Solve 1/4(x-3)-3/4(x-5)=0

$$\frac { 1 } { 4 } ( x - 3 ) - \frac { 3 } { 4 } ( x - 5 ) = 0$$

Solve for x

$$x=6$$

Steps for Solving Linear Equation

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

$$\frac{1}{4}x+\frac{1}{4}\left(-3\right)-\frac{3}{4}\left(x-5\right)=0$$

Multiply \frac{1}{4} and -3 to get \frac{-3}{4}.

$$\frac{1}{4}x+\frac{-3}{4}-\frac{3}{4}\left(x-5\right)=0$$

Fraction \frac{-3}{4} can be rewritten as -\frac{3}{4} by extracting the negative sign.

$$\frac{1}{4}x-\frac{3}{4}-\frac{3}{4}\left(x-5\right)=0$$

Use the distributive property to multiply -\frac{3}{4} by x-5.

$$\frac{1}{4}x-\frac{3}{4}-\frac{3}{4}x-\frac{3}{4}\left(-5\right)=0$$

Express -\frac{3}{4}\left(-5\right) as a single fraction.

$$\frac{1}{4}x-\frac{3}{4}-\frac{3}{4}x+\frac{-3\left(-5\right)}{4}=0$$

Multiply -3 and -5 to get 15.

$$\frac{1}{4}x-\frac{3}{4}-\frac{3}{4}x+\frac{15}{4}=0$$

Combine \frac{1}{4}x and -\frac{3}{4}x to get -\frac{1}{2}x.

$$-\frac{1}{2}x-\frac{3}{4}+\frac{15}{4}=0$$

Since -\frac{3}{4} and \frac{15}{4} have the same denominator, add them by adding their numerators.

$$-\frac{1}{2}x+\frac{-3+15}{4}=0$$

Add -3 and 15 to get 12.

$$-\frac{1}{2}x+\frac{12}{4}=0$$

Divide 12 by 4 to get 3.

$$-\frac{1}{2}x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by -2, the reciprocal of -\frac{1}{2}.

$$x=-3\left(-2\right)$$

Multiply -3 and -2 to get 6.

$$x=6$$