Solve 1*8-8x/1.2-1*3-3x/2=5x-0*4/0.3

$$\frac{ 1 \cdot 8-8x }{ 1.2 } - \frac{ 1 \cdot 3-3x }{ 2 } = \frac{ 5x-0 \cdot 4 }{ 0.3 }$$

Solve for x

$$x=\frac{31}{131}\approx 0.236641221$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\times \frac{1\times 8-8x}{1.2}-\left(1\times 3-3x\right)=2\times \frac{5x-0\times 4}{0.3}$$

Multiply 1 and 8 to get 8.

$$2\times \frac{8-8x}{1.2}-\left(1\times 3-3x\right)=2\times \frac{5x-0\times 4}{0.3}$$

Multiply 1 and 3 to get 3.

$$2\times \frac{8-8x}{1.2}-\left(3-3x\right)=2\times \frac{5x-0\times 4}{0.3}$$

To find the opposite of 3-3x, find the opposite of each term.

$$2\times \frac{8-8x}{1.2}-3-\left(-3x\right)=2\times \frac{5x-0\times 4}{0.3}$$

The opposite of -3x is 3x.

$$2\times \frac{8-8x}{1.2}-3+3x=2\times \frac{5x-0\times 4}{0.3}$$

Multiply 0 and 4 to get 0.

$$2\times \frac{8-8x}{1.2}-3+3x=2\times \frac{5x-0}{0.3}$$

Divide each term of 8-8x by 1.2 to get \frac{8}{1.2}+\frac{-8x}{1.2}.

$$2\left(\frac{8}{1.2}+\frac{-8x}{1.2}\right)-3+3x=2\times \frac{5x-0}{0.3}$$

Expand \frac{8}{1.2} by multiplying both numerator and the denominator by 10.

$$2\left(\frac{80}{12}+\frac{-8x}{1.2}\right)-3+3x=2\times \frac{5x-0}{0.3}$$

Reduce the fraction \frac{80}{12} to lowest terms by extracting and canceling out 4.

$$2\left(\frac{20}{3}+\frac{-8x}{1.2}\right)-3+3x=2\times \frac{5x-0}{0.3}$$

Divide -8x by 1.2 to get -\frac{20}{3}x.

$$2\left(\frac{20}{3}-\frac{20}{3}x\right)-3+3x=2\times \frac{5x-0}{0.3}$$

Use the distributive property to multiply 2 by \frac{20}{3}-\frac{20}{3}x.

$$2\times \frac{20}{3}-\frac{40}{3}x-3+3x=2\times \frac{5x-0}{0.3}$$

Express 2\times \frac{20}{3} as a single fraction.

$$\frac{2\times 20}{3}-\frac{40}{3}x-3+3x=2\times \frac{5x-0}{0.3}$$

Multiply 2 and 20 to get 40.

$$\frac{40}{3}-\frac{40}{3}x-3+3x=2\times \frac{5x-0}{0.3}$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{40}{3}-\frac{40}{3}x-\frac{9}{3}+3x=2\times \frac{5x-0}{0.3}$$

Since \frac{40}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{40-9}{3}-\frac{40}{3}x+3x=2\times \frac{5x-0}{0.3}$$

Subtract 9 from 40 to get 31.

$$\frac{31}{3}-\frac{40}{3}x+3x=2\times \frac{5x-0}{0.3}$$

Combine -\frac{40}{3}x and 3x to get -\frac{31}{3}x.

$$\frac{31}{3}-\frac{31}{3}x=2\times \frac{5x-0}{0.3}$$

Divide each term of 5x-0 by 0.3 to get \frac{5x}{0.3}-\frac{0}{0.3}.

$$\frac{31}{3}-\frac{31}{3}x=2\left(\frac{5x}{0.3}-\frac{0}{0.3}\right)$$

Divide 5x by 0.3 to get \frac{50}{3}x.

$$\frac{31}{3}-\frac{31}{3}x=2\left(\frac{50}{3}x-\frac{0}{0.3}\right)$$

Zero divided by any non-zero number gives zero.

$$\frac{31}{3}-\frac{31}{3}x=2\left(\frac{50}{3}x+0\right)$$

Anything plus zero gives itself.

$$\frac{31}{3}-\frac{31}{3}x=2\times \frac{50}{3}x$$

Multiply 2 and \frac{50}{3} to get \frac{100}{3}.

$$\frac{31}{3}-\frac{31}{3}x=\frac{100}{3}x$$

Subtract \frac{100}{3}x from both sides.

$$\frac{31}{3}-\frac{31}{3}x-\frac{100}{3}x=0$$

Combine -\frac{31}{3}x and -\frac{100}{3}x to get -\frac{131}{3}x.

$$\frac{31}{3}-\frac{131}{3}x=0$$

Subtract \frac{31}{3} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{131}{3}x=-\frac{31}{3}$$

Divide both sides by -\frac{131}{3}.

$$x=\frac{-\frac{31}{3}}{-\frac{131}{3}}$$

Express \frac{-\frac{31}{3}}{-\frac{131}{3}} as a single fraction.

$$x=\frac{-31}{3\left(-\frac{131}{3}\right)}$$

Multiply 3 and -\frac{131}{3} to get -131.

$$x=\frac{-31}{-131}$$

Fraction \frac{-31}{-131} can be simplified to \frac{31}{131} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{31}{131}$$