## Introductory Algebra for College Students (7th Edition)

For the given number to be a solution of the equation, the equation must hold true when we substitute the variable with the given number. $LHS=\displaystyle \frac{1}{3}(26-2)=\frac{24}{3}=\frac{24\div 3}{3\div 3}=\frac{8}{1}=8$ $RHS=\displaystyle \frac{1}{5}(26+4)+3$ $=\displaystyle \frac{30}{5}+3$ $=\displaystyle \frac{30\div 5}{5\div 5}+3$ $=\displaystyle \frac{6}{1}+3$ $=6+3$ $=9$ $LHS\neq RHS$ , so 26 is not a solution.