Question

Determine if the following systems of equations have zero, one or infinitely many solutions. A) 3x + y = -10 3y = -2-10 B) 2x = 4y - 4 4x - 16y = -16 C) 2x-y= -6 x- 4y = 12

Answer 1

To determine if the following systems of equations have zero, one or infinitely many solutions we can graph each linear equation or solve the system by a method like substitution:

A)

$\begin{gathered} 3x+y=-10\text{ (1)} \\ y=-\frac{2}{3}-\frac{10}{3} \\ y=-\frac{32}{3}\text{ (2)} \\ \text{substituing (2) in (1)} \\ 3x+(-\frac{32}{3})=-10 \\ 3x=-10+\frac{32}{3} \\ x=\frac{2}{3\times3}=\frac{2}{9} \\ It\text{ has one solution} \end{gathered}$

B)

$\begin{gathered} 2x=4y-4\text{ (1) }\rightarrow\text{ x=2y-2 (1)} \\ 4x-16y=-16\text{ (2)} \\ Substitute\text{ (1) in (2)} \\ 4(2y-2)-16y=-16 \\ 8y-8-16y=-16 \\ -8y-8=-16 \\ 16-8=8y \\ y=\frac{8}{8}=1 \\ It\text{ has one solution} \end{gathered}$

C)

$\begin{gathered} 2x-y=-6\text{ (1)} \\ x-4y=12\text{ (2)}\rightarrow x=12+4y\text{ (2)} \\ Substitute\text{ (2) in (1)} \\ 2(12+4y)-y=-6 \\ 24+8y-y=-6 \\ 24-7y=-6 \\ 7y=24+6 \\ y=\frac{30}{7} \\ It\text{ has one solution} \end{gathered}$