If a2 = a1, b2 = 2b1, and c1 = a1 + b1, solve for c2 in terms of c1.

[Student1]

Question: If a₂ = a₁, b₂ = 2b₁, and c₁ = a₁ + b₁, solve for c₂ in terms of c₁.



My solution so far: c₂ = a₂ + b₂ = a₁ + 2b₁

I've verified the bold equation with my teacher so I know I'm in the right path. And she said to develop it further.



So I need help trying to convert the equation like this:

c₂ = 2 [(a₁/2) + b₁] so that I can write the 2 [(a₁/2) + b₁] in "n[a₁ + b₁]" terms to make the following true c₂ = n(a₁ + b₁) = n(c₁).

Is that even possible? I can't seem to solve this.

 

[Deleted member 4993]

Question: If a₂ = a₁, b₂ = 2b₁, and c₁ = a₁ + b₁, solve for c₂ in terms of c₁.



My solution so far: c₂ = a₂ + b₂ = a₁ + 2b₁

I've verified the bold equation with my teacher so I know I'm in the right path. And she said to develop it further.



So I need help trying to convert the equation like this:

c₂ = 2 [(a₁/2) + b₁] so that I can write the 2 [(a₁/2) + b₁] in "n[a₁ + b₁]" terms to make the following true c₂ = n(a₁ + b₁) = n(c₁).

Is that even possible? I can't seem to solve this.

I don't see any c₂ there!

 

[mmm4444bot]

I don't see any c₂ there!

It appears in the next paragraph:

c2 = a2 + b2 = a1 + 2b1

I've verified [this] equation with my teacher...

 

[mmm4444bot]

...I need help trying to...write the 2 [(a₁/2) + b₁] in "n[a₁ + b₁]" terms to make the following true c₂ = n(a₁ + b₁) = n(c₁).

Is that even possible?

No. To see why, check some specific cases.

For example, let a1, a2, and b1 each equal 1, then b2 equals 2. With these values, you would need:

c2 = (3/2)(c1)

Let a1 and a2 equal 7, and b1 equal 1, then b2 equals 2. With these values, you would need:

c2 = (9/8)(c1)

You could express n in terms of a1 and a2, but then your answer would express c2 in terms of a1, a2, and c1. I'm not sure what your teacher has in mind.

Have you given us the complete context of the exercise (i.e., diagrams and/or meaning of symbols a,b,c, etc.)?