Solve 2/x - 5/(x - 7) = 2 | 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

$$\frac{2}{x}-\frac{5}{x-7}=2$$

Solve for x

$x=2$
$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$

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 $0,7$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-7\right)$, the least common multiple of $x,x-7$.

$$\left(x-7\right)\times 2-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $x-7$ by $2$.

$$2x-14-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $2x$ by $x-7$.

$$2x-14-x\times 5=2x^{2}-14x$$

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

$$2x-14-x\times 5-2x^{2}=-14x$$

Add $14x$ to both sides.

$$2x-14-x\times 5-2x^{2}+14x=0$$

Combine $2x$ and $14x$ to get $16x$.

$$16x-14-x\times 5-2x^{2}=0$$

Multiply $-1$ and $5$ to get $-5$.

$$16x-14-5x-2x^{2}=0$$

Combine $16x$ and $-5x$ to get $11x$.

$$11x-14-2x^{2}=0$$

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

$$-2x^{2}+11x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx-14$. To find $a$ and $b$, set up a system to be solved.

$$a+b=11$$ $$ab=-2\left(-14\right)=28$$

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 $28$.

$$1,28$$ $$2,14$$ $$4,7$$

Calculate the sum for each pair.

$$1+28=29$$ $$2+14=16$$ $$4+7=11$$

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

$$a=7$$ $$b=4$$

Rewrite $-2x^{2}+11x-14$ as $\left(-2x^{2}+7x\right)+\left(4x-14\right)$.

$$\left(-2x^{2}+7x\right)+\left(4x-14\right)$$

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

$$-x\left(2x-7\right)+2\left(2x-7\right)$$

Factor out common term $2x-7$ by using distributive property.

$$\left(2x-7\right)\left(-x+2\right)$$

To find equation solutions, solve $2x-7=0$ and $-x+2=0$.

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

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $0,7$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-7\right)$, the least common multiple of $x,x-7$.

$$\left(x-7\right)\times 2-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $x-7$ by $2$.

$$2x-14-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $2x$ by $x-7$.

$$2x-14-x\times 5=2x^{2}-14x$$

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

$$2x-14-x\times 5-2x^{2}=-14x$$

Add $14x$ to both sides.

$$2x-14-x\times 5-2x^{2}+14x=0$$

Combine $2x$ and $14x$ to get $16x$.

$$16x-14-x\times 5-2x^{2}=0$$

Multiply $-1$ and $5$ to get $-5$.

$$16x-14-5x-2x^{2}=0$$

Combine $16x$ and $-5x$ to get $11x$.

$$11x-14-2x^{2}=0$$

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.

$$-2x^{2}+11x-14=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $11$ for $b$, and $-14$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

Square $11$.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-14\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-11±\sqrt{121+8\left(-14\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-14$.

$$x=\frac{-11±\sqrt{121-112}}{2\left(-2\right)}$$

Add $121$ to $-112$.

$$x=\frac{-11±\sqrt{9}}{2\left(-2\right)}$$

Take the square root of $9$.

$$x=\frac{-11±3}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-11±3}{-4}$$

Now solve the equation $x=\frac{-11±3}{-4}$ when $±$ is plus. Add $-11$ to $3$.

$$x=-\frac{8}{-4}$$

Divide $-8$ by $-4$.

$$x=2$$

Now solve the equation $x=\frac{-11±3}{-4}$ when $±$ is minus. Subtract $3$ from $-11$.

$$x=-\frac{14}{-4}$$

Reduce the fraction $\frac{-14}{-4}$ to lowest terms by extracting and canceling out $2$.

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

The equation is now solved.

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

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $0,7$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x-7\right)$, the least common multiple of $x,x-7$.

$$\left(x-7\right)\times 2-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $x-7$ by $2$.

$$2x-14-x\times 5=2x\left(x-7\right)$$

Use the distributive property to multiply $2x$ by $x-7$.

$$2x-14-x\times 5=2x^{2}-14x$$

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

$$2x-14-x\times 5-2x^{2}=-14x$$

Add $14x$ to both sides.

$$2x-14-x\times 5-2x^{2}+14x=0$$

Combine $2x$ and $14x$ to get $16x$.

$$16x-14-x\times 5-2x^{2}=0$$

Add $14$ to both sides. Anything plus zero gives itself.

$$16x-x\times 5-2x^{2}=14$$

Multiply $-1$ and $5$ to get $-5$.

$$16x-5x-2x^{2}=14$$

Combine $16x$ and $-5x$ to get $11x$.

$$11x-2x^{2}=14$$

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

$$-2x^{2}+11x=14$$

Divide both sides by $-2$.

$$\frac{-2x^{2}+11x}{-2}=\frac{14}{-2}$$

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

$$x^{2}+\frac{11}{-2}x=\frac{14}{-2}$$

Divide $11$ by $-2$.

$$x^{2}-\frac{11}{2}x=\frac{14}{-2}$$

Divide $14$ by $-2$.

$$x^{2}-\frac{11}{2}x=-7$$

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.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-7+\left(-\frac{11}{4}\right)^{2}$$

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

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-7+\frac{121}{16}$$

Add $-7$ to $\frac{121}{16}$.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{9}{16}$$

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 $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{3}{4}$$ $$x-\frac{11}{4}=-\frac{3}{4}$$

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

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