Solve 5/x - 1/(x - 2) = 1 | 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

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

Solve for x (complex solution)

$x=3-i$
$x=3+i$

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

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

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

$$5x-10-x=x\left(x-2\right)$$

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

$$5x-10-x=x^{2}-2x$$

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

$$5x-10-x-x^{2}=-2x$$

Add $2x$ to both sides.

$$5x-10-x-x^{2}+2x=0$$

Combine $5x$ and $2x$ to get $7x$.

$$7x-10-x-x^{2}=0$$

Combine $7x$ and $-x$ to get $6x$.

$$6x-10-x^{2}=0$$

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.

$$-x^{2}+6x-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $6$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Square $6$.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-6±\sqrt{36+4\left(-10\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-10$.

$$x=\frac{-6±\sqrt{36-40}}{2\left(-1\right)}$$

Add $36$ to $-40$.

$$x=\frac{-6±\sqrt{-4}}{2\left(-1\right)}$$

Take the square root of $-4$.

$$x=\frac{-6±2i}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-6±2i}{-2}$$

Now solve the equation $x=\frac{-6±2i}{-2}$ when $±$ is plus. Add $-6$ to $2i$.

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

Divide $-6+2i$ by $-2$.

$$x=3-i$$

Now solve the equation $x=\frac{-6±2i}{-2}$ when $±$ is minus. Subtract $2i$ from $-6$.

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

Divide $-6-2i$ by $-2$.

$$x=3+i$$

The equation is now solved.

$$x=3-i$$ $$x=3+i$$

Steps for Completing the Square

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

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

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

$$5x-10-x=x\left(x-2\right)$$

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

$$5x-10-x=x^{2}-2x$$

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

$$5x-10-x-x^{2}=-2x$$

Add $2x$ to both sides.

$$5x-10-x-x^{2}+2x=0$$

Combine $5x$ and $2x$ to get $7x$.

$$7x-10-x-x^{2}=0$$

Add $10$ to both sides. Anything plus zero gives itself.

$$7x-x-x^{2}=10$$

Combine $7x$ and $-x$ to get $6x$.

$$6x-x^{2}=10$$

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

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

Divide both sides by $-1$.

$$\frac{-x^{2}+6x}{-1}=\frac{10}{-1}$$

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

$$x^{2}+\frac{6}{-1}x=\frac{10}{-1}$$

Divide $6$ by $-1$.

$$x^{2}-6x=\frac{10}{-1}$$

Divide $10$ by $-1$.

$$x^{2}-6x=-10$$

Divide $-6$, the coefficient of the $x$ term, by $2$ to get $-3$. Then add the square of $-3$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-10+\left(-3\right)^{2}$$

Square $-3$.

$$x^{2}-6x+9=-10+9$$

Add $-10$ to $9$.

$$x^{2}-6x+9=-1$$

Factor $x^{2}-6x+9$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-3\right)^{2}=-1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-1}$$

Simplify.

$$x-3=i$$ $$x-3=-i$$

Add $3$ to both sides of the equation.

$$x=3+i$$ $$x=3-i$$