Solve x+3/4+4/6leq2x+1/3

$$\frac { x + 3 } { 4 } + \frac { 4 } { 6 } \leq \frac { 2 x + 1 } { 3 }$$

Solve for x

$$x\geq \frac{13}{5}$$

Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 4,6,3. Since 12 is positive, the inequality direction remains the same.

$$3\left(x+3\right)+2\times 4\leq 4\left(2x+1\right)$$

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

$$3x+9+2\times 4\leq 4\left(2x+1\right)$$

Multiply 2 and 4 to get 8.

$$3x+9+8\leq 4\left(2x+1\right)$$

Add 9 and 8 to get 17.

$$3x+17\leq 4\left(2x+1\right)$$

Use the distributive property to multiply 4 by 2x+1.

$$3x+17\leq 8x+4$$

Subtract 8x from both sides.

$$3x+17-8x\leq 4$$

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

$$-5x+17\leq 4$$

Subtract 17 from both sides.

$$-5x\leq 4-17$$

Subtract 17 from 4 to get -13.

$$-5x\leq -13$$

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

$$x\geq \frac{-13}{-5}$$

Fraction \frac{-13}{-5} can be simplified to \frac{13}{5} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{13}{5}$$