify and combine like terms.

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Step 2.1.4.1

Simplify each term.

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Step 2.1.4.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.4.1.1.1

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 2 - 1 x - 1 \cdot - 2 = (x + 3) \cdot 4$

Step 2.1.4.1.1.2

Multiply $x$ by $x$ .

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

Step 2.1.4.1.2

Multiply $- 2$ by $2$ .

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

Step 2.1.4.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 2.1.4.1.4

Multiply $- 1$ by $- 2$ .

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

Step 2.1.4.2

Subtract $x$ from $- 4 x$ .

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

Step 2.2

Simplify $(x + 3) \cdot 4$ .

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Step 2.2.1

Apply the distributive property.

$2 x^{2} - 5 x + 2 = x \cdot 4 + 3 \cdot 4$

Step 2.2.2

Simplify the expression.

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Step 2.2.2.1

Move $4$ to the left of $x$ .

$2 x^{2} - 5 x + 2 = 4 \cdot x + 3 \cdot 4$

Step 2.2.2.2

Multiply $3$ by $4$ .

$2 x^{2} - 5 x + 2 = 4 x + 12$

Step 2.3

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

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Step 2.3.1

Subtract $4 x$ from both sides of the equation.

$2 x^{2} - 5 x + 2 - 4 x = 12$

Step 2.3.2

Subtract $4 x$ from $- 5 x$ .

$2 x^{2} - 9 x + 2 = 12$

Step 2.4

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

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Step 2.4.1

Subtract $12$ from both sides of the equation.

$2 x^{2} - 9 x + 2 - 12 = 0$

Step 2.4.2

Subtract $12$ from $2$ .

$2 x^{2} - 9 x - 10 = 0$

Step 2.5

Use the quadratic formula to find the solutions.

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

Step 2.6

Substitute the values $a = 2$ , $b = - 9$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (2 \cdot - 10)}}{2 \cdot 2}$

Step 2.7

Simplify.

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Step 2.7.1

Simplify the numerator.

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Step 2.7.1.1

Raise $- 9$ to the power of $2$ .

$x = \frac{9 \pm \sqrt{81 - 4 \cdot 2 \cdot - 10}}{2 \cdot 2}$

Step 2.7.1.2

Multiply $- 4 \cdot 2 \cdot - 10$ .

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Step 2.7.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{9 \pm \sqrt{81 - 8 \cdot - 10}}{2 \cdot 2}$

Step 2.7.1.2.2

Multiply $- 8$ by $- 10$ .

$x = \frac{9 \pm \sqrt{81 + 80}}{2 \cdot 2}$

Step 2.7.1.3

Add $81$ and $80$ .

$x = \frac{9 \pm \sqrt{161}}{2 \cdot 2}$

Step 2.7.2

Multiply $2$ by $2$ .

$x = \frac{9 \pm \sqrt{161}}{4}$

Step 2.8

The final answer is the combination of both solutions.

$x = \frac{9 + \sqrt{161}}{4}, \frac{9 - \sqrt{161}}{4}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{9 + \sqrt{161}}{4}, \frac{9 - \sqrt{161}}{4}$

Decimal Form:

$x = 5.42214438 \dots, - 0.92214438 \dots$