Solve g (x + 4)/(3x - 4) = (2x - 1)/(2x + 1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for x

$x=5$
$x=0$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-\frac{1}{2},\frac{4}{3}$ since division by zero is not defined. Multiply both sides of the equation by $\left(3x-4\right)\left(2x+1\right)$, the least common multiple of $3x-4,2x+1$.

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

Use the distributive property to multiply $2x+1$ by $x+4$ and combine like terms.

$$2x^{2}+9x+4=\left(3x-4\right)\left(2x-1\right)$$

Use the distributive property to multiply $3x-4$ by $2x-1$ and combine like terms.

$$2x^{2}+9x+4=6x^{2}-11x+4$$

Subtract $6x^{2}$ from both sides.

$$2x^{2}+9x+4-6x^{2}=-11x+4$$

Combine $2x^{2}$ and $-6x^{2}$ to get $-4x^{2}$.

$$-4x^{2}+9x+4=-11x+4$$

Add $11x$ to both sides.

$$-4x^{2}+9x+4+11x=4$$

Combine $9x$ and $11x$ to get $20x$.

$$-4x^{2}+20x+4=4$$

Subtract $4$ from both sides.

$$-4x^{2}+20x+4-4=0$$

Subtract $4$ from $4$ to get $0$.

$$-4x^{2}+20x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-4$ for $a$, $20$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-20±\sqrt{20^{2}}}{2\left(-4\right)}$$

Take the square root of $20^{2}$.

$$x=\frac{-20±20}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{-20±20}{-8}$$

Now solve the equation $x=\frac{-20±20}{-8}$ when $±$ is plus. Add $-20$ to $20$.

$$x=\frac{0}{-8}$$

Divide $0$ by $-8$.

$$x=0$$

Now solve the equation $x=\frac{-20±20}{-8}$ when $±$ is minus. Subtract $20$ from $-20$.

$$x=-\frac{40}{-8}$$

Divide $-40$ by $-8$.

$$x=5$$

The equation is now solved.

$$x=0$$ $$x=5$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-\frac{1}{2},\frac{4}{3}$ since division by zero is not defined. Multiply both sides of the equation by $\left(3x-4\right)\left(2x+1\right)$, the least common multiple of $3x-4,2x+1$.

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

Use the distributive property to multiply $2x+1$ by $x+4$ and combine like terms.

$$2x^{2}+9x+4=\left(3x-4\right)\left(2x-1\right)$$

Use the distributive property to multiply $3x-4$ by $2x-1$ and combine like terms.

$$2x^{2}+9x+4=6x^{2}-11x+4$$

Subtract $6x^{2}$ from both sides.

$$2x^{2}+9x+4-6x^{2}=-11x+4$$

Combine $2x^{2}$ and $-6x^{2}$ to get $-4x^{2}$.

$$-4x^{2}+9x+4=-11x+4$$

Add $11x$ to both sides.

$$-4x^{2}+9x+4+11x=4$$

Combine $9x$ and $11x$ to get $20x$.

$$-4x^{2}+20x+4=4$$

Subtract $4$ from both sides.

$$-4x^{2}+20x=4-4$$

Subtract $4$ from $4$ to get $0$.

$$-4x^{2}+20x=0$$

Divide both sides by $-4$.

$$\frac{-4x^{2}+20x}{-4}=\frac{0}{-4}$$

Dividing by $-4$ undoes the multiplication by $-4$.

$$x^{2}+\frac{20}{-4}x=\frac{0}{-4}$$

Divide $20$ by $-4$.

$$x^{2}-5x=\frac{0}{-4}$$

Divide $0$ by $-4$.

$$x^{2}-5x=0$$

Divide $-5$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{2}$. Then add the square of $-\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}$$

Square $-\frac{5}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Factor $x^{2}-5x+\frac{25}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{5}{2}$$ $$x-\frac{5}{2}=-\frac{5}{2}$$

Add $\frac{5}{2}$ to both sides of the equation.

$$x=5$$ $$x=0$$