Solve 1/(2x - 10) + 5/(2x - 2) = 1 3 | 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

$$\frac { 1 } { 2 x - 10 } + \frac { 5 } { 2 x - 2 } = 1$$

Solve for x

$x=6$
$x=3$

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 $1,5$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-5\right)\left(x-1\right)$, the least common multiple of $2x-10,2x-2$.

$$x-1+\left(x-5\right)\times 5=2\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply $x-5$ by $5$.

$$x-1+5x-25=2\left(x-5\right)\left(x-1\right)$$

Combine $x$ and $5x$ to get $6x$.

$$6x-1-25=2\left(x-5\right)\left(x-1\right)$$

Subtract $25$ from $-1$ to get $-26$.

$$6x-26=2\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply $2$ by $x-5$.

$$6x-26=\left(2x-10\right)\left(x-1\right)$$

Use the distributive property to multiply $2x-10$ by $x-1$ and combine like terms.

$$6x-26=2x^{2}-12x+10$$

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

$$6x-26-2x^{2}=-12x+10$$

Add $12x$ to both sides.

$$6x-26-2x^{2}+12x=10$$

Combine $6x$ and $12x$ to get $18x$.

$$18x-26-2x^{2}=10$$

Subtract $10$ from both sides.

$$18x-26-2x^{2}-10=0$$

Subtract $10$ from $-26$ to get $-36$.

$$18x-36-2x^{2}=0$$

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.

$$-2x^{2}+18x-36=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $18$ for $b$, and $-36$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-2\right)\left(-36\right)}}{2\left(-2\right)}$$

Square $18$.

$$x=\frac{-18±\sqrt{324-4\left(-2\right)\left(-36\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-18±\sqrt{324+8\left(-36\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-36$.

$$x=\frac{-18±\sqrt{324-288}}{2\left(-2\right)}$$

Add $324$ to $-288$.

$$x=\frac{-18±\sqrt{36}}{2\left(-2\right)}$$

Take the square root of $36$.

$$x=\frac{-18±6}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-18±6}{-4}$$

Now solve the equation $x=\frac{-18±6}{-4}$ when $±$ is plus. Add $-18$ to $6$.

$$x=-\frac{12}{-4}$$

Divide $-12$ by $-4$.

$$x=3$$

Now solve the equation $x=\frac{-18±6}{-4}$ when $±$ is minus. Subtract $6$ from $-18$.

$$x=-\frac{24}{-4}$$

Divide $-24$ by $-4$.

$$x=6$$

The equation is now solved.

$$x=3$$ $$x=6$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $1,5$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-5\right)\left(x-1\right)$, the least common multiple of $2x-10,2x-2$.

$$x-1+\left(x-5\right)\times 5=2\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply $x-5$ by $5$.

$$x-1+5x-25=2\left(x-5\right)\left(x-1\right)$$

Combine $x$ and $5x$ to get $6x$.

$$6x-1-25=2\left(x-5\right)\left(x-1\right)$$

Subtract $25$ from $-1$ to get $-26$.

$$6x-26=2\left(x-5\right)\left(x-1\right)$$

Use the distributive property to multiply $2$ by $x-5$.

$$6x-26=\left(2x-10\right)\left(x-1\right)$$

Use the distributive property to multiply $2x-10$ by $x-1$ and combine like terms.

$$6x-26=2x^{2}-12x+10$$

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

$$6x-26-2x^{2}=-12x+10$$

Add $12x$ to both sides.

$$6x-26-2x^{2}+12x=10$$

Combine $6x$ and $12x$ to get $18x$.

$$18x-26-2x^{2}=10$$

Add $26$ to both sides.

$$18x-2x^{2}=10+26$$

Add $10$ and $26$ to get $36$.

$$18x-2x^{2}=36$$

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 $x^{2}+bx=c$.

$$-2x^{2}+18x=36$$

Divide both sides by $-2$.

$$\frac{-2x^{2}+18x}{-2}=\frac{36}{-2}$$

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

$$x^{2}+\frac{18}{-2}x=\frac{36}{-2}$$

Divide $18$ by $-2$.

$$x^{2}-9x=\frac{36}{-2}$$

Divide $36$ by $-2$.

$$x^{2}-9x=-18$$

Divide $-9$, the coefficient of the $x$ term, by $2$ to get $-\frac{9}{2}$. Then add the square of $-\frac{9}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square $-\frac{9}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-18+\frac{81}{4}$$

Add $-18$ to $\frac{81}{4}$.

$$x^{2}-9x+\frac{81}{4}=\frac{9}{4}$$

Factor $x^{2}-9x+\frac{81}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{9}{2}=\frac{3}{2}$$ $$x-\frac{9}{2}=-\frac{3}{2}$$

Add $\frac{9}{2}$ to both sides of the equation.

$$x=6$$ $$x=3$$