Algebra Examples

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

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.

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

Step 2

Solve the equation for $x$ .

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Step 2.1

Simplify $(x + 1) (4 x - 1)$ .

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Step 2.1.1

Rewrite.

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

Step 2.1.2

Simplify by adding zeros.

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

Step 2.1.3

Expand $(x + 1) (4 x - 1)$ using the FOIL Method.

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Step 2.1.3.1

Apply the distributive property.

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

Step 2.1.3.2

Apply the distributive property.

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

Step 2.1.3.3

Apply the distributive property.

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

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

Rewrite using the commutative property of multiplication.

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

Step 2.1.4.1.2

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

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Step 2.1.4.1.2.1

Move $x$ .

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

Step 2.1.4.1.2.2

Multiply $x$ by $x$ .

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

Step 2.1.4.1.3

Move $- 1$ to the left of $x$ .

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

Step 2.1.4.1.4

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

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

Step 2.1.4.1.5

Multiply $4 x$ by $1$ .

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

Step 2.1.4.1.6

Multiply $- 1$ by $1$ .

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

Step 2.1.4.2

Add $- x$ and $4 x$ .

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

Step 2.2

Simplify $(2 x - 1) \cdot 3$ .

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Step 2.2.1

Apply the distributive property.

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

Step 2.2.2

Multiply.

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Step 2.2.2.1

Multiply $3$ by $2$ .

$4 x^{2} + 3 x - 1 = 6 x - 1 \cdot 3$

Step 2.2.2.2

Multiply $- 1$ by $3$ .

$4 x^{2} + 3 x - 1 = 6 x - 3$

Step 2.3

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

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Step 2.3.1

Subtract $6 x$ from both sides of the equation.

$4 x^{2} + 3 x - 1 - 6 x = - 3$

Step 2.3.2

Subtract $6 x$ from $3 x$ .

$4 x^{2} - 3 x - 1 = - 3$

Step 2.4

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

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Step 2.4.1

Add $3$ to both sides of the equation.

$4 x^{2} - 3 x - 1 + 3 = 0$

Step 2.4.2

Add $- 1$ and $3$ .

$4 x^{2} - 3 x + 2 = 0$

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 $a = 4$ , $b = - 3$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (4 \cdot 2)}}{2 \cdot 4}$

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 4 \cdot 2}}{2 \cdot 4}$

Step 2.7.1.2

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

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Step 2.7.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{3 \pm \sqrt{9 - 16 \cdot 2}}{2 \cdot 4}$

Step 2.7.1.2.2

Multiply $- 16$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 32}}{2 \cdot 4}$

Step 2.7.1.3

Subtract $32$ from $9$ .

$x = \frac{3 \pm \sqrt{- 23}}{2 \cdot 4}$

Step 2.7.1.4

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 4}$

Step 2.7.1.5

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 4}$

Step 2.7.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{3 \pm i \sqrt{23}}{2 \cdot 4}$

Step 2.7.2

Multiply $2$ by $4$ .

$x = \frac{3 \pm i \sqrt{23}}{8}$

Step 2.8

The final answer is the combination of both solutions.

$x = \frac{3 + i \sqrt{23}}{8}, \frac{3 - i \sqrt{23}}{8}$