Solve ii. 3x + 4/(6x - 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

$$3 x + \frac { 4 } { 6 x - 2 } = - 1$$

Solve for x (complex solution)

$x=\frac{1}{3}i\approx 0.333333333i$
$x=-\frac{1}{3}i\approx -0.333333333i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Variable $x$ cannot be equal to $\frac{1}{3}$ since division by zero is not defined. Multiply both sides of the equation by $2\left(3x-1\right)$.

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

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

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

Use the distributive property to multiply $6x$ by $3x-1$.

$$18x^{2}-6x+4=-2\left(3x-1\right)$$

Use the distributive property to multiply $-2$ by $3x-1$.

$$18x^{2}-6x+4=-6x+2$$

Add $6x$ to both sides.

$$18x^{2}-6x+4+6x=2$$

Combine $-6x$ and $6x$ to get $0$.

$$18x^{2}+4=2$$

Subtract $4$ from both sides.

$$18x^{2}=2-4$$

Subtract $4$ from $2$ to get $-2$.

$$18x^{2}=-2$$

Divide both sides by $18$.

$$x^{2}=\frac{-2}{18}$$

Reduce the fraction $\frac{-2}{18}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}=-\frac{1}{9}$$

The equation is now solved.

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

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $\frac{1}{3}$ since division by zero is not defined. Multiply both sides of the equation by $2\left(3x-1\right)$.

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

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

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

Use the distributive property to multiply $6x$ by $3x-1$.

$$18x^{2}-6x+4=-2\left(3x-1\right)$$

Use the distributive property to multiply $-2$ by $3x-1$.

$$18x^{2}-6x+4=-6x+2$$

Add $6x$ to both sides.

$$18x^{2}-6x+4+6x=2$$

Combine $-6x$ and $6x$ to get $0$.

$$18x^{2}+4=2$$

Subtract $2$ from both sides.

$$18x^{2}+4-2=0$$

Subtract $2$ from $4$ to get $2$.

$$18x^{2}+2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $18$ for $a$, $0$ for $b$, and $2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\times 18\times 2}}{2\times 18}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\times 18\times 2}}{2\times 18}$$

Multiply $-4$ times $18$.

$$x=\frac{0±\sqrt{-72\times 2}}{2\times 18}$$

Multiply $-72$ times $2$.

$$x=\frac{0±\sqrt{-144}}{2\times 18}$$

Take the square root of $-144$.

$$x=\frac{0±12i}{2\times 18}$$

Multiply $2$ times $18$.

$$x=\frac{0±12i}{36}$$

Now solve the equation $x=\frac{0±12i}{36}$ when $±$ is plus.

$$x=\frac{1}{3}i$$

Now solve the equation $x=\frac{0±12i}{36}$ when $±$ is minus.

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

The equation is now solved.

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