Solve 1/5(3x+8)-1/4(x+1)geq1-2/3(x+2)

$$\frac { 1 } { 5 } ( 3 x + 8 ) - \frac { 1 } { 4 } ( x + 1 ) \geq 1 - \frac { 2 } { 3 } ( x + 2 )$$

Solve for x

$$x\geq -\frac{101}{61}$$

Solution Steps

Use the distributive property to multiply \frac{1}{5} by 3x+8.

$$\frac{1}{5}\times 3x+\frac{1}{5}\times 8-\frac{1}{4}\left(x+1\right)\geq 1-\frac{2}{3}\left(x+2\right)$$

Multiply \frac{1}{5} and 3 to get \frac{3}{5}.

$$\frac{3}{5}x+\frac{1}{5}\times 8-\frac{1}{4}\left(x+1\right)\geq 1-\frac{2}{3}\left(x+2\right)$$

Multiply \frac{1}{5} and 8 to get \frac{8}{5}.

$$\frac{3}{5}x+\frac{8}{5}-\frac{1}{4}\left(x+1\right)\geq 1-\frac{2}{3}\left(x+2\right)$$

Use the distributive property to multiply -\frac{1}{4} by x+1.

$$\frac{3}{5}x+\frac{8}{5}-\frac{1}{4}x-\frac{1}{4}\geq 1-\frac{2}{3}\left(x+2\right)$$

Combine \frac{3}{5}x and -\frac{1}{4}x to get \frac{7}{20}x.

$$\frac{7}{20}x+\frac{8}{5}-\frac{1}{4}\geq 1-\frac{2}{3}\left(x+2\right)$$

Least common multiple of 5 and 4 is 20. Convert \frac{8}{5} and \frac{1}{4} to fractions with denominator 20.

$$\frac{7}{20}x+\frac{32}{20}-\frac{5}{20}\geq 1-\frac{2}{3}\left(x+2\right)$$

Since \frac{32}{20} and \frac{5}{20} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{20}x+\frac{32-5}{20}\geq 1-\frac{2}{3}\left(x+2\right)$$

Subtract 5 from 32 to get 27.

$$\frac{7}{20}x+\frac{27}{20}\geq 1-\frac{2}{3}\left(x+2\right)$$

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

$$\frac{7}{20}x+\frac{27}{20}\geq 1-\frac{2}{3}x-\frac{2}{3}\times 2$$

Express -\frac{2}{3}\times 2 as a single fraction.

$$\frac{7}{20}x+\frac{27}{20}\geq 1-\frac{2}{3}x+\frac{-2\times 2}{3}$$

Multiply -2 and 2 to get -4.

$$\frac{7}{20}x+\frac{27}{20}\geq 1-\frac{2}{3}x+\frac{-4}{3}$$

Fraction \frac{-4}{3} can be rewritten as -\frac{4}{3} by extracting the negative sign.

$$\frac{7}{20}x+\frac{27}{20}\geq 1-\frac{2}{3}x-\frac{4}{3}$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{7}{20}x+\frac{27}{20}\geq \frac{3}{3}-\frac{2}{3}x-\frac{4}{3}$$

Since \frac{3}{3} and \frac{4}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{20}x+\frac{27}{20}\geq \frac{3-4}{3}-\frac{2}{3}x$$

Subtract 4 from 3 to get -1.

$$\frac{7}{20}x+\frac{27}{20}\geq -\frac{1}{3}-\frac{2}{3}x$$

Add \frac{2}{3}x to both sides.

$$\frac{7}{20}x+\frac{27}{20}+\frac{2}{3}x\geq -\frac{1}{3}$$

Combine \frac{7}{20}x and \frac{2}{3}x to get \frac{61}{60}x.

$$\frac{61}{60}x+\frac{27}{20}\geq -\frac{1}{3}$$

Subtract \frac{27}{20} from both sides.

$$\frac{61}{60}x\geq -\frac{1}{3}-\frac{27}{20}$$

Least common multiple of 3 and 20 is 60. Convert -\frac{1}{3} and \frac{27}{20} to fractions with denominator 60.

$$\frac{61}{60}x\geq -\frac{20}{60}-\frac{81}{60}$$

Since -\frac{20}{60} and \frac{81}{60} have the same denominator, subtract them by subtracting their numerators.

$$\frac{61}{60}x\geq \frac{-20-81}{60}$$

Subtract 81 from -20 to get -101.

$$\frac{61}{60}x\geq -\frac{101}{60}$$

Multiply both sides by \frac{60}{61}, the reciprocal of \frac{61}{60}. Since \frac{61}{60} is positive, the inequality direction remains the same.

$$x\geq -\frac{101}{60}\times \frac{60}{61}$$

Multiply -\frac{101}{60} times \frac{60}{61} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{-101\times 60}{60\times 61}$$

Cancel out 60 in both numerator and denominator.

$$x\geq \frac{-101}{61}$$

Fraction \frac{-101}{61} can be rewritten as -\frac{101}{61} by extracting the negative sign.

$$x\geq -\frac{101}{61}$$