Solve (x + 1)/2 = (x(x + 2))/3 = 0 x + 1 ) = 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

$$\left. \begin{array} { l } { \frac { x + 1 } { 2 } = \frac { x ( x + 2 ) } { 3 } = 0 } \\ { x + 1 = 2 } \end{array} \right.$$

Solve for x

$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Multiply both sides of the equation by $6$, the least common multiple of $2,3$.

$$3\left(x+1\right)=2x\left(x+2\right)$$

Use the distributive property to multiply $3$ by $x+1$.

$$3x+3=2x\left(x+2\right)$$

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

$$3x+3=2x^{2}+4x$$

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

$$3x+3-2x^{2}=4x$$

Subtract $4x$ from both sides.

$$3x+3-2x^{2}-4x=0$$

Combine $3x$ and $-4x$ to get $-x$.

$$-x+3-2x^{2}=0$$

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

$$-2x^{2}-x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-2x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-1$$ $$ab=-2\times 3=-6$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-6$.

$$1,-6$$ $$2,-3$$

Calculate the sum for each pair.

$$1-6=-5$$ $$2-3=-1$$

The solution is the pair that gives sum $-1$.

$$a=2$$ $$b=-3$$

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

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

Factor out $2x$ in the first and $3$ in the second group.

$$2x\left(-x+1\right)+3\left(-x+1\right)$$

Factor out common term $-x+1$ by using distributive property.

$$\left(-x+1\right)\left(2x+3\right)$$

To find equation solutions, solve $-x+1=0$ and $2x+3=0$.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by $6$, the least common multiple of $2,3$.

$$3\left(x+1\right)=2x\left(x+2\right)$$

Use the distributive property to multiply $3$ by $x+1$.

$$3x+3=2x\left(x+2\right)$$

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

$$3x+3=2x^{2}+4x$$

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

$$3x+3-2x^{2}=4x$$

Subtract $4x$ from both sides.

$$3x+3-2x^{2}-4x=0$$

Combine $3x$ and $-4x$ to get $-x$.

$$-x+3-2x^{2}=0$$

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.

$$-2x^{2}-x+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-1$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\times 3}}{2\left(-2\right)}$$

Multiply $8$ times $3$.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-2\right)}$$

Add $1$ to $24$.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of $25$.

$$x=\frac{-\left(-1\right)±5}{2\left(-2\right)}$$

The opposite of $-1$ is $1$.

$$x=\frac{1±5}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{1±5}{-4}$$

Now solve the equation $x=\frac{1±5}{-4}$ when $±$ is plus. Add $1$ to $5$.

$$x=\frac{6}{-4}$$

Reduce the fraction $\frac{6}{-4}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{1±5}{-4}$ when $±$ is minus. Subtract $5$ from $1$.

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

Divide $-4$ by $-4$.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by $6$, the least common multiple of $2,3$.

$$3\left(x+1\right)=2x\left(x+2\right)$$

Use the distributive property to multiply $3$ by $x+1$.

$$3x+3=2x\left(x+2\right)$$

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

$$3x+3=2x^{2}+4x$$

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

$$3x+3-2x^{2}=4x$$

Subtract $4x$ from both sides.

$$3x+3-2x^{2}-4x=0$$

Combine $3x$ and $-4x$ to get $-x$.

$$-x+3-2x^{2}=0$$

Subtract $3$ from both sides. Anything subtracted from zero gives its negation.

$$-x-2x^{2}=-3$$

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

$$-2x^{2}-x=-3$$

Divide both sides by $-2$.

$$\frac{-2x^{2}-x}{-2}=-\frac{3}{-2}$$

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

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

Divide $-1$ by $-2$.

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

Divide $-3$ by $-2$.

$$x^{2}+\frac{1}{2}x=\frac{3}{2}$$

Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{3}{2}+\left(\frac{1}{4}\right)^{2}$$

Square $\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add $\frac{3}{2}$ to $\frac{1}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{5}{4}$$ $$x+\frac{1}{4}=-\frac{5}{4}$$

Subtract $\frac{1}{4}$ from both sides of the equation.

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