Answer:
The solution of the system of equations will be:
[tex]x=0,\:y=3[/tex]
And the system of equations has ONLY ONE solution.
Step-by-step explanation:
Given the system of the equations
[tex]\begin{bmatrix}3x+6y=18\\ 3y=-5x+9\end{bmatrix}[/tex]
Arrange equation variables for elimination
[tex]\begin{bmatrix}3x+6y=18\\ 5x+3y=9\end{bmatrix}[/tex]
[tex]\mathrm{Multiply\:}3x+6y=18\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:15x+30y=90[/tex]
[tex]\mathrm{Multiply\:}5x+3y=9\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:15x+9y=27[/tex]
[tex]\begin{bmatrix}15x+30y=90\\ 15x+9y=27\end{bmatrix}[/tex]
[tex]15x+9y=27[/tex]
[tex]-[/tex]
[tex]\underline{15x+30y=90}[/tex]
[tex]-21y=-63[/tex]
so the system of the equations becomes
[tex]\begin{bmatrix}15x+30y=90\\ -21y=-63\end{bmatrix}[/tex]
solve -21y = -63
[tex]-21y=-63[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}-21[/tex]
[tex]\frac{-21y}{-21}=\frac{-63}{-21}[/tex]
[tex]y=3[/tex]
[tex]\mathrm{For\:}15x+30y=90\mathrm{\:plug\:in\:}y=3[/tex]
[tex]15x+30\cdot \:3=90[/tex]
[tex]15x+90=90[/tex]
subtract 90 from both sides
[tex]15x+90-90=90-90[/tex]
[tex]15x=0[/tex]
Divide both sides by 15
[tex]\frac{15x}{15}=\frac{0}{15}[/tex]
[tex]x = 0[/tex]
as
[tex]x = 0[/tex], [tex]y=3[/tex]
so, the system of equations contains only one solution.
Therefore, the solution of the system of equations will be:
[tex]x=0,\:y=3[/tex]
And the system of equations has ONLY ONE solution.
