Solve 2(3/2x-21/10)+17/10geq2(12/5x-frac{7}{2})

$$2 ( \frac { 3 } { 2 } x - \frac { 21 } { 10 } ) + \frac { 17 } { 10 } \geq 2 ( \frac { 12 } { 5 } x - \frac { 7 } { 2 } )$$

Solve for x

$$x\leq \frac{5}{2}$$

Solution Steps

Use the distributive property to multiply 2 by \frac{3}{2}x-\frac{21}{10}.

$$2\times \frac{3}{2}x+2\left(-\frac{21}{10}\right)+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Cancel out 2 and 2.

$$3x+2\left(-\frac{21}{10}\right)+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Express 2\left(-\frac{21}{10}\right) as a single fraction.

$$3x+\frac{2\left(-21\right)}{10}+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Multiply 2 and -21 to get -42.

$$3x+\frac{-42}{10}+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Reduce the fraction \frac{-42}{10} to lowest terms by extracting and canceling out 2.

$$3x-\frac{21}{5}+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Least common multiple of 5 and 10 is 10. Convert -\frac{21}{5} and \frac{17}{10} to fractions with denominator 10.

$$3x-\frac{42}{10}+\frac{17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Since -\frac{42}{10} and \frac{17}{10} have the same denominator, add them by adding their numerators.

$$3x+\frac{-42+17}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Add -42 and 17 to get -25.

$$3x+\frac{-25}{10}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Reduce the fraction \frac{-25}{10} to lowest terms by extracting and canceling out 5.

$$3x-\frac{5}{2}\geq 2\left(\frac{12}{5}x-\frac{7}{2}\right)$$

Use the distributive property to multiply 2 by \frac{12}{5}x-\frac{7}{2}.

$$3x-\frac{5}{2}\geq 2\times \frac{12}{5}x+2\left(-\frac{7}{2}\right)$$

Express 2\times \frac{12}{5} as a single fraction.

$$3x-\frac{5}{2}\geq \frac{2\times 12}{5}x+2\left(-\frac{7}{2}\right)$$

Multiply 2 and 12 to get 24.

$$3x-\frac{5}{2}\geq \frac{24}{5}x+2\left(-\frac{7}{2}\right)$$

Cancel out 2 and 2.

$$3x-\frac{5}{2}\geq \frac{24}{5}x-7$$

Subtract \frac{24}{5}x from both sides.

$$3x-\frac{5}{2}-\frac{24}{5}x\geq -7$$

Combine 3x and -\frac{24}{5}x to get -\frac{9}{5}x.

$$-\frac{9}{5}x-\frac{5}{2}\geq -7$$

Add \frac{5}{2} to both sides.

$$-\frac{9}{5}x\geq -7+\frac{5}{2}$$

Convert -7 to fraction -\frac{14}{2}.

$$-\frac{9}{5}x\geq -\frac{14}{2}+\frac{5}{2}$$

Since -\frac{14}{2} and \frac{5}{2} have the same denominator, add them by adding their numerators.

$$-\frac{9}{5}x\geq \frac{-14+5}{2}$$

Add -14 and 5 to get -9.

$$-\frac{9}{5}x\geq -\frac{9}{2}$$

Multiply both sides by -\frac{5}{9}, the reciprocal of -\frac{9}{5}. Since -\frac{9}{5} is negative, the inequality direction is changed.

$$x\leq -\frac{9}{2}\left(-\frac{5}{9}\right)$$

Multiply -\frac{9}{2} times -\frac{5}{9} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-9\left(-5\right)}{2\times 9}$$

Do the multiplications in the fraction \frac{-9\left(-5\right)}{2\times 9}.

$$x\leq \frac{45}{18}$$

Reduce the fraction \frac{45}{18} to lowest terms by extracting and canceling out 9.

$$x\leq \frac{5}{2}$$