Solve 2-5/2+8=z-5/2+13

$$\frac { 2 - 5 } { 2 + 8 } = \frac { z - 5 } { 2 + 13 }$$

Solve for z

$$z=\frac{1}{2}=0.5$$

Steps for Solving Linear Equation

Subtract 5 from 2 to get -3.

$$\frac{-3}{2+8}=\frac{z-5}{2+13}$$

Add 2 and 8 to get 10.

$$\frac{-3}{10}=\frac{z-5}{2+13}$$

Fraction \frac{-3}{10} can be rewritten as -\frac{3}{10} by extracting the negative sign.

$$-\frac{3}{10}=\frac{z-5}{2+13}$$

Add 2 and 13 to get 15.

$$-\frac{3}{10}=\frac{z-5}{15}$$

Divide each term of z-5 by 15 to get \frac{1}{15}z-\frac{1}{3}.

$$-\frac{3}{10}=\frac{1}{15}z-\frac{1}{3}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{15}z-\frac{1}{3}=-\frac{3}{10}$$

Add \frac{1}{3} to both sides.

$$\frac{1}{15}z=-\frac{3}{10}+\frac{1}{3}$$

Least common multiple of 10 and 3 is 30. Convert -\frac{3}{10} and \frac{1}{3} to fractions with denominator 30.

$$\frac{1}{15}z=-\frac{9}{30}+\frac{10}{30}$$

Since -\frac{9}{30} and \frac{10}{30} have the same denominator, add them by adding their numerators.

$$\frac{1}{15}z=\frac{-9+10}{30}$$

Add -9 and 10 to get 1.

$$\frac{1}{15}z=\frac{1}{30}$$

Multiply both sides by 15, the reciprocal of \frac{1}{15}.

$$z=\frac{1}{30}\times 15$$

Multiply \frac{1}{30} and 15 to get \frac{15}{30}.

$$z=\frac{15}{30}$$

Reduce the fraction \frac{15}{30} to lowest terms by extracting and canceling out 15.

$$z=\frac{1}{2}$$