To answer this question, we have to evaluate each equation at the x-entry of each point. If the result of the evaluation is equal to the y-entry of the point, then it is a solution.
1.- (3,4), evaluating the first equation at 2, we get:
[tex]\begin{gathered} 6\cdot3+4y=34, \\ 18+4y=34. \end{gathered}[/tex]Solving for y, we get:
[tex]\begin{gathered} 4y=34-18, \\ y=\frac{16}{4}=4. \end{gathered}[/tex]Therefore the point is a solution to the first equation. Also, point (3,2) is not a solution to the first equation.
Evaluating the second equation at x=3, we get:
[tex]15-2y=15.[/tex]Solving for y, we get:
[tex]\begin{gathered} -2y=0, \\ y=0. \end{gathered}[/tex]Therefore, points (3,4) and (3,2) are not solutions to the second equation.
Evaluating the first equation at x=4, we get:
[tex]24+4y=34.[/tex]Solving for y, we get:
[tex]\begin{gathered} 4y=34-24=10, \\ y=\frac{10}{4}=2.5. \end{gathered}[/tex]Therefore, point (4,2.5) is a solution to the first equation.
Evaluating the second equation at x=4, we get:
[tex]20-2y=15.[/tex]Solving for y, we get:
[tex]\begin{gathered} 5=2y, \\ y=\frac{5}{2}=2.5. \end{gathered}[/tex]Therefore, (4,2.5) is a solution to both equations.
Evaluating the first equation at x=5, we get:
[tex]30+4y=34.[/tex]Solving for y, we get:
[tex]\begin{gathered} 4y=4, \\ y=1. \end{gathered}[/tex]Therefore, point (5,5) is not a solution to the first equation.
Evaluating the second equation at x=5, we get:
[tex]25-2y=15.[/tex]Solving for y, we get:
[tex]\begin{gathered} 10=2y, \\ y=5. \end{gathered}[/tex]Therefore, (5,5) satisfies the second equation.
Answer:
(3,4) First equation.
(4,2.5) Both equations.
(5,5) Second equation.
(3,2) Neither.
