Solve 1/2[1/2(1/2x-3)-3]-3=0

$$\frac { 1 } { 2 } [ \frac { 1 } { 2 } ( \frac { 1 } { 2 } x - 3 ) - 3 ] - 3 = 0$$

Solve for x

$$x=42$$

Steps for Solving Linear Equation

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

$$\frac{1}{2}\left(\frac{1}{2}\times \frac{1}{2}x+\frac{1}{2}\left(-3\right)-3\right)-3=0$$

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

$$\frac{1}{2}\left(\frac{1\times 1}{2\times 2}x+\frac{1}{2}\left(-3\right)-3\right)-3=0$$

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

$$\frac{1}{2}\left(\frac{1}{4}x+\frac{1}{2}\left(-3\right)-3\right)-3=0$$

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

$$\frac{1}{2}\left(\frac{1}{4}x+\frac{-3}{2}-3\right)-3=0$$

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

$$\frac{1}{2}\left(\frac{1}{4}x-\frac{3}{2}-3\right)-3=0$$

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

$$\frac{1}{2}\left(\frac{1}{4}x-\frac{3}{2}-\frac{6}{2}\right)-3=0$$

Since -\frac{3}{2} and \frac{6}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}\left(\frac{1}{4}x+\frac{-3-6}{2}\right)-3=0$$

Subtract 6 from -3 to get -9.

$$\frac{1}{2}\left(\frac{1}{4}x-\frac{9}{2}\right)-3=0$$

Use the distributive property to multiply \frac{1}{2} by \frac{1}{4}x-\frac{9}{2}.

$$\frac{1}{2}\times \frac{1}{4}x+\frac{1}{2}\left(-\frac{9}{2}\right)-3=0$$

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

$$\frac{1\times 1}{2\times 4}x+\frac{1}{2}\left(-\frac{9}{2}\right)-3=0$$

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

$$\frac{1}{8}x+\frac{1}{2}\left(-\frac{9}{2}\right)-3=0$$

Multiply \frac{1}{2} times -\frac{9}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{8}x+\frac{1\left(-9\right)}{2\times 2}-3=0$$

Do the multiplications in the fraction \frac{1\left(-9\right)}{2\times 2}.

$$\frac{1}{8}x+\frac{-9}{4}-3=0$$

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

$$\frac{1}{8}x-\frac{9}{4}-3=0$$

Convert 3 to fraction \frac{12}{4}.

$$\frac{1}{8}x-\frac{9}{4}-\frac{12}{4}=0$$

Since -\frac{9}{4} and \frac{12}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{8}x+\frac{-9-12}{4}=0$$

Subtract 12 from -9 to get -21.

$$\frac{1}{8}x-\frac{21}{4}=0$$

Add \frac{21}{4} to both sides. Anything plus zero gives itself.

$$\frac{1}{8}x=\frac{21}{4}$$

Multiply both sides by 8, the reciprocal of \frac{1}{8}.

$$x=\frac{21}{4}\times 8$$

Express \frac{21}{4}\times 8 as a single fraction.

$$x=\frac{21\times 8}{4}$$

Multiply 21 and 8 to get 168.

$$x=\frac{168}{4}$$

Divide 168 by 4 to get 42.

$$x=42$$