Option 1: The system has no solution
Option 2: The system has one solution
Option 3: The system has infinitely many solution.
Explanation:
Option 1: The equations are [tex]2 x+y=-12[/tex] and [tex]-4 x-2 y=30[/tex]
Solving using substitution method,
Let us substitute [tex]y=-12-2x[/tex] in [tex]-4 x-2 y=30[/tex]
[tex]\begin{array}{r}{-4 x-2(-12-2 x)=30} \\{-4 x+24+4 x=30} \\{24=30}\end{array}[/tex]
Since, both sides of the equation are not equal. The system has no solution.
Option 2: The equations are [tex]\frac{1}{3} y=\frac{7}{3} x+\frac{5}{3}[/tex] and [tex]x-3 y=5[/tex]
Solving using substitution method,
Let us substitute [tex]x=5+3y[/tex] in [tex]\frac{1}{3} y=\frac{7}{3} x+\frac{5}{3}[/tex]
[tex]\begin{aligned}\frac{1}{3} y &=\frac{7}{3}(5+3 y)+\frac{5}{3} \\\frac{1}{3} y &=\frac{35}{3}+7 y+\frac{5}{3} \\\frac{1}{3} y-7 y &=\frac{40}{3} \\-20 y &=40 \\y &=-2\end{aligned}[/tex]
substituting y=-2 in [tex]x=5+3y[/tex], we get,
[tex]\begin{aligned}x &=5+3(-2) \\&=5-6 \\&=-1\end{aligned}[/tex]
Thus, the system has one solution.
Option 3: The equations are [tex]y=x-7[/tex] and [tex]3 x-3 y=21[/tex]
Solving using substitution method,
Let us substitute [tex]y=x-7[/tex] in [tex]3 x-3 y=21[/tex]
[tex]\begin{aligned}3 x-3(x-7) &=21 \\3 x-3 x+21 &=21 \\21 &=21\end{aligned}[/tex]
Since, both sides of the equation are equal. Thus, the system has infinitely many solution.
