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
$x-4y=4$
$-x-4y=4$
○ the system has no solution.
● the system has a unique solution:
$(x,y)=(0,-1)$
○ the system has infinitely many solutions.
they must satisfy the following equation:
$y=\square$

system b
$x-2y=6$
$-x+2y-6=0$
● 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: Solve System A via elimination

Add the two equations:
$$(x-4y)+(-x-4y)=4+4$$
$$-8y=8$$

Step2: Solve for $y$

$$y=\frac{8}{-8}=-1$$

Step3: Substitute $y=-1$ to find $x$

Use $x-4y=4$:
$$x-4(-1)=4$$
$$x+4=4$$
$$x=0$$

Step4: Analyze System B

Rewrite the second equation:
$$-x+2y-6=0 \implies -x+2y=6 \implies x-2y=-6$$
Compare to first equation $x-2y=6$: $6
eq-6$, so no solution.

Answer:

System A:

The system has a unique solution:
$(x, y) = (0, -1)$

System B:

The system has no solution.