Solve (x-2)/(x-1)=(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

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

Solve for x

$x=4$
$x=0$

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 $-2,1$ 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 $x-1,x+2$.

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

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$x^{2}-4=\left(x-1\right)\times 4$$

Use the distributive property to multiply $x-1$ by $4$.

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

Subtract $4x$ from both sides.

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

Add $4$ to both sides.

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

Add $-4$ and $4$ to get $0$.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-4$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

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

Take the square root of $\left(-4\right)^{2}$.

$$x=\frac{-\left(-4\right)±4}{2}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±4}{2}$$

Now solve the equation $x=\frac{4±4}{2}$ when $±$ is plus. Add $4$ to $4$.

$$x=\frac{8}{2}$$

Divide $8$ by $2$.

$$x=4$$

Now solve the equation $x=\frac{4±4}{2}$ when $±$ is minus. Subtract $4$ from $4$.

$$x=\frac{0}{2}$$

Divide $0$ by $2$.

$$x=0$$

The equation is now solved.

$$x=4$$ $$x=0$$

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 $x-1,x+2$.

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

Consider $\left(x+2\right)\left(x-2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $2$.

$$x^{2}-4=\left(x-1\right)\times 4$$

Use the distributive property to multiply $x-1$ by $4$.

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

Subtract $4x$ from both sides.

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

Add $4$ to both sides.

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

Add $-4$ and $4$ to get $0$.

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

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

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

Square $-2$.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2$$ $$x-2=-2$$

Add $2$ to both sides of the equation.

$$x=4$$ $$x=0$$