Solve ( ) (x + 2)/(x + 4) = 8/(3x - 8) TYP गर्न | 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

$$\frac { x + 2 } { x + 4 } = \frac { 8 } { 3 x - 8 }$$

Solve for x

$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$
$x=6$

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 $-4,\frac{8}{3}$ since division by zero is not defined. Multiply both sides of the equation by $\left(3x-8\right)\left(x+4\right)$, the least common multiple of $x+4,3x-8$.

$$\left(3x-8\right)\left(x+2\right)=\left(x+4\right)\times 8$$

Use the distributive property to multiply $3x-8$ by $x+2$ and combine like terms.

$$3x^{2}-2x-16=\left(x+4\right)\times 8$$

Use the distributive property to multiply $x+4$ by $8$.

$$3x^{2}-2x-16=8x+32$$

Subtract $8x$ from both sides.

$$3x^{2}-2x-16-8x=32$$

Combine $-2x$ and $-8x$ to get $-10x$.

$$3x^{2}-10x-16=32$$

Subtract $32$ from both sides.

$$3x^{2}-10x-16-32=0$$

Subtract $32$ from $-16$ to get $-48$.

$$3x^{2}-10x-48=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-10$ for $b$, and $-48$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-48\right)}}{2\times 3}$$

Square $-10$.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-48\right)}}{2\times 3}$$

Multiply $-4$ times $3$.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\left(-48\right)}}{2\times 3}$$

Multiply $-12$ times $-48$.

$$x=\frac{-\left(-10\right)±\sqrt{100+576}}{2\times 3}$$

Add $100$ to $576$.

$$x=\frac{-\left(-10\right)±\sqrt{676}}{2\times 3}$$

Take the square root of $676$.

$$x=\frac{-\left(-10\right)±26}{2\times 3}$$

The opposite of $-10$ is $10$.

$$x=\frac{10±26}{2\times 3}$$

Multiply $2$ times $3$.

$$x=\frac{10±26}{6}$$

Now solve the equation $x=\frac{10±26}{6}$ when $±$ is plus. Add $10$ to $26$.

$$x=\frac{36}{6}$$

Divide $36$ by $6$.

$$x=6$$

Now solve the equation $x=\frac{10±26}{6}$ when $±$ is minus. Subtract $26$ from $10$.

$$x=-\frac{16}{6}$$

Reduce the fraction $\frac{-16}{6}$ to lowest terms by extracting and canceling out $2$.

$$x=-\frac{8}{3}$$

The equation is now solved.

$$x=6$$ $$x=-\frac{8}{3}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-4,\frac{8}{3}$ since division by zero is not defined. Multiply both sides of the equation by $\left(3x-8\right)\left(x+4\right)$, the least common multiple of $x+4,3x-8$.

$$\left(3x-8\right)\left(x+2\right)=\left(x+4\right)\times 8$$

Use the distributive property to multiply $3x-8$ by $x+2$ and combine like terms.

$$3x^{2}-2x-16=\left(x+4\right)\times 8$$

Use the distributive property to multiply $x+4$ by $8$.

$$3x^{2}-2x-16=8x+32$$

Subtract $8x$ from both sides.

$$3x^{2}-2x-16-8x=32$$

Combine $-2x$ and $-8x$ to get $-10x$.

$$3x^{2}-10x-16=32$$

Add $16$ to both sides.

$$3x^{2}-10x=32+16$$

Add $32$ and $16$ to get $48$.

$$3x^{2}-10x=48$$

Divide both sides by $3$.

$$\frac{3x^{2}-10x}{3}=\frac{48}{3}$$

Dividing by $3$ undoes the multiplication by $3$.

$$x^{2}-\frac{10}{3}x=\frac{48}{3}$$

Divide $48$ by $3$.

$$x^{2}-\frac{10}{3}x=16$$

Divide $-\frac{10}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{3}$. Then add the square of $-\frac{5}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=16+\left(-\frac{5}{3}\right)^{2}$$

Square $-\frac{5}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=16+\frac{25}{9}$$

Add $16$ to $\frac{25}{9}$.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{169}{9}$$

Factor $x^{2}-\frac{10}{3}x+\frac{25}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{169}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{169}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{13}{3}$$ $$x-\frac{5}{3}=-\frac{13}{3}$$

Add $\frac{5}{3}$ to both sides of the equation.

$$x=6$$ $$x=-\frac{8}{3}$$