match the system of equations with the solution set.hint: solve algebraically using substitution method.A. no solutionB. infinite solutionsC. (-8/3, 5)D. (2, 1)

We will solve all the systems by substitution method .
System 1.
By substituting the second equation into the first one, we get
[tex]x-3(\frac{1}{3}x-2)=6[/tex]which gives
[tex]\begin{gathered} x-x+6=6 \\ 6=6 \end{gathered}[/tex]this means that the given equations are the same. Then, the answer is B: infinite solutions.
System 2.
By substituting the first equation into the second one, we have
[tex]6x+3(-2x+3)=-5[/tex]which gives
[tex]\begin{gathered} 6x-6x+9=-5 \\ 9=-5 \end{gathered}[/tex]but this result is an absurd. This means that the equations represent parallel lines. Then, the answer is option A: no solution.
System 3.
By substituting the first equation into the second one, we obtain
[tex]-\frac{3}{2}x+1=-\frac{3}{4}x+3[/tex]by moving -3/4x to the left hand side and +1 to the right hand side, we get
[tex]-\frac{3}{2}x+\frac{3}{4}x=3-1[/tex]By combining similar terms, we have
[tex]-\frac{3}{4}x=2[/tex]this leads to
[tex]x=-\frac{4\times2}{3}[/tex]then, x is given by
[tex]x=-\frac{8}{3}[/tex]Now, we can substitute this result into the first equation and get
[tex]y=-\frac{3}{2}(-\frac{8}{3})+1[/tex]which leads to
[tex]\begin{gathered} y=4+1 \\ y=5 \end{gathered}[/tex]then, the answer is option C: (-8/3, 5)
System 4.
By substituting the second equation into the first one, we get
[tex]-5x+(2x-3)=-9[/tex]By combing similar terms, we have
[tex]\begin{gathered} -3x-3=-9 \\ -3x=-9+3 \\ -3x=-6 \\ x=\frac{-6}{-3} \\ x=2 \end{gathered}[/tex]By substituting this result into the second equation, we have
[tex]\begin{gathered} y=2(2)-3 \\ y=4-3 \\ y=1 \end{gathered}[/tex]then, the answer is option D