[tex]\tt5( x - 3 ) = 5x - 15 \\ \tt5x - 15 = 5x - 15 \\ \tt5x - 5x = - 15 + 15 \\ \tt\bcancel{5x} - \bcancel{5x} = \bcancel{- 15} + \bcancel{15} \\ \boxed{\tt0 = 0}[/tex]
- The answer is 0.
- LHS = RHS, hence verified.
Value of x (substitute 0 in the place of x).
[tex]\tt5( (0) - 3 ) = 5(0) - 15 \\ \tt0 - 15 = 0 - 15 \\ \tt \boxed{\tt -15 = -15}[/tex]
