Solve 3/4[4/3(1/2x-1/4)-8]-frac{3}{2}x=1

$$\frac { 3 } { 4 } [ \frac { 4 } { 3 } ( \frac { 1 } { 2 } x - \frac { 1 } { 4 } ) - 8 ] - \frac { 3 } { 2 } x = 1$$

Solve for x

$$x = -\frac{29}{4} = -7\frac{1}{4} = -7.25$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{4}{3} by \frac{1}{2}x-\frac{1}{4}.

$$\frac{3}{4}\left(\frac{4}{3}\times \frac{1}{2}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)-\frac{3}{2}x=1$$

Multiply \frac{4}{3} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{4}\left(\frac{4\times 1}{3\times 2}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)-\frac{3}{2}x=1$$

Do the multiplications in the fraction \frac{4\times 1}{3\times 2}.

$$\frac{3}{4}\left(\frac{4}{6}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)-\frac{3}{2}x=1$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

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

Multiply \frac{4}{3} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{4\left(-1\right)}{3\times 4}-8\right)-\frac{3}{2}x=1$$

Cancel out 4 in both numerator and denominator.

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

Fraction \frac{-1}{3} can be rewritten as -\frac{1}{3} by extracting the negative sign.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{1}{3}-8\right)-\frac{3}{2}x=1$$

Convert 8 to fraction \frac{24}{3}.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{1}{3}-\frac{24}{3}\right)-\frac{3}{2}x=1$$

Since -\frac{1}{3} and \frac{24}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{-1-24}{3}\right)-\frac{3}{2}x=1$$

Subtract 24 from -1 to get -25.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{25}{3}\right)-\frac{3}{2}x=1$$

Use the distributive property to multiply \frac{3}{4} by \frac{2}{3}x-\frac{25}{3}.

$$\frac{3}{4}\times \frac{2}{3}x+\frac{3}{4}\left(-\frac{25}{3}\right)-\frac{3}{2}x=1$$

Multiply \frac{3}{4} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 2}{4\times 3}x+\frac{3}{4}\left(-\frac{25}{3}\right)-\frac{3}{2}x=1$$

Cancel out 3 in both numerator and denominator.

$$\frac{2}{4}x+\frac{3}{4}\left(-\frac{25}{3}\right)-\frac{3}{2}x=1$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{1}{2}x+\frac{3}{4}\left(-\frac{25}{3}\right)-\frac{3}{2}x=1$$

Multiply \frac{3}{4} times -\frac{25}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x+\frac{3\left(-25\right)}{4\times 3}-\frac{3}{2}x=1$$

Cancel out 3 in both numerator and denominator.

$$\frac{1}{2}x+\frac{-25}{4}-\frac{3}{2}x=1$$

Fraction \frac{-25}{4} can be rewritten as -\frac{25}{4} by extracting the negative sign.

$$\frac{1}{2}x-\frac{25}{4}-\frac{3}{2}x=1$$

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

$$-x-\frac{25}{4}=1$$

Add \frac{25}{4} to both sides.

$$-x=1+\frac{25}{4}$$

Convert 1 to fraction \frac{4}{4}.

$$-x=\frac{4}{4}+\frac{25}{4}$$

Since \frac{4}{4} and \frac{25}{4} have the same denominator, add them by adding their numerators.

$$-x=\frac{4+25}{4}$$

Add 4 and 25 to get 29.

$$-x=\frac{29}{4}$$

Multiply both sides by -1.

$$x=-\frac{29}{4}$$