Solve 1/3(53-y)+3/7y=21

$$\frac { 1 } { 3 } ( 53 - y ) + \frac { 3 } { 7 } y = 21$$

Solve for y

$$y=35$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{3} by 53-y.

$$\frac{1}{3}\times 53+\frac{1}{3}\left(-1\right)y+\frac{3}{7}y=21$$

Multiply \frac{1}{3} and 53 to get \frac{53}{3}.

$$\frac{53}{3}+\frac{1}{3}\left(-1\right)y+\frac{3}{7}y=21$$

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

$$\frac{53}{3}-\frac{1}{3}y+\frac{3}{7}y=21$$

Combine -\frac{1}{3}y and \frac{3}{7}y to get \frac{2}{21}y.

$$\frac{53}{3}+\frac{2}{21}y=21$$

Subtract \frac{53}{3} from both sides.

$$\frac{2}{21}y=21-\frac{53}{3}$$

Convert 21 to fraction \frac{63}{3}.

$$\frac{2}{21}y=\frac{63}{3}-\frac{53}{3}$$

Since \frac{63}{3} and \frac{53}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2}{21}y=\frac{63-53}{3}$$

Subtract 53 from 63 to get 10.

$$\frac{2}{21}y=\frac{10}{3}$$

Multiply both sides by \frac{21}{2}, the reciprocal of \frac{2}{21}.

$$y=\frac{10}{3}\times \frac{21}{2}$$

Multiply \frac{10}{3} times \frac{21}{2} by multiplying numerator times numerator and denominator times denominator.

$$y=\frac{10\times 21}{3\times 2}$$

Do the multiplications in the fraction \frac{10\times 21}{3\times 2}.

$$y=\frac{210}{6}$$

Divide 210 by 6 to get 35.

$$y=35$$