Thomas R. answered • 05/15/18

A.S. in math with a minor in physics

Okay, this is a modest linear system problem. In other words, you have more than one equation, but they are all straight line relationships.

First off, each tricycle or bicycle has one seat, so you can simply add them:

**B + C = 28**

When it comes to wheels, however, you must multiply the number of bicycles by two and the tricycles by three:

**2B + 3C = 79**

You didn't specify if you are required to use addition or substitution, so let's run with substitution in this case. The first equation is easier to solve for one variable, and let's pick B on a whim:

**B + C = 28**

**-C -C**

**B = 28 - C**

Now substitute in the second equation:

**2B + 3C = 79**

**2(28 - C) + 3c = 79**

**56 - 2C + 3C = 79**

**56 + C = 79**

**-56 -56**

**C = 23**

Now, we go in circles, plugging back into the equation we solved earlier:

**B = 28 - C**

**B = 28 - 23**

**B = 5**