How to solve nd (3x)/4 - 1 > x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve nd (3x)/4 - 1 > x | Equatix

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$$\frac { 3 x } { 4 } - 1 \gt x$$

$x<-4$

Subtract $x$ from both sides.

$$\frac{3x}{4}-1-x>0$$

Add $1$ to both sides. Anything plus zero gives itself.

$$\frac{3x}{4}-x>1$$

Multiply both sides of the equation by $4$. Since $4$ is positive, the inequality direction remains the same.

$$3x-4x>4$$

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

$$-x>4$$

Divide both sides by $-1$. Since $-1$ is negative, the inequality direction is changed.

$$x<-4$$