Question

Kevin sold 2 unicycles and 1 bicycle for a total of $110. Greg sold 4 unicycles and 3 bicycles for a total cost of $268. What is the cost of 1 bicycle?

Answer 1

\[\huge 2\color{royalblue}{U}+1\color{orangered}{B}=110\]\[\huge 4\color{royalblue}{U}+3\color{orangered}{B}=268\]

I came up with these equations based on the information they gave us, does that make sense?

U represents the price of 1 Unicycle, B the price of 1 Bicycle.

Answer 2

We have a system of equations, and we want to solve for B.

Answer 3

There are TWO useful methods for solving a system at this level,
SUBSTITUTION and ELIMINATION.

Are you familiar or comfortable with either method? :)

Answer 4

Substitution yes but Ive always been very bad at word problems can you help @zepdrix

Answer 5

Read my first post.
Are you having any trouble understanding how I set up those equations?

Answer 6

yes @zepdrix

Answer 7

Wow why are your responses ssoooooooo slow? :\ lol

Answer 8

Casue im Doing other things haha.. Im sorry ):

Answer 9

We want to establish what our Unknowns are.
In this problem, our unknowns are "the price of a unicycle" and "the price of a bicycle".

So we'll use letters to represents those unknowns.
It's a little different than problems you may have done in the past because, in this case, U does NOT represent the NUMBER of unicycles. It instead represents the price of one unicycle.

So we'll assign our unknowns these letters,
Price of a Unicycle -> U
Price of a Bicycle -> B

Answer 10

Kevin sold 2 Unicycles, so the money he made from those 2 unicycles is,

2 times "the price of a unicycle", right?
Or using our variable,

2U

Answer 11

Kevin made this much money from Unicycles and Bicycles.

2U+1B

2Unicycles x The Price of a Unicycle, PLUS 1Bicycle x The Price of a Bicycle.
And it added up to $110.

Answer 12

right

Answer 13

I think maybe that's the tough part, realizing that the letter does not represent a number of Bicycles, it represents a price.

Answer 14

Wait so its 110?

Answer 15

Because the possible chpices are
27

Answer 16

31
48
79

Answer 17

@zepdrix

Answer 18

I was just explaining how to get the 2 equations :x

We have to solve for B still.
You said you like to use Substitution? Ok let's try that.
Solving the first equation for U gives us,\[\large 2\color{royalblue}{U}+1\color{orangered}{B}=110 \qquad \rightarrow \qquad \color{royalblue}{U}=55-\frac{1}{2}\color{orangered}{B}\]

Substituting this into the other equation gives us,\[\large 4\color{royalblue}{U}+3\color{orangered}{B}=268 \qquad \rightarrow \qquad4\left(55-\frac{1}{2}\color{orangered}{B}\right)+3\color{orangered}{B}=268\]

Answer 19

Distribute the 4 to each term in the brackets,\[\large 220-2\color{orangered}{B}+3\color{orangered}{B}=268\]

And then solve for B! :)

Answer 20

B= 48?

Answer 21

Yay good job.