Solve 4+x+3/3geqx-1

$$4 + \frac { x + 3 } { 3 } \geq x - 1$$

Solve for x

$$x\leq 9$$

Solution Steps

Multiply both sides of the equation by 3. Since 3 is positive, the inequality direction remains the same.

$$12+x+3\geq 3x-3$$

Add 12 and 3 to get 15.

$$15+x\geq 3x-3$$

Subtract 3x from both sides.

$$15+x-3x\geq -3$$

Combine x and -3x to get -2x.

$$15-2x\geq -3$$

Subtract 15 from both sides.

$$-2x\geq -3-15$$

Subtract 15 from -3 to get -18.

$$-2x\geq -18$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x\leq \frac{-18}{-2}$$

Divide -18 by -2 to get 9.

$$x\leq 9$$