How to solve 4(x-3)-8 <= 5-x

Q: How to solve 4(x-3)-8 <= 5-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 4(x-3)-8 <= 5-x | Equatix

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Question

$$4(x-3)-8 \leq 5-x$$

Solve for x

$x\leq 5$

Solution Steps

Use the distributive property to multiply $4$ by $x-3$.

$$4x-12-8\leq 5-x$$

Subtract $8$ from $-12$ to get $-20$.

$$4x-20\leq 5-x$$

Add $x$ to both sides.

$$4x-20+x\leq 5$$

Combine $4x$ and $x$ to get $5x$.

$$5x-20\leq 5$$

Add $20$ to both sides.

$$5x\leq 5+20$$

Add $5$ and $20$ to get $25$.

$$5x\leq 25$$

Divide both sides by $5$. Since $5$ is positive, the inequality direction remains the same.

$$x\leq \frac{25}{5}$$

Divide $25$ by $5$ to get $5$.

$$x\leq 5$$

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