We are given the following equation
[tex]2(x+3)=6+2x_{}[/tex]We are asked to determine if the above equation has "One Solution", "No Solutions", or "Infinite Many Solutions"
Let us first open the brackets on the left side of the equation
[tex]\begin{gathered} 2(x+3)=6+2x_{} \\ 2x+6=6+2x_{} \end{gathered}[/tex]As you can see the L.H.S and R.H.S of the equation is the same.
Whenever an equation has L.H.S = R.H.S then we get "Infinite Many Solutions"
Let us verify if there are many such solutions
Let us substitute x = 1
[tex]\begin{gathered} 2(1)+6=6+2(1) \\ 2+6=6+2_{} \\ 8=8 \end{gathered}[/tex]As you can see the equation is satisfied.
Now Let us substitute x = 2
[tex]\begin{gathered} 2(2)+6=6+2(2) \\ 4+6=6+4 \\ 10=10 \end{gathered}[/tex]No matter what value you substitute, the equation will always be
