Given:

And

Required:
To find the two possible values of c.
Explanation:
Consider

So

And also given

Now from (1) and (2), we get


Now put a in (1) we get

We can interpret that either of a or b are equal to 3 or 5.
When a=3 and b=5, we have

When a=5 and b=3, we have

Final Answer:
The option D is correct.
31 and 41
Answer:
Substitute y=x-2y=x−2 into y=-0.5x+4y=−0.5x+4.
x-2=-0.5x+4x−2=−0.5x+4
2 Solve for xx in x-2=-0.5x+4x−2=−0.5x+4.
x=4x=4
3 Substitute x=4x=4 into y=x-2y=x−2.
y=2y=2
4 Therefore,
\begin{aligned}&x=4\\&y=2\end{aligned}
x=4
y=2
Step-by-step explanation:
Answer:
e
Step-by-step explanation:
e