Solve (3x - 1)/x - (2x)/(2x - 1) - 7/6 = 0 | 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

$$\frac{3x-1}{x}-\frac{2x}{2x-1}-\frac{7}{6}=0$$

Solve for x

$x=2$
$x=\frac{3}{10}=0.3$

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 $0,\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(2x-1\right)$, the least common multiple of $x,2x-1,6$.

$$\left(12x-6\right)\left(3x-1\right)-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Use the distributive property to multiply $12x-6$ by $3x-1$ and combine like terms.

$$36x^{2}-30x+6-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $x$ and $x$ to get $x^{2}$.

$$36x^{2}-30x+6-6x^{2}\times 2+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $2$ to get $12$.

$$36x^{2}-30x+6-12x^{2}+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $-\frac{7}{6}$ to get $-7$.

$$36x^{2}-30x+6-12x^{2}-7x\left(2x-1\right)=0$$

Use the distributive property to multiply $-7x$ by $2x-1$.

$$36x^{2}-30x+6-12x^{2}-14x^{2}+7x=0$$

Combine $36x^{2}$ and $-14x^{2}$ to get $22x^{2}$.

$$22x^{2}-30x+6-12x^{2}+7x=0$$

Combine $-30x$ and $7x$ to get $-23x$.

$$22x^{2}-23x+6-12x^{2}=0$$

Combine $22x^{2}$ and $-12x^{2}$ to get $10x^{2}$.

$$10x^{2}-23x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $10x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-23$$ $$ab=10\times 6=60$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $60$.

$$-1,-60$$ $$-2,-30$$ $$-3,-20$$ $$-4,-15$$ $$-5,-12$$ $$-6,-10$$

Calculate the sum for each pair.

$$-1-60=-61$$ $$-2-30=-32$$ $$-3-20=-23$$ $$-4-15=-19$$ $$-5-12=-17$$ $$-6-10=-16$$

The solution is the pair that gives sum $-23$.

$$a=-20$$ $$b=-3$$

Rewrite $10x^{2}-23x+6$ as $\left(10x^{2}-20x\right)+\left(-3x+6\right)$.

$$\left(10x^{2}-20x\right)+\left(-3x+6\right)$$

Factor out $10x$ in the first and $-3$ in the second group.

$$10x\left(x-2\right)-3\left(x-2\right)$$

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

$$\left(x-2\right)\left(10x-3\right)$$

To find equation solutions, solve $x-2=0$ and $10x-3=0$.

$$x=2$$ $$x=\frac{3}{10}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $0,\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(2x-1\right)$, the least common multiple of $x,2x-1,6$.

$$\left(12x-6\right)\left(3x-1\right)-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Use the distributive property to multiply $12x-6$ by $3x-1$ and combine like terms.

$$36x^{2}-30x+6-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $x$ and $x$ to get $x^{2}$.

$$36x^{2}-30x+6-6x^{2}\times 2+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $2$ to get $12$.

$$36x^{2}-30x+6-12x^{2}+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $-\frac{7}{6}$ to get $-7$.

$$36x^{2}-30x+6-12x^{2}-7x\left(2x-1\right)=0$$

Use the distributive property to multiply $-7x$ by $2x-1$.

$$36x^{2}-30x+6-12x^{2}-14x^{2}+7x=0$$

Combine $36x^{2}$ and $-14x^{2}$ to get $22x^{2}$.

$$22x^{2}-30x+6-12x^{2}+7x=0$$

Combine $-30x$ and $7x$ to get $-23x$.

$$22x^{2}-23x+6-12x^{2}=0$$

Combine $22x^{2}$ and $-12x^{2}$ to get $10x^{2}$.

$$10x^{2}-23x+6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $10$ for $a$, $-23$ for $b$, and $6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 10\times 6}}{2\times 10}$$

Square $-23$.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 10\times 6}}{2\times 10}$$

Multiply $-4$ times $10$.

$$x=\frac{-\left(-23\right)±\sqrt{529-40\times 6}}{2\times 10}$$

Multiply $-40$ times $6$.

$$x=\frac{-\left(-23\right)±\sqrt{529-240}}{2\times 10}$$

Add $529$ to $-240$.

$$x=\frac{-\left(-23\right)±\sqrt{289}}{2\times 10}$$

Take the square root of $289$.

$$x=\frac{-\left(-23\right)±17}{2\times 10}$$

The opposite of $-23$ is $23$.

$$x=\frac{23±17}{2\times 10}$$

Multiply $2$ times $10$.

$$x=\frac{23±17}{20}$$

Now solve the equation $x=\frac{23±17}{20}$ when $±$ is plus. Add $23$ to $17$.

$$x=\frac{40}{20}$$

Divide $40$ by $20$.

$$x=2$$

Now solve the equation $x=\frac{23±17}{20}$ when $±$ is minus. Subtract $17$ from $23$.

$$x=\frac{6}{20}$$

Reduce the fraction $\frac{6}{20}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{3}{10}$$

The equation is now solved.

$$x=2$$ $$x=\frac{3}{10}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $0,\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(2x-1\right)$, the least common multiple of $x,2x-1,6$.

$$\left(12x-6\right)\left(3x-1\right)-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Use the distributive property to multiply $12x-6$ by $3x-1$ and combine like terms.

$$36x^{2}-30x+6-6x\times 2x+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $x$ and $x$ to get $x^{2}$.

$$36x^{2}-30x+6-6x^{2}\times 2+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $2$ to get $12$.

$$36x^{2}-30x+6-12x^{2}+6x\left(2x-1\right)\left(-\frac{7}{6}\right)=0$$

Multiply $6$ and $-\frac{7}{6}$ to get $-7$.

$$36x^{2}-30x+6-12x^{2}-7x\left(2x-1\right)=0$$

Use the distributive property to multiply $-7x$ by $2x-1$.

$$36x^{2}-30x+6-12x^{2}-14x^{2}+7x=0$$

Combine $36x^{2}$ and $-14x^{2}$ to get $22x^{2}$.

$$22x^{2}-30x+6-12x^{2}+7x=0$$

Combine $-30x$ and $7x$ to get $-23x$.

$$22x^{2}-23x+6-12x^{2}=0$$

Subtract $6$ from both sides. Anything subtracted from zero gives its negation.

$$22x^{2}-23x-12x^{2}=-6$$

Combine $22x^{2}$ and $-12x^{2}$ to get $10x^{2}$.

$$10x^{2}-23x=-6$$

Divide both sides by $10$.

$$\frac{10x^{2}-23x}{10}=-\frac{6}{10}$$

Dividing by $10$ undoes the multiplication by $10$.

$$x^{2}-\frac{23}{10}x=-\frac{6}{10}$$

Reduce the fraction $\frac{-6}{10}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}-\frac{23}{10}x=-\frac{3}{5}$$

Divide $-\frac{23}{10}$, the coefficient of the $x$ term, by $2$ to get $-\frac{23}{20}$. Then add the square of $-\frac{23}{20}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{10}x+\left(-\frac{23}{20}\right)^{2}=-\frac{3}{5}+\left(-\frac{23}{20}\right)^{2}$$

Square $-\frac{23}{20}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=-\frac{3}{5}+\frac{529}{400}$$

Add $-\frac{3}{5}$ to $\frac{529}{400}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{10}x+\frac{529}{400}=\frac{289}{400}$$

Factor $x^{2}-\frac{23}{10}x+\frac{529}{400}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{23}{20}\right)^{2}=\frac{289}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{20}\right)^{2}}=\sqrt{\frac{289}{400}}$$

Simplify.

$$x-\frac{23}{20}=\frac{17}{20}$$ $$x-\frac{23}{20}=-\frac{17}{20}$$

Add $\frac{23}{20}$ to both sides of the equation.

$$x=2$$ $$x=\frac{3}{10}$$