Solve 50(-2/3y+68/3)+40y=1240

$$50 ( - \frac { 2 } { 3 } y + \frac { 68 } { 3 } ) + 40 y = 1240$$

Solve for y

$$y=16$$

Steps for Solving Linear Equation

Use the distributive property to multiply 50 by -\frac{2}{3}y+\frac{68}{3}.

$$50\left(-\frac{2}{3}\right)y+50\times \frac{68}{3}+40y=1240$$

Express 50\left(-\frac{2}{3}\right) as a single fraction.

$$\frac{50\left(-2\right)}{3}y+50\times \frac{68}{3}+40y=1240$$

Multiply 50 and -2 to get -100.

$$\frac{-100}{3}y+50\times \frac{68}{3}+40y=1240$$

Fraction \frac{-100}{3} can be rewritten as -\frac{100}{3} by extracting the negative sign.

$$-\frac{100}{3}y+50\times \frac{68}{3}+40y=1240$$

Express 50\times \frac{68}{3} as a single fraction.

$$-\frac{100}{3}y+\frac{50\times 68}{3}+40y=1240$$

Multiply 50 and 68 to get 3400.

$$-\frac{100}{3}y+\frac{3400}{3}+40y=1240$$

Combine -\frac{100}{3}y and 40y to get \frac{20}{3}y.

$$\frac{20}{3}y+\frac{3400}{3}=1240$$

Subtract \frac{3400}{3} from both sides.

$$\frac{20}{3}y=1240-\frac{3400}{3}$$

Convert 1240 to fraction \frac{3720}{3}.

$$\frac{20}{3}y=\frac{3720}{3}-\frac{3400}{3}$$

Since \frac{3720}{3} and \frac{3400}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{20}{3}y=\frac{3720-3400}{3}$$

Subtract 3400 from 3720 to get 320.

$$\frac{20}{3}y=\frac{320}{3}$$

Multiply both sides by \frac{3}{20}, the reciprocal of \frac{20}{3}.

$$y=\frac{320}{3}\times \frac{3}{20}$$

Multiply \frac{320}{3} times \frac{3}{20} by multiplying numerator times numerator and denominator times denominator.

$$y=\frac{320\times 3}{3\times 20}$$

Cancel out 3 in both numerator and denominator.

$$y=\frac{320}{20}$$

Divide 320 by 20 to get 16.

$$y=16$$