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Question 1:
Adding 2 to both sides of the equation gives
[tex]|x|=3x+4[/tex]Now, the absolute value decomposes the above equation into two separate equations
[tex]\begin{gathered} x=3x+4 \\ x=-3x-4 \end{gathered}[/tex]The first equation gives
[tex]\begin{gathered} x=3x+4 \\ -2x=4 \\ \boxed{x=-2} \end{gathered}[/tex]The second equation gives
[tex]\begin{gathered} x=-3x-4 \\ 4x=-4 \\ \boxed{x=-1} \end{gathered}[/tex]The graph of the system is
We see that the solutions exists at x =-1.
Question 2.
Adding 2 to both sides of the equation gives
[tex]|2x-1|=-x+2[/tex]Decomposing the absolute value on LHS gives us two equations
[tex]\begin{gathered} -(2x-1)=-x+2 \\ 2x-1=-x+2 \end{gathered}[/tex]Solving the first equation gives
[tex]\begin{gathered} 2x-1=x-2 \\ x=-1 \end{gathered}[/tex]Solving the second equation gives
[tex]\begin{gathered} 2x-1=-x+2 \\ 3x=3 \\ x=1 \end{gathered}[/tex]Hence, the solution to the equation is
[tex]x=\pm1[/tex]The graph of the solutions is
We see that the solution is at x = -1 and x = 1; hence, our solution is confirmed.
