Question

Solve each system using substitution. Tell whether the system has one solution, infinitely solutions, or no solution.

Answer 1

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 $2 x+y=-12$ and $-4 x-2 y=30$

Solving using substitution method,

Let us substitute $y=-12-2x$ in $-4 x-2 y=30$

$\begin{array}{r}{-4 x-2(-12-2 x)=30} \\{-4 x+24+4 x=30} \\{24=30}\end{array}$

Since, both sides of the equation are not equal. The system has no solution.

Option 2: The equations are $\frac{1}{3} y=\frac{7}{3} x+\frac{5}{3}$ and $x-3 y=5$

Solving using substitution method,

Let us substitute $x=5+3y$ in $\frac{1}{3} y=\frac{7}{3} x+\frac{5}{3}$

$\begin{aligned}\frac{1}{3} y &=\frac{7}{3}(5+3 y)+\frac{5}{3} \\\frac{1}{3} y &=\frac{35}{3}+7 y+\frac{5}{3} \\\frac{1}{3} y-7 y &=\frac{40}{3} \\-20 y &=40 \\y &=-2\end{aligned}$

substituting y=-2 in $x=5+3y$, we get,

$\begin{aligned}x &=5+3(-2) \\&=5-6 \\&=-1\end{aligned}$

Thus, the system has one solution.

Option 3: The equations are $y=x-7$ and $3 x-3 y=21$

Solving using substitution method,

Let us substitute $y=x-7$ in $3 x-3 y=21$

$\begin{aligned}3 x-3(x-7) &=21 \\3 x-3 x+21 &=21 \\21 &=21\end{aligned}$

Since, both sides of the equation are equal. Thus, the system has infinitely many solution.