Solve (3)/(x+2)-(1)/(x)=(4)/(15) | 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

$$\frac{ 3 }{ x+2 } - \frac{ 1 }{ x } = \frac{ 4 }{ 15 }$$

Solve for x

$x=3$
$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$

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 $-2,0$ since division by zero is not defined. Multiply both sides of the equation by $15x\left(x+2\right)$, the least common multiple of $x+2,x,15$.

$$15x\times 3-\left(15x+30\right)=4x\left(x+2\right)$$

Multiply $15$ and $3$ to get $45$.

$$45x-\left(15x+30\right)=4x\left(x+2\right)$$

To find the opposite of $15x+30$, find the opposite of each term.

$$45x-15x-30=4x\left(x+2\right)$$

Combine $45x$ and $-15x$ to get $30x$.

$$30x-30=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$30x-30=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

$$30x-30-4x^{2}=8x$$

Subtract $8x$ from both sides.

$$30x-30-4x^{2}-8x=0$$

Combine $30x$ and $-8x$ to get $22x$.

$$22x-30-4x^{2}=0$$

Divide both sides by $2$.

$$11x-15-2x^{2}=0$$

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

$$-2x^{2}+11x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.

$$a+b=11$$ $$ab=-2\left(-15\right)=30$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $30$.

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

Calculate the sum for each pair.

$$1+30=31$$ $$2+15=17$$ $$3+10=13$$ $$5+6=11$$

The solution is the pair that gives sum $11$.

$$a=6$$ $$b=5$$

Rewrite $-2x^{2}+11x-15$ as $\left(-2x^{2}+6x\right)+\left(5x-15\right)$.

$$\left(-2x^{2}+6x\right)+\left(5x-15\right)$$

Factor out $2x$ in the first and $-5$ in the second group.

$$2x\left(-x+3\right)-5\left(-x+3\right)$$

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

$$\left(-x+3\right)\left(2x-5\right)$$

To find equation solutions, solve $-x+3=0$ and $2x-5=0$.

$$x=3$$ $$x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-2,0$ since division by zero is not defined. Multiply both sides of the equation by $15x\left(x+2\right)$, the least common multiple of $x+2,x,15$.

$$15x\times 3-\left(15x+30\right)=4x\left(x+2\right)$$

Multiply $15$ and $3$ to get $45$.

$$45x-\left(15x+30\right)=4x\left(x+2\right)$$

To find the opposite of $15x+30$, find the opposite of each term.

$$45x-15x-30=4x\left(x+2\right)$$

Combine $45x$ and $-15x$ to get $30x$.

$$30x-30=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$30x-30=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

$$30x-30-4x^{2}=8x$$

Subtract $8x$ from both sides.

$$30x-30-4x^{2}-8x=0$$

Combine $30x$ and $-8x$ to get $22x$.

$$22x-30-4x^{2}=0$$

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.

$$-4x^{2}+22x-30=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-4$ for $a$, $22$ for $b$, and $-30$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-22±\sqrt{22^{2}-4\left(-4\right)\left(-30\right)}}{2\left(-4\right)}$$

Square $22$.

$$x=\frac{-22±\sqrt{484-4\left(-4\right)\left(-30\right)}}{2\left(-4\right)}$$

Multiply $-4$ times $-4$.

$$x=\frac{-22±\sqrt{484+16\left(-30\right)}}{2\left(-4\right)}$$

Multiply $16$ times $-30$.

$$x=\frac{-22±\sqrt{484-480}}{2\left(-4\right)}$$

Add $484$ to $-480$.

$$x=\frac{-22±\sqrt{4}}{2\left(-4\right)}$$

Take the square root of $4$.

$$x=\frac{-22±2}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{-22±2}{-8}$$

Now solve the equation $x=\frac{-22±2}{-8}$ when $±$ is plus. Add $-22$ to $2$.

$$x=-\frac{20}{-8}$$

Reduce the fraction $\frac{-20}{-8}$ to lowest terms by extracting and canceling out $4$.

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

Now solve the equation $x=\frac{-22±2}{-8}$ when $±$ is minus. Subtract $2$ from $-22$.

$$x=-\frac{24}{-8}$$

Divide $-24$ by $-8$.

$$x=3$$

The equation is now solved.

$$x=\frac{5}{2}$$ $$x=3$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-2,0$ since division by zero is not defined. Multiply both sides of the equation by $15x\left(x+2\right)$, the least common multiple of $x+2,x,15$.

$$15x\times 3-\left(15x+30\right)=4x\left(x+2\right)$$

Multiply $15$ and $3$ to get $45$.

$$45x-\left(15x+30\right)=4x\left(x+2\right)$$

To find the opposite of $15x+30$, find the opposite of each term.

$$45x-15x-30=4x\left(x+2\right)$$

Combine $45x$ and $-15x$ to get $30x$.

$$30x-30=4x\left(x+2\right)$$

Use the distributive property to multiply $4x$ by $x+2$.

$$30x-30=4x^{2}+8x$$

Subtract $4x^{2}$ from both sides.

$$30x-30-4x^{2}=8x$$

Subtract $8x$ from both sides.

$$30x-30-4x^{2}-8x=0$$

Combine $30x$ and $-8x$ to get $22x$.

$$22x-30-4x^{2}=0$$

Add $30$ to both sides. Anything plus zero gives itself.

$$22x-4x^{2}=30$$

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

$$-4x^{2}+22x=30$$

Divide both sides by $-4$.

$$\frac{-4x^{2}+22x}{-4}=\frac{30}{-4}$$

Dividing by $-4$ undoes the multiplication by $-4$.

$$x^{2}+\frac{22}{-4}x=\frac{30}{-4}$$

Reduce the fraction $\frac{22}{-4}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{11}{2}x=\frac{30}{-4}$$

Reduce the fraction $\frac{30}{-4}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{11}{2}x=-\frac{15}{2}$$

Divide $-\frac{11}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{11}{4}$. Then add the square of $-\frac{11}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{15}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square $-\frac{11}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{15}{2}+\frac{121}{16}$$

Add $-\frac{15}{2}$ to $\frac{121}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{1}{16}$$

Factor $x^{2}-\frac{11}{2}x+\frac{121}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{1}{4}$$ $$x-\frac{11}{4}=-\frac{1}{4}$$

Add $\frac{11}{4}$ to both sides of the equation.

$$x=3$$ $$x=\frac{5}{2}$$