Solve 2x/3-x-3/6leq3left(x+1right)

$$\frac{ 2x }{ 3 } - \frac{ x-3 }{ 6 } \leq 3 \left( x+1 \right)$$

Solve for x

$$x\geq -1$$

Solution Steps

Multiply both sides of the equation by 6, the least common multiple of 3,6. Since 6 is positive, the inequality direction remains the same.

$$2\times 2x-\left(x-3\right)\leq 18\left(x+1\right)$$

Multiply 2 and 2 to get 4.

$$4x-\left(x-3\right)\leq 18\left(x+1\right)$$

To find the opposite of x-3, find the opposite of each term.

$$4x-x-\left(-3\right)\leq 18\left(x+1\right)$$

The opposite of -3 is 3.

$$4x-x+3\leq 18\left(x+1\right)$$

Combine 4x and -x to get 3x.

$$3x+3\leq 18\left(x+1\right)$$

Use the distributive property to multiply 18 by x+1.

$$3x+3\leq 18x+18$$

Subtract 18x from both sides.

$$3x+3-18x\leq 18$$

Combine 3x and -18x to get -15x.

$$-15x+3\leq 18$$

Subtract 3 from both sides.

$$-15x\leq 18-3$$

Subtract 3 from 18 to get 15.

$$-15x\leq 15$$

Divide both sides by -15. Since -15 is negative, the inequality direction is changed.

$$x\geq \frac{15}{-15}$$

Divide 15 by -15 to get -1.

$$x\geq -1$$