Question

Solve the inequalities for real x for the following equation: $\frac{(2x-1)}{3} \geq \frac{(3x-2)}{4} -\frac{(2-x)}{5}$

Answer 1

$\dfrac{(2x-1)}{3} \geq \dfrac{(3x-2)}{4} -\dfrac{(2-x)}{5}$

Make into single fractions:

$\dfrac{(2x-1)}{3} \geq \dfrac{5(3x-2)}{20} -\dfrac{4(2-x)}{20}$

$\dfrac{(2x-1)}{3} \geq \dfrac{5(3x-2) - 4(2 - x)}{20}$

Distribute 5 and 4:

$\dfrac{(2x-1)}{3} \geq \dfrac{15x-10 - 8 + 4x}{20}$

Combine like terms:

$\dfrac{(2x-1)}{3} \geq \dfrac{19x-18 }{20}$

Multiply by 20 and 3:

$20(2x - 1) \geq 3(19x - 18)$

Distribute the 20 and 3:

$40x - 20 \geq 57x - 54$

Rearrange:

$40x - 57x \geq -54+ 20$

Combine like terms:

$-17x \geq -34$

Divide both sides by -17:

$x \leq 2$