Solve 5x - 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

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

Solve for x

$x = \frac{\sqrt{101} + 11}{10} \approx 2.104987562$
$x=\frac{11-\sqrt{101}}{10}\approx 0.095012438$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $x-2$.

$$5x\left(x-2\right)-1=x-2$$

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

$$5x^{2}-10x-1=x-2$$

Subtract $x$ from both sides.

$$5x^{2}-10x-1-x=-2$$

Combine $-10x$ and $-x$ to get $-11x$.

$$5x^{2}-11x-1=-2$$

Add $2$ to both sides.

$$5x^{2}-11x-1+2=0$$

Add $-1$ and $2$ to get $1$.

$$5x^{2}-11x+1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-11$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 5}}{2\times 5}$$

Square $-11$.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 5}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-11\right)±\sqrt{121-20}}{2\times 5}$$

Add $121$ to $-20$.

$$x=\frac{-\left(-11\right)±\sqrt{101}}{2\times 5}$$

The opposite of $-11$ is $11$.

$$x=\frac{11±\sqrt{101}}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{11±\sqrt{101}}{10}$$

Now solve the equation $x=\frac{11±\sqrt{101}}{10}$ when $±$ is plus. Add $11$ to $\sqrt{101}$.

$$x=\frac{\sqrt{101}+11}{10}$$

Now solve the equation $x=\frac{11±\sqrt{101}}{10}$ when $±$ is minus. Subtract $\sqrt{101}$ from $11$.

$$x=\frac{11-\sqrt{101}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{101}+11}{10}$$ $$x=\frac{11-\sqrt{101}}{10}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $2$ since division by zero is not defined. Multiply both sides of the equation by $x-2$.

$$5x\left(x-2\right)-1=x-2$$

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

$$5x^{2}-10x-1=x-2$$

Subtract $x$ from both sides.

$$5x^{2}-10x-1-x=-2$$

Combine $-10x$ and $-x$ to get $-11x$.

$$5x^{2}-11x-1=-2$$

Add $1$ to both sides.

$$5x^{2}-11x=-2+1$$

Add $-2$ and $1$ to get $-1$.

$$5x^{2}-11x=-1$$

Divide both sides by $5$.

$$\frac{5x^{2}-11x}{5}=-\frac{1}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}-\frac{11}{5}x=-\frac{1}{5}$$

Divide $-\frac{11}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{11}{10}$. Then add the square of $-\frac{11}{10}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{1}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square $-\frac{11}{10}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{1}{5}+\frac{121}{100}$$

Add $-\frac{1}{5}$ to $\frac{121}{100}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{101}{100}$$

Factor $x^{2}-\frac{11}{5}x+\frac{121}{100}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{101}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{101}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{\sqrt{101}}{10}$$ $$x-\frac{11}{10}=-\frac{\sqrt{101}}{10}$$

Add $\frac{11}{10}$ to both sides of the equation.

$$x=\frac{\sqrt{101}+11}{10}$$ $$x=\frac{11-\sqrt{101}}{10}$$