Question

two systems of equations are given below.
for each system, choose the best description of its solution.
if applicable, give the solution.

system a
$x+3y=6$
$-x-3y=6$
● the system has no solution.
○ the system has a unique solution:
$(x,y)=(\square,\square)$
○ the system has infinitely many solutions.
they must satisfy the following equation:
$y=\square$

system b
$x+3y=9$
$-x-3y+9=0$
● the system has no solution.
○ the system has a unique solution:
$(x,y)=(\square,\square)$
○ the system has infinitely many solutions.
they must satisfy the following equation:
$y=\square$

Explanation:

Step1: Add System A equations

Add $x+3y=6$ and $-x-3y=6$:
$$(x+3y)+(-x-3y)=6+6$$
$$0=12$$

Step2: Analyze System A result

$0=12$ is a false statement, so no solution exists.

Step3: Rewrite System B's second equation

Rearrange $-x-3y+9=0$:
$$-x-3y=-9$$
Multiply by $-1$:
$$x+3y=9$$

Step4: Analyze System B equations

Both equations are identical ($x+3y=9$), so there are infinitely many solutions. Solve for $y$:
$$3y=9-x$$
$$y=\frac{9-x}{3}=3-\frac{x}{3}$$

Answer:

System A:

The system has no solution.

System B:

The system has infinitely many solutions.
They must satisfy the following equation:
$y=3-\frac{x}{3}$