Solve i ) (x + 2)/(x - 1) - (4 - x)/(2x) = 2 1/3 | 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

$$\frac { x + 2 } { x - 1 } - \frac { 4 - x } { 2 x } = 2 \frac { 1 } { 3 }$$

Solve for x

$x=-\frac{4}{5}=-0.8$
$x=3$

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 $0,1$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x-1\right)$, the least common multiple of $x-1,2x,3$.

$$6x\left(x+2\right)-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $6x$ by $x+2$.

$$6x^{2}+12x-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $3x-3$ by $4-x$ and combine like terms.

$$6x^{2}+12x-\left(15x-3x^{2}-12\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

To find the opposite of $15x-3x^{2}-12$, find the opposite of each term.

$$6x^{2}+12x-15x+3x^{2}+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Combine $12x$ and $-15x$ to get $-3x$.

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

Combine $6x^{2}$ and $3x^{2}$ to get $9x^{2}$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Multiply $2$ and $3$ to get $6$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(6+1\right)$$

Add $6$ and $1$ to get $7$.

$$9x^{2}-3x+12=2x\left(x-1\right)\times 7$$

Multiply $2$ and $7$ to get $14$.

$$9x^{2}-3x+12=14x\left(x-1\right)$$

Use the distributive property to multiply $14x$ by $x-1$.

$$9x^{2}-3x+12=14x^{2}-14x$$

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

$$9x^{2}-3x+12-14x^{2}=-14x$$

Combine $9x^{2}$ and $-14x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x+12=-14x$$

Add $14x$ to both sides.

$$-5x^{2}-3x+12+14x=0$$

Combine $-3x$ and $14x$ to get $11x$.

$$-5x^{2}+11x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-5x^{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.

$$a+b=11$$ $$ab=-5\times 12=-60$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the 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.

$$-1+60=59$$ $$-2+30=28$$ $$-3+20=17$$ $$-4+15=11$$ $$-5+12=7$$ $$-6+10=4$$

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

$$a=15$$ $$b=-4$$

Rewrite $-5x^{2}+11x+12$ as $\left(-5x^{2}+15x\right)+\left(-4x+12\right)$.

$$\left(-5x^{2}+15x\right)+\left(-4x+12\right)$$

Factor out $5x$ in the first and $4$ in the second group.

$$5x\left(-x+3\right)+4\left(-x+3\right)$$

Factor out common term $-x+3$ by using distributive property.

$$\left(-x+3\right)\left(5x+4\right)$$

To find equation solutions, solve $-x+3=0$ and $5x+4=0$.

$$x=3$$ $$x=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $0,1$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x-1\right)$, the least common multiple of $x-1,2x,3$.

$$6x\left(x+2\right)-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $6x$ by $x+2$.

$$6x^{2}+12x-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $3x-3$ by $4-x$ and combine like terms.

$$6x^{2}+12x-\left(15x-3x^{2}-12\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

To find the opposite of $15x-3x^{2}-12$, find the opposite of each term.

$$6x^{2}+12x-15x+3x^{2}+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Combine $12x$ and $-15x$ to get $-3x$.

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

Combine $6x^{2}$ and $3x^{2}$ to get $9x^{2}$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Multiply $2$ and $3$ to get $6$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(6+1\right)$$

Add $6$ and $1$ to get $7$.

$$9x^{2}-3x+12=2x\left(x-1\right)\times 7$$

Multiply $2$ and $7$ to get $14$.

$$9x^{2}-3x+12=14x\left(x-1\right)$$

Use the distributive property to multiply $14x$ by $x-1$.

$$9x^{2}-3x+12=14x^{2}-14x$$

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

$$9x^{2}-3x+12-14x^{2}=-14x$$

Combine $9x^{2}$ and $-14x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x+12=-14x$$

Add $14x$ to both sides.

$$-5x^{2}-3x+12+14x=0$$

Combine $-3x$ and $14x$ to get $11x$.

$$-5x^{2}+11x+12=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $11$ for $b$, and $12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-5\right)\times 12}}{2\left(-5\right)}$$

Square $11$.

$$x=\frac{-11±\sqrt{121-4\left(-5\right)\times 12}}{2\left(-5\right)}$$

Multiply $-4$ times $-5$.

$$x=\frac{-11±\sqrt{121+20\times 12}}{2\left(-5\right)}$$

Multiply $20$ times $12$.

$$x=\frac{-11±\sqrt{121+240}}{2\left(-5\right)}$$

Add $121$ to $240$.

$$x=\frac{-11±\sqrt{361}}{2\left(-5\right)}$$

Take the square root of $361$.

$$x=\frac{-11±19}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{-11±19}{-10}$$

Now solve the equation $x=\frac{-11±19}{-10}$ when $±$ is plus. Add $-11$ to $19$.

$$x=\frac{8}{-10}$$

Reduce the fraction $\frac{8}{-10}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-11±19}{-10}$ when $±$ is minus. Subtract $19$ from $-11$.

$$x=-\frac{30}{-10}$$

Divide $-30$ by $-10$.

$$x=3$$

The equation is now solved.

$$x=-\frac{4}{5}$$ $$x=3$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $0,1$ since division by zero is not defined. Multiply both sides of the equation by $6x\left(x-1\right)$, the least common multiple of $x-1,2x,3$.

$$6x\left(x+2\right)-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $6x$ by $x+2$.

$$6x^{2}+12x-\left(3x-3\right)\left(4-x\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

Use the distributive property to multiply $3x-3$ by $4-x$ and combine like terms.

$$6x^{2}+12x-\left(15x-3x^{2}-12\right)=2x\left(x-1\right)\left(2\times 3+1\right)$$

To find the opposite of $15x-3x^{2}-12$, find the opposite of each term.

$$6x^{2}+12x-15x+3x^{2}+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Combine $12x$ and $-15x$ to get $-3x$.

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

Combine $6x^{2}$ and $3x^{2}$ to get $9x^{2}$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(2\times 3+1\right)$$

Multiply $2$ and $3$ to get $6$.

$$9x^{2}-3x+12=2x\left(x-1\right)\left(6+1\right)$$

Add $6$ and $1$ to get $7$.

$$9x^{2}-3x+12=2x\left(x-1\right)\times 7$$

Multiply $2$ and $7$ to get $14$.

$$9x^{2}-3x+12=14x\left(x-1\right)$$

Use the distributive property to multiply $14x$ by $x-1$.

$$9x^{2}-3x+12=14x^{2}-14x$$

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

$$9x^{2}-3x+12-14x^{2}=-14x$$

Combine $9x^{2}$ and $-14x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-3x+12=-14x$$

Add $14x$ to both sides.

$$-5x^{2}-3x+12+14x=0$$

Combine $-3x$ and $14x$ to get $11x$.

$$-5x^{2}+11x+12=0$$

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

$$-5x^{2}+11x=-12$$

Divide both sides by $-5$.

$$\frac{-5x^{2}+11x}{-5}=-\frac{12}{-5}$$

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

$$x^{2}+\frac{11}{-5}x=-\frac{12}{-5}$$

Divide $11$ by $-5$.

$$x^{2}-\frac{11}{5}x=-\frac{12}{-5}$$

Divide $-12$ by $-5$.

$$x^{2}-\frac{11}{5}x=\frac{12}{5}$$

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.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=\frac{12}{5}+\left(-\frac{11}{10}\right)^{2}$$

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

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{12}{5}+\frac{121}{100}$$

Add $\frac{12}{5}$ to $\frac{121}{100}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{361}{100}$$

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

$$\left(x-\frac{11}{10}\right)^{2}=\frac{361}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{361}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{19}{10}$$ $$x-\frac{11}{10}=-\frac{19}{10}$$

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

$$x=3$$ $$x=-\frac{4}{5}$$