Solve (5x - 8)/(x - 1) = (7x - 4)/(x + 2) | 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

$$\frac{5x-8}{x-1}=\frac{7x-4}{x+2}$$

Solve for x

$x=4$
$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$

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 $-2,1$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-1\right)\left(x+2\right)$, the least common multiple of $x-1,x+2$.

$$\left(x+2\right)\left(5x-8\right)=\left(x-1\right)\left(7x-4\right)$$

Use the distributive property to multiply $x+2$ by $5x-8$ and combine like terms.

$$5x^{2}+2x-16=\left(x-1\right)\left(7x-4\right)$$

Use the distributive property to multiply $x-1$ by $7x-4$ and combine like terms.

$$5x^{2}+2x-16=7x^{2}-11x+4$$

Subtract $7x^{2}$ from both sides.

$$5x^{2}+2x-16-7x^{2}=-11x+4$$

Combine $5x^{2}$ and $-7x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+2x-16=-11x+4$$

Add $11x$ to both sides.

$$-2x^{2}+2x-16+11x=4$$

Combine $2x$ and $11x$ to get $13x$.

$$-2x^{2}+13x-16=4$$

Subtract $4$ from both sides.

$$-2x^{2}+13x-16-4=0$$

Subtract $4$ from $-16$ to get $-20$.

$$-2x^{2}+13x-20=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $13$ for $b$, and $-20$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-2\right)\left(-20\right)}}{2\left(-2\right)}$$

Square $13$.

$$x=\frac{-13±\sqrt{169-4\left(-2\right)\left(-20\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-13±\sqrt{169+8\left(-20\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-20$.

$$x=\frac{-13±\sqrt{169-160}}{2\left(-2\right)}$$

Add $169$ to $-160$.

$$x=\frac{-13±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of $9$.

$$x=\frac{-13±3}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-13±3}{-4}$$

Now solve the equation $x=\frac{-13±3}{-4}$ when $±$ is plus. Add $-13$ to $3$.

$$x=-\frac{10}{-4}$$

Reduce the fraction $\frac{-10}{-4}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-13±3}{-4}$ when $±$ is minus. Subtract $3$ from $-13$.

$$x=-\frac{16}{-4}$$

Divide $-16$ by $-4$.

$$x=4$$

The equation is now solved.

$$x=\frac{5}{2}$$ $$x=4$$

Steps for Completing the Square

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

$$\left(x+2\right)\left(5x-8\right)=\left(x-1\right)\left(7x-4\right)$$

Use the distributive property to multiply $x+2$ by $5x-8$ and combine like terms.

$$5x^{2}+2x-16=\left(x-1\right)\left(7x-4\right)$$

Use the distributive property to multiply $x-1$ by $7x-4$ and combine like terms.

$$5x^{2}+2x-16=7x^{2}-11x+4$$

Subtract $7x^{2}$ from both sides.

$$5x^{2}+2x-16-7x^{2}=-11x+4$$

Combine $5x^{2}$ and $-7x^{2}$ to get $-2x^{2}$.

$$-2x^{2}+2x-16=-11x+4$$

Add $11x$ to both sides.

$$-2x^{2}+2x-16+11x=4$$

Combine $2x$ and $11x$ to get $13x$.

$$-2x^{2}+13x-16=4$$

Add $16$ to both sides.

$$-2x^{2}+13x=4+16$$

Add $4$ and $16$ to get $20$.

$$-2x^{2}+13x=20$$

Divide both sides by $-2$.

$$\frac{-2x^{2}+13x}{-2}=\frac{20}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$x^{2}+\frac{13}{-2}x=\frac{20}{-2}$$

Divide $13$ by $-2$.

$$x^{2}-\frac{13}{2}x=\frac{20}{-2}$$

Divide $20$ by $-2$.

$$x^{2}-\frac{13}{2}x=-10$$

Divide $-\frac{13}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{13}{4}$. Then add the square of $-\frac{13}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-10+\left(-\frac{13}{4}\right)^{2}$$

Square $-\frac{13}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-10+\frac{169}{16}$$

Add $-10$ to $\frac{169}{16}$.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{9}{16}$$

Factor $x^{2}-\frac{13}{2}x+\frac{169}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{3}{4}$$ $$x-\frac{13}{4}=-\frac{3}{4}$$

Add $\frac{13}{4}$ to both sides of the equation.

$$x=4$$ $$x=\frac{5}{2}$$