Solve 3x+1/7-2-4x/3geq-5x-4/14+7x/6

$$\frac{ 3x+1 }{ 7 } - \frac{ 2-4x }{ 3 } \geq \frac { -5x-4 } { 14 } + \frac{ 7x }{ 6 }$$

Solve for x

$$x\geq \frac{1}{4}$$

Solution Steps

Multiply both sides of the equation by 42, the least common multiple of 7,3,14,6. Since 42 is positive, the inequality direction remains the same.

$$6\left(3x+1\right)-14\left(2-4x\right)\geq 3\left(-5x-4\right)+7\times 7x$$

Use the distributive property to multiply 6 by 3x+1.

$$18x+6-14\left(2-4x\right)\geq 3\left(-5x-4\right)+7\times 7x$$

Use the distributive property to multiply -14 by 2-4x.

$$18x+6-28+56x\geq 3\left(-5x-4\right)+7\times 7x$$

Subtract 28 from 6 to get -22.

$$18x-22+56x\geq 3\left(-5x-4\right)+7\times 7x$$

Combine 18x and 56x to get 74x.

$$74x-22\geq 3\left(-5x-4\right)+7\times 7x$$

Use the distributive property to multiply 3 by -5x-4.

$$74x-22\geq -15x-12+7\times 7x$$

Multiply 7 and 7 to get 49.

$$74x-22\geq -15x-12+49x$$

Combine -15x and 49x to get 34x.

$$74x-22\geq 34x-12$$

Subtract 34x from both sides.

$$74x-22-34x\geq -12$$

Combine 74x and -34x to get 40x.

$$40x-22\geq -12$$

Add 22 to both sides.

$$40x\geq -12+22$$

Add -12 and 22 to get 10.

$$40x\geq 10$$

Divide both sides by 40. Since 40 is positive, the inequality direction remains the same.

$$x\geq \frac{10}{40}$$

Reduce the fraction \frac{10}{40} to lowest terms by extracting and canceling out 10.

$$x\geq \frac{1}{4}$$