Solve 10/11(740-50y)+55y=730

$$\frac{ 10 }{ 11 } (740-50y)+55y=730$$

Solve for y

$$y=6$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{10}{11} by 740-50y.

$$\frac{10}{11}\times 740+\frac{10}{11}\left(-50\right)y+55y=730$$

Express \frac{10}{11}\times 740 as a single fraction.

$$\frac{10\times 740}{11}+\frac{10}{11}\left(-50\right)y+55y=730$$

Multiply 10 and 740 to get 7400.

$$\frac{7400}{11}+\frac{10}{11}\left(-50\right)y+55y=730$$

Express \frac{10}{11}\left(-50\right) as a single fraction.

$$\frac{7400}{11}+\frac{10\left(-50\right)}{11}y+55y=730$$

Multiply 10 and -50 to get -500.

$$\frac{7400}{11}+\frac{-500}{11}y+55y=730$$

Fraction \frac{-500}{11} can be rewritten as -\frac{500}{11} by extracting the negative sign.

$$\frac{7400}{11}-\frac{500}{11}y+55y=730$$

Combine -\frac{500}{11}y and 55y to get \frac{105}{11}y.

$$\frac{7400}{11}+\frac{105}{11}y=730$$

Subtract \frac{7400}{11} from both sides.

$$\frac{105}{11}y=730-\frac{7400}{11}$$

Convert 730 to fraction \frac{8030}{11}.

$$\frac{105}{11}y=\frac{8030}{11}-\frac{7400}{11}$$

Since \frac{8030}{11} and \frac{7400}{11} have the same denominator, subtract them by subtracting their numerators.

$$\frac{105}{11}y=\frac{8030-7400}{11}$$

Subtract 7400 from 8030 to get 630.

$$\frac{105}{11}y=\frac{630}{11}$$

Multiply both sides by \frac{11}{105}, the reciprocal of \frac{105}{11}.

$$y=\frac{630}{11}\times \frac{11}{105}$$

Multiply \frac{630}{11} times \frac{11}{105} by multiplying numerator times numerator and denominator times denominator.

$$y=\frac{630\times 11}{11\times 105}$$

Cancel out 11 in both numerator and denominator.

$$y=\frac{630}{105}$$

Divide 630 by 105 to get 6.

$$y=6$$