## Dollar Dealing

Hector says to Larissa, “If you give me $2, we will have an equal amount of money.” Larissa responds, “That’s true, but on the other hand if you give me $2, I will have twice as much money as you.”

Hector says to Larissa, “If you give me $2, we will have an equal amount of money.” Larissa responds, “That’s true, but on the other hand if you give me $2, I will have twice as much money as you.”

Categories : Free Scenarios

© Drexel University 2016. All Rights Reserved

The Math Forum is a research and educational enterprise of the Drexel University School of Education.

Hector had $10, Larissa had $14.

Sorry Rex, sometimes posts with dollar signs in them get caught in the blog’s spam filters. I’d love to hear how you got started figuring that out — maybe some of the students who weren’t sure what to make of the story would learn from reading your ideas!

Well, then, let’s leave out the dollar signs.

Start with Hector having H dollars and Larissa having L dollars (I will leave dollars out of it from now on and am using caps because lower-case Ls look like 1s).

First claim: Hector gets 2 from Larissa and they are then even.

So L – 2 = H + 2

Second claim: If Larissa gets 2 from Hector, she’ll have double what he has (AFTER the exchange).

So L + 2 = 2(H – 2) = 2H – 4

You can solve this system of two linear equations in two variables several ways, including graphically, but I’ll stick to algebra.

I’ll use a somewhat non-standard or hybrid approach out of sheer laziness.

[numbers in brackets are labels for the different statements/equations]

I see that I can add L – 2 = H + 2 [1]

and L + 2 = 2H – 4 [2]

which leaves 2L = 3H – 2 [3]

Dividing both sides of [3] by 2 yields

L = 3H/2 – 1 [4]

Substituting the right side of [4] for L in [1] gives:

3H/2 – 1 -2 = H + 2 [5]

Multiply both sides of [5] by 2 and get

3H – 2 – 4 = 2H + 4 [6]

Combining terms on the left side of [6] yields

3H – 6 = 2H + 4] [7]

In a bold move, we’ll subtract 2H from both sides AND add 6 to both sides of [7].

We get H = 10.

Now we can substitute 10 for H in either of the two original equations to get

L = 14.

==================

Another take:

L – 2 = H + 2 [1]

L + 2 = 2H – 4 [2]

Solve [1] for L in terms of H by adding 2 to both sides and getting

L = H + 4 [3] (this is more direct than my previous meandering approach).

Now substitute the right side of [3] for L in [2], resulting in

H + 4 + 2 = 2H – 4 [4]

We can now combine like terms on the left of [4] to get

H + 6 = 2H – 4 [5]

Boldly subtracting H from both sides and adding 4 to both sides of [5] yields

10 = H; substitute 10 for H in, say, 1, and get that L = 14.

==============================================

Third take (elimination rather than hybrid or straight substitution):

For convenience, reverse the order of the two original equations and start with

L + 2 = 2H – 4 [1]

L – 2 = H + 2 [2]

Subtract [2] from [1] leaving

4 = H – 6 [3]

Now add 6 to both sides of [3] giving

10 = H

Substitute 10 for H in [2]; add 2 to both sides and get, again, L = 14.

No matter how we do it, H = 10 and L = 14.

There really are other ways to do this that work and make mathematical sense, but that’s enough for now.