Solve x-7x+8/6leq1+x/2

$$x - \frac { 7 x + 8 } { 6 } \leq 1 + \frac { x } { 2 }$$

Solve for x

$$x\geq -\frac{7}{2}$$

Solution Steps

Multiply both sides of the equation by 6, the least common multiple of 6,2. Since 6 is positive, the inequality direction remains the same.

$$6x-\left(7x+8\right)\leq 6+3x$$

To find the opposite of 7x+8, find the opposite of each term.

$$6x-7x-8\leq 6+3x$$

Combine 6x and -7x to get -x.

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

Subtract 3x from both sides.

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

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

$$-4x-8\leq 6$$

Add 8 to both sides.

$$-4x\leq 6+8$$

Add 6 and 8 to get 14.

$$-4x\leq 14$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\geq \frac{14}{-4}$$

Reduce the fraction \frac{14}{-4} to lowest terms by extracting and canceling out 2.

$$x\geq -\frac{7}{2}$$