How to solve 3x+11 <= 6x+8

Q: How to solve 3x+11 <= 6x+8

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 3x+11 <= 6x+8 | Equatix

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Question

$$3x+11 \leq 6x+8$$

Solve for x

$x\geq 1$

Solution Steps

Subtract $6x$ from both sides.

$$3x+11-6x\leq 8$$

Combine $3x$ and $-6x$ to get $-3x$.

$$-3x+11\leq 8$$

Subtract $11$ from both sides.

$$-3x\leq 8-11$$

Subtract $11$ from $8$ to get $-3$.

$$-3x\leq -3$$

Divide both sides by $-3$. Since $-3$ is negative, the inequality direction is changed.

$$x\geq \frac{-3}{-3}$$

Divide $-3$ by $-3$ to get $1$.

$$x\geq 1$$

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