Solve 0.02x+0.01/0.03-0.6x+1/0.6=2quad-61/4x=

$$\frac { 0.02 x + 0.01 } { 0.03 } - \frac { 0.6 x + 1 } { 0.6 } = 2 \quad - \frac { 61 } { 4 } x =$$

Solve for x

$$x=\frac{40}{179}\approx 0.223463687$$

Steps for Solving Linear Equation

Divide each term of 0.02x+0.01 by 0.03 to get \frac{0.02x}{0.03}+\frac{0.01}{0.03}.

$$\frac{0.02x}{0.03}+\frac{0.01}{0.03}-\frac{0.6x+1}{0.6}=2-\frac{61}{4}x$$

Divide 0.02x by 0.03 to get \frac{2}{3}x.

$$\frac{2}{3}x+\frac{0.01}{0.03}-\frac{0.6x+1}{0.6}=2-\frac{61}{4}x$$

Expand \frac{0.01}{0.03} by multiplying both numerator and the denominator by 100.

$$\frac{2}{3}x+\frac{1}{3}-\frac{0.6x+1}{0.6}=2-\frac{61}{4}x$$

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

$$\frac{2}{3}x+\frac{1}{3}-\left(\frac{0.6x}{0.6}+\frac{1}{0.6}\right)=2-\frac{61}{4}x$$

Cancel out 0.6 and 0.6.

$$\frac{2}{3}x+\frac{1}{3}-\left(x+\frac{1}{0.6}\right)=2-\frac{61}{4}x$$

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

$$\frac{2}{3}x+\frac{1}{3}-\left(x+\frac{10}{6}\right)=2-\frac{61}{4}x$$

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

$$\frac{2}{3}x+\frac{1}{3}-\left(x+\frac{5}{3}\right)=2-\frac{61}{4}x$$

To find the opposite of x+\frac{5}{3}, find the opposite of each term.

$$\frac{2}{3}x+\frac{1}{3}-x-\frac{5}{3}=2-\frac{61}{4}x$$

Combine \frac{2}{3}x and -x to get -\frac{1}{3}x.

$$-\frac{1}{3}x+\frac{1}{3}-\frac{5}{3}=2-\frac{61}{4}x$$

Since \frac{1}{3} and \frac{5}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{1}{3}x+\frac{1-5}{3}=2-\frac{61}{4}x$$

Subtract 5 from 1 to get -4.

$$-\frac{1}{3}x-\frac{4}{3}=2-\frac{61}{4}x$$

Add \frac{61}{4}x to both sides.

$$-\frac{1}{3}x-\frac{4}{3}+\frac{61}{4}x=2$$

Combine -\frac{1}{3}x and \frac{61}{4}x to get \frac{179}{12}x.

$$\frac{179}{12}x-\frac{4}{3}=2$$

Add \frac{4}{3} to both sides.

$$\frac{179}{12}x=2+\frac{4}{3}$$

Convert 2 to fraction \frac{6}{3}.

$$\frac{179}{12}x=\frac{6}{3}+\frac{4}{3}$$

Since \frac{6}{3} and \frac{4}{3} have the same denominator, add them by adding their numerators.

$$\frac{179}{12}x=\frac{6+4}{3}$$

Add 6 and 4 to get 10.

$$\frac{179}{12}x=\frac{10}{3}$$

Multiply both sides by \frac{12}{179}, the reciprocal of \frac{179}{12}.

$$x=\frac{10}{3}\times \frac{12}{179}$$

Multiply \frac{10}{3} times \frac{12}{179} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{10\times 12}{3\times 179}$$

Do the multiplications in the fraction \frac{10\times 12}{3\times 179}.

$$x=\frac{120}{537}$$

Reduce the fraction \frac{120}{537} to lowest terms by extracting and canceling out 3.

$$x=\frac{40}{179}$$