Solve 2y+1/0.6=10-1-2y/0.3

$$\frac { 2 y + 1 } { 0.6 } = 10 - \frac { 1 - 2 y } { 0.3 }$$

Solve for y

$$y=-1.5$$

Steps for Solving Linear Equation

Divide each term of 2y+1 by 0.6 to get \frac{2y}{0.6}+\frac{1}{0.6}.

$$\frac{2y}{0.6}+\frac{1}{0.6}=10-\frac{1-2y}{0.3}$$

Divide 2y by 0.6 to get \frac{10}{3}y.

$$\frac{10}{3}y+\frac{1}{0.6}=10-\frac{1-2y}{0.3}$$

Expand \frac{1}{0.6} by multiplying both numerator and the denominator by 10.

$$\frac{10}{3}y+\frac{10}{6}=10-\frac{1-2y}{0.3}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$\frac{10}{3}y+\frac{5}{3}=10-\frac{1-2y}{0.3}$$

Divide each term of 1-2y by 0.3 to get \frac{1}{0.3}+\frac{-2y}{0.3}.

$$\frac{10}{3}y+\frac{5}{3}=10-\left(\frac{1}{0.3}+\frac{-2y}{0.3}\right)$$

Expand \frac{1}{0.3} by multiplying both numerator and the denominator by 10.

$$\frac{10}{3}y+\frac{5}{3}=10-\left(\frac{10}{3}+\frac{-2y}{0.3}\right)$$

Divide -2y by 0.3 to get -\frac{20}{3}y.

$$\frac{10}{3}y+\frac{5}{3}=10-\left(\frac{10}{3}-\frac{20}{3}y\right)$$

To find the opposite of \frac{10}{3}-\frac{20}{3}y, find the opposite of each term.

$$\frac{10}{3}y+\frac{5}{3}=10-\frac{10}{3}-\left(-\frac{20}{3}y\right)$$

The opposite of -\frac{20}{3}y is \frac{20}{3}y.

$$\frac{10}{3}y+\frac{5}{3}=10-\frac{10}{3}+\frac{20}{3}y$$

Convert 10 to fraction \frac{30}{3}.

$$\frac{10}{3}y+\frac{5}{3}=\frac{30}{3}-\frac{10}{3}+\frac{20}{3}y$$

Since \frac{30}{3} and \frac{10}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{10}{3}y+\frac{5}{3}=\frac{30-10}{3}+\frac{20}{3}y$$

Subtract 10 from 30 to get 20.

$$\frac{10}{3}y+\frac{5}{3}=\frac{20}{3}+\frac{20}{3}y$$

Subtract \frac{20}{3}y from both sides.

$$\frac{10}{3}y+\frac{5}{3}-\frac{20}{3}y=\frac{20}{3}$$

Combine \frac{10}{3}y and -\frac{20}{3}y to get -\frac{10}{3}y.

$$-\frac{10}{3}y+\frac{5}{3}=\frac{20}{3}$$

Subtract \frac{5}{3} from both sides.

$$-\frac{10}{3}y=\frac{20}{3}-\frac{5}{3}$$

Since \frac{20}{3} and \frac{5}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{10}{3}y=\frac{20-5}{3}$$

Subtract 5 from 20 to get 15.

$$-\frac{10}{3}y=\frac{15}{3}$$

Divide 15 by 3 to get 5.

$$-\frac{10}{3}y=5$$

Divide both sides by -\frac{10}{3}.

$$y=\frac{5}{-\frac{10}{3}}$$