To do this you can subtract 5 from both sides of the equation
[tex]\begin{gathered} 5x+5=2x+20 \\ 5x+5-5=2x+20-5 \\ 5x=2x+15 \end{gathered}[/tex]Now you can subtract 2x from both sides of the equation
[tex]\begin{gathered} 5x=2x+15 \\ 5x-2x=2x+15-2x \\ 3x=15 \end{gathered}[/tex]Finally, you can divide by 3 both sides of the equation
[tex]\begin{gathered} \frac{3x}{3}=\frac{15}{3} \\ x=5 \end{gathered}[/tex]On the other hand, to check you replace the value of x found in the original expression
[tex]\begin{gathered} 5x+5=2x+20 \\ 5(5)+5=2(5)+20 \\ 25+5=10+20 \\ 30=30 \end{gathered}[/tex]Since a true equality is obtained, then it is true that the value of x is 5.
