Solve 1/(2 - x) - 1 = 1/(x - 2) - (6 - x)/(3x ^ 2 - 12) | 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

$$\frac{1}{2-x}-1=\frac{1}{x-2}-\frac{6-x}{3x^{2}-12}$$

Solve for x

$x=-3$
$x=\frac{2}{3}\approx 0.666666667$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to any of the values $-2,2$ since division by zero is not defined. Multiply both sides of the equation by $3\left(x-2\right)\left(x+2\right)$, the least common multiple of $2-x,x-2,3x^{2}-12$.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-\left(6-x\right)$$

Multiply $3$ and $-1$ to get $-3$.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3$ by $x-2$.

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3x+6$ by $x+2$ and combine like terms.

$$-6-3x-3x^{2}+12=3x+6-\left(6-x\right)$$

Add $-6$ and $12$ to get $6$.

$$6-3x-3x^{2}=3x+6-\left(6-x\right)$$

To find the opposite of $6-x$, find the opposite of each term.

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

Subtract $6$ from $6$ to get $0$.

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

Combine $3x$ and $x$ to get $4x$.

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

Subtract $4x$ from both sides.

$$6-3x-3x^{2}-4x=0$$

Combine $-3x$ and $-4x$ to get $-7x$.

$$6-7x-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-7x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-3x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-7$$ $$ab=-3\times 6=-18$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-18$.

$$1,-18$$ $$2,-9$$ $$3,-6$$

Calculate the sum for each pair.

$$1-18=-17$$ $$2-9=-7$$ $$3-6=-3$$

The solution is the pair that gives sum $-7$.

$$a=2$$ $$b=-9$$

Rewrite $-3x^{2}-7x+6$ as $\left(-3x^{2}+2x\right)+\left(-9x+6\right)$.

$$\left(-3x^{2}+2x\right)+\left(-9x+6\right)$$

Factor out $-x$ in the first and $-3$ in the second group.

$$-x\left(3x-2\right)-3\left(3x-2\right)$$

Factor out common term $3x-2$ by using distributive property.

$$\left(3x-2\right)\left(-x-3\right)$$

To find equation solutions, solve $3x-2=0$ and $-x-3=0$.

$$x=\frac{2}{3}$$ $$x=-3$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-2,2$ since division by zero is not defined. Multiply both sides of the equation by $3\left(x-2\right)\left(x+2\right)$, the least common multiple of $2-x,x-2,3x^{2}-12$.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-\left(6-x\right)$$

Multiply $3$ and $-1$ to get $-3$.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3$ by $x-2$.

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3x+6$ by $x+2$ and combine like terms.

$$-6-3x-3x^{2}+12=3x+6-\left(6-x\right)$$

Add $-6$ and $12$ to get $6$.

$$6-3x-3x^{2}=3x+6-\left(6-x\right)$$

To find the opposite of $6-x$, find the opposite of each term.

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

Subtract $6$ from $6$ to get $0$.

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

Combine $3x$ and $x$ to get $4x$.

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

Subtract $4x$ from both sides.

$$6-3x-3x^{2}-4x=0$$

Combine $-3x$ and $-4x$ to get $-7x$.

$$6-7x-3x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-3x^{2}-7x+6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-3$ for $a$, $-7$ for $b$, and $6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Multiply $-4$ times $-3$.

$$x=\frac{-\left(-7\right)±\sqrt{49+12\times 6}}{2\left(-3\right)}$$

Multiply $12$ times $6$.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\left(-3\right)}$$

Add $49$ to $72$.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2\left(-3\right)}$$

Take the square root of $121$.

$$x=\frac{-\left(-7\right)±11}{2\left(-3\right)}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±11}{2\left(-3\right)}$$

Multiply $2$ times $-3$.

$$x=\frac{7±11}{-6}$$

Now solve the equation $x=\frac{7±11}{-6}$ when $±$ is plus. Add $7$ to $11$.

$$x=\frac{18}{-6}$$

Divide $18$ by $-6$.

$$x=-3$$

Now solve the equation $x=\frac{7±11}{-6}$ when $±$ is minus. Subtract $11$ from $7$.

$$x=-\frac{4}{-6}$$

Reduce the fraction $\frac{-4}{-6}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{2}{3}$$

The equation is now solved.

$$x=-3$$ $$x=\frac{2}{3}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-2,2$ since division by zero is not defined. Multiply both sides of the equation by $3\left(x-2\right)\left(x+2\right)$, the least common multiple of $2-x,x-2,3x^{2}-12$.

$$-6-3x+3\left(x-2\right)\left(x+2\right)\left(-1\right)=3x+6-\left(6-x\right)$$

Multiply $3$ and $-1$ to get $-3$.

$$-6-3x-3\left(x-2\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3$ by $x-2$.

$$-6-3x+\left(-3x+6\right)\left(x+2\right)=3x+6-\left(6-x\right)$$

Use the distributive property to multiply $-3x+6$ by $x+2$ and combine like terms.

$$-6-3x-3x^{2}+12=3x+6-\left(6-x\right)$$

Add $-6$ and $12$ to get $6$.

$$6-3x-3x^{2}=3x+6-\left(6-x\right)$$

To find the opposite of $6-x$, find the opposite of each term.

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

Subtract $6$ from $6$ to get $0$.

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

Combine $3x$ and $x$ to get $4x$.

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

Subtract $4x$ from both sides.

$$6-3x-3x^{2}-4x=0$$

Combine $-3x$ and $-4x$ to get $-7x$.

$$6-7x-3x^{2}=0$$

Subtract $6$ from both sides. Anything subtracted from zero gives its negation.

$$-7x-3x^{2}=-6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-3x^{2}-7x=-6$$

Divide both sides by $-3$.

$$\frac{-3x^{2}-7x}{-3}=-\frac{6}{-3}$$

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

$$x^{2}+\left(-\frac{7}{-3}\right)x=-\frac{6}{-3}$$

Divide $-7$ by $-3$.

$$x^{2}+\frac{7}{3}x=-\frac{6}{-3}$$

Divide $-6$ by $-3$.

$$x^{2}+\frac{7}{3}x=2$$

Divide $\frac{7}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{7}{6}$. Then add the square of $\frac{7}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=2+\left(\frac{7}{6}\right)^{2}$$

Square $\frac{7}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=2+\frac{49}{36}$$

Add $2$ to $\frac{49}{36}$.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{121}{36}$$

Factor $x^{2}+\frac{7}{3}x+\frac{49}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{11}{6}$$ $$x+\frac{7}{6}=-\frac{11}{6}$$

Subtract $\frac{7}{6}$ from both sides of the equation.

$$x=\frac{2}{3}$$ $$x=-3$$