Solve (1)/(x)+(1)/(x+5)=(1)/(6) | 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

$$\frac{ 1 }{ x } + \frac{ 1 }{ x+5 } = \frac{ 1 }{ 6 }$$

Solve for x

$x=-3$
$x=10$

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 $-5,0$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x+5\right)$, the least common multiple of $x,x+5,6$.

$$6x+30+6x=x\left(x+5\right)$$

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

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

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

$$12x+30-x^{2}=5x$$

Subtract $5x$ from both sides.

$$12x+30-x^{2}-5x=0$$

Combine $12x$ and $-5x$ to get $7x$.

$$7x+30-x^{2}=0$$

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

$$-x^{2}+7x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+30$. To find $a$ and $b$, set up a system to be solved.

$$a+b=7$$ $$ab=-30=-30$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-30$.

$$-1,30$$ $$-2,15$$ $$-3,10$$ $$-5,6$$

Calculate the sum for each pair.

$$-1+30=29$$ $$-2+15=13$$ $$-3+10=7$$ $$-5+6=1$$

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

$$a=10$$ $$b=-3$$

Rewrite $-x^{2}+7x+30$ as $\left(-x^{2}+10x\right)+\left(-3x+30\right)$.

$$\left(-x^{2}+10x\right)+\left(-3x+30\right)$$

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

$$-x\left(x-10\right)-3\left(x-10\right)$$

Factor out common term $x-10$ by using distributive property.

$$\left(x-10\right)\left(-x-3\right)$$

To find equation solutions, solve $x-10=0$ and $-x-3=0$.

$$x=10$$ $$x=-3$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-5,0$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x+5\right)$, the least common multiple of $x,x+5,6$.

$$6x+30+6x=x\left(x+5\right)$$

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

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

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

$$12x+30-x^{2}=5x$$

Subtract $5x$ from both sides.

$$12x+30-x^{2}-5x=0$$

Combine $12x$ and $-5x$ to get $7x$.

$$7x+30-x^{2}=0$$

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.

$$-x^{2}+7x+30=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $7$ for $b$, and $30$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Square $7$.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-7±\sqrt{49+4\times 30}}{2\left(-1\right)}$$

Multiply $4$ times $30$.

$$x=\frac{-7±\sqrt{49+120}}{2\left(-1\right)}$$

Add $49$ to $120$.

$$x=\frac{-7±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of $169$.

$$x=\frac{-7±13}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-7±13}{-2}$$

Now solve the equation $x=\frac{-7±13}{-2}$ when $±$ is plus. Add $-7$ to $13$.

$$x=\frac{6}{-2}$$

Divide $6$ by $-2$.

$$x=-3$$

Now solve the equation $x=\frac{-7±13}{-2}$ when $±$ is minus. Subtract $13$ from $-7$.

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

Divide $-20$ by $-2$.

$$x=10$$

The equation is now solved.

$$x=-3$$ $$x=10$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-5,0$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x+5\right)$, the least common multiple of $x,x+5,6$.

$$6x+30+6x=x\left(x+5\right)$$

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

$$12x+30=x\left(x+5\right)$$

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

$$12x+30=x^{2}+5x$$

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

$$12x+30-x^{2}=5x$$

Subtract $5x$ from both sides.

$$12x+30-x^{2}-5x=0$$

Combine $12x$ and $-5x$ to get $7x$.

$$7x+30-x^{2}=0$$

Subtract $30$ from both sides. Anything subtracted from zero gives its negation.

$$7x-x^{2}=-30$$

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

$$-x^{2}+7x=-30$$

Divide both sides by $-1$.

$$\frac{-x^{2}+7x}{-1}=-\frac{30}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{7}{-1}x=-\frac{30}{-1}$$

Divide $7$ by $-1$.

$$x^{2}-7x=-\frac{30}{-1}$$

Divide $-30$ by $-1$.

$$x^{2}-7x=30$$

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

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=30+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=30+\frac{49}{4}$$

Add $30$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{169}{4}$$

Factor $x^{2}-7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{13}{2}$$ $$x-\frac{7}{2}=-\frac{13}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$x=10$$ $$x=-3$$