To solve both systems of equations, we will use the addition method.
A) Adding the given equations, we get:
[tex]2x+3y+(-2x+4y)=7+14.[/tex]
Simplifying the above equation, we get:
[tex]7y=21.[/tex]
Dividing by 7:
[tex]y=\frac{21}{7}=3.[/tex]
Substituting y=3 in the first equation of the system and solving for x, we get:
[tex]\begin{gathered} 2x+9=7, \\ 2x=-2, \\ x=-1. \end{gathered}[/tex]
Answer part A:
[tex]\begin{gathered} y=3, \\ x=-1. \end{gathered}[/tex]
B) Adding the two equations of the system, we get:
[tex]2x+3y+(3x-3y)=7+3.[/tex]
Simplifying the above equation, we get:
[tex]5x=10.[/tex]
Dividing by 5, we get:
[tex]\begin{gathered} 5x=10, \\ x=\frac{10}{5}, \\ x=2. \end{gathered}[/tex]
Substituting x=2 in the second equation of the system, we get:
[tex]\begin{gathered} 3(2)-3y=3, \\ 6-3y=3. \end{gathered}[/tex]
Subtracting 6 from both sides of the equation, we get:
[tex]\begin{gathered} 6-3y-6=3-6, \\ -3y=-3. \end{gathered}[/tex]
Dividing by -3, we get:
[tex]\begin{gathered} \frac{-3y}{-3}=\frac{-3}{-3}, \\ y=1. \end{gathered}[/tex]
Answer part B:
[tex]\begin{gathered} x=2, \\ y=1. \end{gathered}[/tex]
