Solve 6*sqrt{12-3}*4r=4/4+3/8

$$6 \cdot \sqrt { 12 - 3 } \cdot 4 r = \frac { 4 } { 4 } + \frac { 3 } { 8 }$$

Solve for r

$$r=\frac{11}{576}\approx 0.019097222$$

Steps for Solving Linear Equation

Multiply 4 and 6 to get 24.

$$24\sqrt{12-3}r=\frac{4}{4}+\frac{3}{8}$$

Subtract 3 from 12 to get 9.

$$24\sqrt{9}r=\frac{4}{4}+\frac{3}{8}$$

Calculate the square root of 9 and get 3.

$$24\times 3r=\frac{4}{4}+\frac{3}{8}$$

Multiply 24 and 3 to get 72.

$$72r=\frac{4}{4}+\frac{3}{8}$$

Divide 4 by 4 to get 1.

$$72r=1+\frac{3}{8}$$

Convert 1 to fraction \frac{8}{8}.

$$72r=\frac{8}{8}+\frac{3}{8}$$

Since \frac{8}{8} and \frac{3}{8} have the same denominator, add them by adding their numerators.

$$72r=\frac{8+3}{8}$$

Add 8 and 3 to get 11.

$$72r=\frac{11}{8}$$

Divide both sides by 72.

$$r=\frac{\frac{11}{8}}{72}$$

Express \frac{\frac{11}{8}}{72} as a single fraction.

$$r=\frac{11}{8\times 72}$$

Multiply 8 and 72 to get 576.

$$r=\frac{11}{576}$$