Solve 3/4(x-3)-5x+2=3-1/3(2-5x)+32-4x

$$\frac { 3 } { 4 } ( x - 3 ) - 5 x + 2 = 3 - \frac { 1 } { 3 } ( 2 - 5 x ) + 32 - 4 x$$

Solve for x

$$x = -\frac{415}{23} = -18\frac{1}{23} \approx -18.043478261$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{3}{4} by x-3.

$$\frac{3}{4}x+\frac{3}{4}\left(-3\right)-5x+2=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Express \frac{3}{4}\left(-3\right) as a single fraction.

$$\frac{3}{4}x+\frac{3\left(-3\right)}{4}-5x+2=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Multiply 3 and -3 to get -9.

$$\frac{3}{4}x+\frac{-9}{4}-5x+2=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Fraction \frac{-9}{4} can be rewritten as -\frac{9}{4} by extracting the negative sign.

$$\frac{3}{4}x-\frac{9}{4}-5x+2=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Combine \frac{3}{4}x and -5x to get -\frac{17}{4}x.

$$-\frac{17}{4}x-\frac{9}{4}+2=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Convert 2 to fraction \frac{8}{4}.

$$-\frac{17}{4}x-\frac{9}{4}+\frac{8}{4}=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Since -\frac{9}{4} and \frac{8}{4} have the same denominator, add them by adding their numerators.

$$-\frac{17}{4}x+\frac{-9+8}{4}=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Add -9 and 8 to get -1.

$$-\frac{17}{4}x-\frac{1}{4}=3-\frac{1}{3}\left(2-5x\right)+32-4x$$

Use the distributive property to multiply -\frac{1}{3} by 2-5x.

$$-\frac{17}{4}x-\frac{1}{4}=3-\frac{1}{3}\times 2-\frac{1}{3}\left(-5\right)x+32-4x$$

Express -\frac{1}{3}\times 2 as a single fraction.

$$-\frac{17}{4}x-\frac{1}{4}=3+\frac{-2}{3}-\frac{1}{3}\left(-5\right)x+32-4x$$

Fraction \frac{-2}{3} can be rewritten as -\frac{2}{3} by extracting the negative sign.

$$-\frac{17}{4}x-\frac{1}{4}=3-\frac{2}{3}-\frac{1}{3}\left(-5\right)x+32-4x$$

Express -\frac{1}{3}\left(-5\right) as a single fraction.

$$-\frac{17}{4}x-\frac{1}{4}=3-\frac{2}{3}+\frac{-\left(-5\right)}{3}x+32-4x$$

Multiply -1 and -5 to get 5.

$$-\frac{17}{4}x-\frac{1}{4}=3-\frac{2}{3}+\frac{5}{3}x+32-4x$$

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

$$-\frac{17}{4}x-\frac{1}{4}=\frac{9}{3}-\frac{2}{3}+\frac{5}{3}x+32-4x$$

Since \frac{9}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{9-2}{3}+\frac{5}{3}x+32-4x$$

Subtract 2 from 9 to get 7.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{7}{3}+\frac{5}{3}x+32-4x$$

Convert 32 to fraction \frac{96}{3}.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{7}{3}+\frac{5}{3}x+\frac{96}{3}-4x$$

Since \frac{7}{3} and \frac{96}{3} have the same denominator, add them by adding their numerators.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{7+96}{3}+\frac{5}{3}x-4x$$

Add 7 and 96 to get 103.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{103}{3}+\frac{5}{3}x-4x$$

Combine \frac{5}{3}x and -4x to get -\frac{7}{3}x.

$$-\frac{17}{4}x-\frac{1}{4}=\frac{103}{3}-\frac{7}{3}x$$

Add \frac{7}{3}x to both sides.

$$-\frac{17}{4}x-\frac{1}{4}+\frac{7}{3}x=\frac{103}{3}$$

Combine -\frac{17}{4}x and \frac{7}{3}x to get -\frac{23}{12}x.

$$-\frac{23}{12}x-\frac{1}{4}=\frac{103}{3}$$

Add \frac{1}{4} to both sides.

$$-\frac{23}{12}x=\frac{103}{3}+\frac{1}{4}$$

Least common multiple of 3 and 4 is 12. Convert \frac{103}{3} and \frac{1}{4} to fractions with denominator 12.

$$-\frac{23}{12}x=\frac{412}{12}+\frac{3}{12}$$

Since \frac{412}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$-\frac{23}{12}x=\frac{412+3}{12}$$

Add 412 and 3 to get 415.

$$-\frac{23}{12}x=\frac{415}{12}$$

Multiply both sides by -\frac{12}{23}, the reciprocal of -\frac{23}{12}.

$$x=\frac{415}{12}\left(-\frac{12}{23}\right)$$

Multiply \frac{415}{12} times -\frac{12}{23} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{415\left(-12\right)}{12\times 23}$$

Do the multiplications in the fraction \frac{415\left(-12\right)}{12\times 23}.

$$x=\frac{-4980}{276}$$

Reduce the fraction \frac{-4980}{276} to lowest terms by extracting and canceling out 12.

$$x=-\frac{415}{23}$$