Answer:
From the above calculations, it is clear that the consistent system has exactly one solution i.e. x = -1, y =2, therefore, it is independent.
Step-by-step explanation:
Given the system of equations
[tex]\begin{bmatrix}2x-y=-4\\ 3x+y=-1\end{bmatrix}[/tex]
[tex]\mathrm{Multiply\:}2x-y=-4\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:6x-3y=-12[/tex]
[tex]\mathrm{Multiply\:}3x+y=-1\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:6x+2y=-2[/tex]
[tex]\begin{bmatrix}6x-3y=-12\\ 6x+2y=-2\end{bmatrix}[/tex]
[tex]6x+2y=-2[/tex]
[tex]-[/tex]
[tex]\underline{6x-3y=-12}[/tex]
[tex]5y=10[/tex]
so the system of equations becomes
[tex]\begin{bmatrix}6x-3y=-12\\ 5y=10\end{bmatrix}[/tex]
solving 5y = 10 for y
[tex]5y=10[/tex]
Divide both sides by 5
[tex]\frac{5y}{5}=\frac{10}{5}[/tex]
simplify
[tex]y=2[/tex]
[tex]\mathrm{For\:}6x-3y=-12\mathrm{\:plug\:in\:}y=2[/tex]
[tex]6x-3\cdot \:2=-12[/tex]
[tex]6x-6=-12[/tex]
Adding 6 to both sides
[tex]6x-6+6=-12+6[/tex]
[tex]6x=-6[/tex]
Divide both sides by 6
[tex]\frac{6x}{6}=\frac{-6}{6}[/tex]
[tex]x=-1[/tex]
Therefore, the solution to the system of equations be:
[tex]x=-1,\:y=2[/tex]
From the above calculations, it is clear that the consistent system has exactly one solution i.e. x = -1, y =2, therefore, it is independent.
