+0

In the equation above, a, b, and c are constans. If the equation has infinitely many solutions, which of the following must be equal to c?

0
767
1

In the equation above, a, b, and c are constans. If the equation has infinitely many solutions, which of the following must be equal to c?

2(x+b) = ax+c

A) a

B) b

C) 2a

D) 2b

I know the answer is D but can anyone explain it? Thanks!!!!!!!!!!

AND THIS ONE!

If 3x-6y = 9z, which of the following expressions is equivalent to x^2 - 4xy + 4y^2

A) 9z

B) 3z^2

C) 9z^2

D) 81^2

I honestly suck at theses problems with mutiple constants invovled in an equation. Does anybody have tips solving these types of problems? It will make my day. Thank YOU!!!!

Aug 8, 2018

#1
+1

For the first question:

When an equation has infinitely many solutions, it usually can be reduced to 0 = 0. So, what I usually do is try to get as close as possible, like this:

\(2(x+b)=ax+c\)

\(2x+2b=ax+c\)

Well... this was as close as I got. After this, think, what value does c have to be for this equation to become 0 = 0? If you think about it, the two expressions are already very similar. a could be equal to 2, and c could be equal to 2b. To test this, substitute c for 2b to see if it leads to any contradictions:

\(2x+2b=ax+2b\)

\(2x=ax\)

\(2=a\)

This is what we assumed before, that a = 2. Therefore, c = 2b, the answer is d.

For the second question:

If you look at the expression \(x^2-4xy+4y^2\), it is a perfect square trinomial, equal to \((x-2y)^2\). Also, if you simplify \(3x-6y=9z\), it becomes \(x-2y=3z\). Substituting that into our PST, we get \((3z)^2=9z^2\). The answer is c. Aug 11, 2018