# Help with a algebra word problem!!!!

• Oct 3rd 2012, 05:50 PM
kellym
Help with a algebra word problem!!!!
the problem is:
Cycle Center has bicycles and tricycles in the storeroom.There are at least two of each,and there are more bicycles then tricycles.There are 23 wheels altogether.How many bicycles and tricycles are in the storeroom?
• Oct 3rd 2012, 06:25 PM
wtrmlncrawl191
Re: Help with a algebra word problem!!!!
You know you have the equation: 23 = 3*Tricycles + 2*Bicycles
Since you know each has at least two, take these out.
This leaves you with:
13 = 3*(Remaining Tricycles) + 2*(Remaining Bicycles)
Since the number is small, trial and error works best. You have 2 possible combinations to make 13.
1. 1 tricycle, 5 bicycles
2. 3 tricycles, 2 bicycles

Since we know there are more bicycles, option number 1 is correct. Now add back in the first 2, leaving you with:
3 tricycles and 7 bicycles.
• Oct 3rd 2012, 07:11 PM
Soroban
Re: Help with a algebra word problem!!!!
Hello, kellym!

This problem is so small, we can solve it without Algebra.

Quote:

Cycle Center has bicycles and tricycles in the storeroom.
There are at least two of each, and there are more bicycles then tricycles.
There are 23 wheels altogether.
How many bicycles and tricycles are in the storeroom?

Let $B$ = number of bicycles: $B \ge 2.$
Let $T$ = number of tricycles: $T \ge 2.$
. . And: $B > T.$

$B$ bicycles have $2B$ wheels.
$T$ tricycles have $3T$ wheels.
. . There are 23 wheels: . $2B + 3T \:=\:23$

Solve for $B\!:\;B \:=\:\frac{23-3T}{2} \;=\;11 - T - \frac{T-1}{2}$

From the fraction, we see that $T$ must be odd.

There are only four possible cases:

. . $\begin{array}{ccc} \text{Case} & T & B \\ \hline (1) & 1 & 10 \\ (2) & 3 & 7 \\ (3) & 5 & 4 \\ (4) & 7 & 1\end{array}$

Since $T \ge 2$, we can eliminate case (1).
Since $B > T$, we can eliminate cases (3) and (4).

Therefore: . $T = 3,\:B = 7$
• Oct 4th 2012, 09:03 AM
kellym
Re: Help with a algebra word problem!!!!
Thank you soo much for your help !