Solve (x + 2)/x + (x + 4)/(x + 2) = 14/3 | 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

$$\frac{x+2}{x}+\frac{x+4}{x+2}=\frac{14}{3}$$

Solve for x

$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$
$x=1$

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 $3x\left(x+2\right)$, the least common multiple of $x,x+2,3$.

$$\left(3x+6\right)\left(x+2\right)+3x\left(x+4\right)=14x\left(x+2\right)$$

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

$$3x^{2}+12x+12+3x\left(x+4\right)=14x\left(x+2\right)$$

Use the distributive property to multiply $3x$ by $x+4$.

$$3x^{2}+12x+12+3x^{2}+12x=14x\left(x+2\right)$$

Combine $3x^{2}$ and $3x^{2}$ to get $6x^{2}$.

$$6x^{2}+12x+12+12x=14x\left(x+2\right)$$

Combine $12x$ and $12x$ to get $24x$.

$$6x^{2}+24x+12=14x\left(x+2\right)$$

Use the distributive property to multiply $14x$ by $x+2$.

$$6x^{2}+24x+12=14x^{2}+28x$$

Subtract $14x^{2}$ from both sides.

$$6x^{2}+24x+12-14x^{2}=28x$$

Combine $6x^{2}$ and $-14x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+24x+12=28x$$

Subtract $28x$ from both sides.

$$-8x^{2}+24x+12-28x=0$$

Combine $24x$ and $-28x$ to get $-4x$.

$$-8x^{2}-4x+12=0$$

Divide both sides by $4$.

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

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

$$a+b=-1$$ $$ab=-2\times 3=-6$$

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 $-6$.

$$1,-6$$ $$2,-3$$

Calculate the sum for each pair.

$$1-6=-5$$ $$2-3=-1$$

The solution is the pair that gives sum $-1$.

$$a=2$$ $$b=-3$$

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

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

Factor out $2x$ in the first and $3$ in the second group.

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

Factor out common term $-x+1$ by using distributive property.

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

To find equation solutions, solve $-x+1=0$ and $2x+3=0$.

$$x=1$$ $$x=-\frac{3}{2}$$

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 $3x\left(x+2\right)$, the least common multiple of $x,x+2,3$.

$$\left(3x+6\right)\left(x+2\right)+3x\left(x+4\right)=14x\left(x+2\right)$$

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

$$3x^{2}+12x+12+3x\left(x+4\right)=14x\left(x+2\right)$$

Use the distributive property to multiply $3x$ by $x+4$.

$$3x^{2}+12x+12+3x^{2}+12x=14x\left(x+2\right)$$

Combine $3x^{2}$ and $3x^{2}$ to get $6x^{2}$.

$$6x^{2}+12x+12+12x=14x\left(x+2\right)$$

Combine $12x$ and $12x$ to get $24x$.

$$6x^{2}+24x+12=14x\left(x+2\right)$$

Use the distributive property to multiply $14x$ by $x+2$.

$$6x^{2}+24x+12=14x^{2}+28x$$

Subtract $14x^{2}$ from both sides.

$$6x^{2}+24x+12-14x^{2}=28x$$

Combine $6x^{2}$ and $-14x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+24x+12=28x$$

Subtract $28x$ from both sides.

$$-8x^{2}+24x+12-28x=0$$

Combine $24x$ and $-28x$ to get $-4x$.

$$-8x^{2}-4x+12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-8$ for $a$, $-4$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-8\right)\times 12}}{2\left(-8\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-8\right)\times 12}}{2\left(-8\right)}$$

Multiply $-4$ times $-8$.

$$x=\frac{-\left(-4\right)±\sqrt{16+32\times 12}}{2\left(-8\right)}$$

Multiply $32$ times $12$.

$$x=\frac{-\left(-4\right)±\sqrt{16+384}}{2\left(-8\right)}$$

Add $16$ to $384$.

$$x=\frac{-\left(-4\right)±\sqrt{400}}{2\left(-8\right)}$$

Take the square root of $400$.

$$x=\frac{-\left(-4\right)±20}{2\left(-8\right)}$$

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

$$x=\frac{4±20}{2\left(-8\right)}$$

Multiply $2$ times $-8$.

$$x=\frac{4±20}{-16}$$

Now solve the equation $x=\frac{4±20}{-16}$ when $±$ is plus. Add $4$ to $20$.

$$x=\frac{24}{-16}$$

Reduce the fraction $\frac{24}{-16}$ to lowest terms by extracting and canceling out $8$.

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

Now solve the equation $x=\frac{4±20}{-16}$ when $±$ is minus. Subtract $20$ from $4$.

$$x=-\frac{16}{-16}$$

Divide $-16$ by $-16$.

$$x=1$$

The equation is now solved.

$$x=-\frac{3}{2}$$ $$x=1$$

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 $3x\left(x+2\right)$, the least common multiple of $x,x+2,3$.

$$\left(3x+6\right)\left(x+2\right)+3x\left(x+4\right)=14x\left(x+2\right)$$

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

$$3x^{2}+12x+12+3x\left(x+4\right)=14x\left(x+2\right)$$

Use the distributive property to multiply $3x$ by $x+4$.

$$3x^{2}+12x+12+3x^{2}+12x=14x\left(x+2\right)$$

Combine $3x^{2}$ and $3x^{2}$ to get $6x^{2}$.

$$6x^{2}+12x+12+12x=14x\left(x+2\right)$$

Combine $12x$ and $12x$ to get $24x$.

$$6x^{2}+24x+12=14x\left(x+2\right)$$

Use the distributive property to multiply $14x$ by $x+2$.

$$6x^{2}+24x+12=14x^{2}+28x$$

Subtract $14x^{2}$ from both sides.

$$6x^{2}+24x+12-14x^{2}=28x$$

Combine $6x^{2}$ and $-14x^{2}$ to get $-8x^{2}$.

$$-8x^{2}+24x+12=28x$$

Subtract $28x$ from both sides.

$$-8x^{2}+24x+12-28x=0$$

Combine $24x$ and $-28x$ to get $-4x$.

$$-8x^{2}-4x+12=0$$

Subtract $12$ from both sides. Anything subtracted from zero gives its negation.

$$-8x^{2}-4x=-12$$

Divide both sides by $-8$.

$$\frac{-8x^{2}-4x}{-8}=-\frac{12}{-8}$$

Dividing by $-8$ undoes the multiplication by $-8$.

$$x^{2}+\left(-\frac{4}{-8}\right)x=-\frac{12}{-8}$$

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

$$x^{2}+\frac{1}{2}x=-\frac{12}{-8}$$

Reduce the fraction $\frac{-12}{-8}$ to lowest terms by extracting and canceling out $4$.

$$x^{2}+\frac{1}{2}x=\frac{3}{2}$$

Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{3}{2}+\left(\frac{1}{4}\right)^{2}$$

Square $\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add $\frac{3}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{5}{4}$$ $$x+\frac{1}{4}=-\frac{5}{4}$$

Subtract $\frac{1}{4}$ from both sides of the equation.

$$x=1$$ $$x=-\frac{3}{2}$$