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
$4x-y=-8$
$-4x+y-8=0$
$circ$ the system has no solution.
$circ$ the system has a unique solution:
$(x,y)=\left(\square,\square\
ight)$
$circ$ the system has infinitely many solutions.
they must satisfy the following equation:
$y=\square$

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

Explanation:

Step1: Rewrite System A equations

First equation: $4x - y = -8$
Second equation: $-4x + y = 8$ (rewritten from $-4x + y - 8 = 0$)

Step2: Add System A equations

Add the two equations:
$$(4x - y) + (-4x + y) = -8 + 8$$
$$0 = 0$$
This is a contradiction in terms of solution (the lines are parallel and distinct, so no overlap).

Step3: Solve System B via elimination

Add the two equations of System B:
$$(3x + y) + (3x - y) = 9 + 9$$
$$6x = 18$$

Step4: Solve for x in System B

$$x = \frac{18}{6} = 3$$

Step5: Substitute x into System B equation

Substitute $x=3$ into $3x + y = 9$:
$$3(3) + y = 9$$
$$9 + y = 9$$
$$y = 0$$

Answer:

System A:

The system has no solution.

System B:

The system has a unique solution:
$(x, y) = (3, 0)$