Solve 3+x/4geq2+2x/3

$$\frac { 3 + x } { 4 } \geq \frac { 2 + 2 x } { 3 }$$

Solve for x

$$x\leq \frac{1}{5}$$

Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 4,3. Since 12 is positive, the inequality direction remains the same.

$$3\left(3+x\right)\geq 4\left(2+2x\right)$$

Use the distributive property to multiply 3 by 3+x.

$$9+3x\geq 4\left(2+2x\right)$$

Use the distributive property to multiply 4 by 2+2x.

$$9+3x\geq 8+8x$$

Subtract 8x from both sides.

$$9+3x-8x\geq 8$$

Combine 3x and -8x to get -5x.

$$9-5x\geq 8$$

Subtract 9 from both sides.

$$-5x\geq 8-9$$

Subtract 9 from 8 to get -1.

$$-5x\geq -1$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x\leq \frac{-1}{-5}$$

Fraction \frac{-1}{-5} can be simplified to \frac{1}{5} by removing the negative sign from both the numerator and the denominator.

$$x\leq \frac{1}{5}$$