Set up Algebra equation

[chrishowk]

I am at work and am trying to solve a problem, but I am having trouble remembering how I should set this up.
The problem:
How many of each type of transactions(X and Y) can I do in 7,200 seconds if X transactions occur 55% of the time and take 16 seconds and Y transactions occur 45% of the time and take 24 seconds?

 

[JeffM]

I am at work and am trying to solve a problem, but I am having trouble remembering how I should set this up.
The problem:
How many of each type of transactions(X and Y) can I do in 7,200 seconds if X transactions occur 55% of the time and take 16 seconds and Y transactions occur 45% of the time and take 24 seconds?

You have two unknowns so you need two equations.

x = number of X transactions

y = number of Y transactions

\(\displaystyle 16x + 24y = 7200.\) This one is obvious. I assume you found it yourself.

\(\displaystyle \dfrac{x}{y} = \dfrac{55}{45}.\) This is the one that may have been hard to see.

\(\displaystyle \dfrac{x}{y} = \dfrac{55}{45} = \dfrac{11}{9} \implies y = \dfrac{9x}{11}.\)

\(\displaystyle 16x + 24 * \dfrac{9x}{11} = 7200 \implies \dfrac{11 * 16x}{11} + \dfrac{24 * 9x}{11} = \dfrac{176x + 216x}{11} = \dfrac{392x}{11} = 7200 \implies\)

\(\displaystyle 392x = 7200 * 11 = 79200 \implies x \approx 202.\)

\(\displaystyle 16 * 202 + 24y = 7200 \implies 24y = 7200 - 3232 = 3968 \implies y \approx 165.\)

Let's check.

\(\displaystyle 16 * 202 + 24 * 165 = 3232 + 3960 = 7192 \approx 7200.\)

\(\displaystyle \dfrac{202}{202 + 165} = \dfrac{202}{367} \approx 55\%\)

The answers are approximate because only integer answers are meaningful.

 

[chrishowk]

You have two unknowns so you need two equations.

x = number of X transactions

y = number of Y transactions

\(\displaystyle 16x + 24y = 7200.\) This one is obvious. I assume you found it yourself.

\(\displaystyle \dfrac{x}{y} = \dfrac{55}{45}.\) This is the one that may have been hard to see.

\(\displaystyle \dfrac{x}{y} = \dfrac{55}{45} = \dfrac{11}{9} \implies y = \dfrac{9x}{11}.\)

\(\displaystyle 16x + 24 * \dfrac{9x}{11} = 7200 \implies\dfrac{11 * 16x}{11}+ \dfrac{24 * 9x}{11} = \dfrac{176x + 216x}{11} = \dfrac{392x}{11} = 7200 \implies\)

\(\displaystyle 392x = 7200 * 11 = 79200 \implies x \approx 202.\)

\(\displaystyle 16 * 202 + 24y = 7200 \implies 24y = 7200 - 3232 = 3968 \implies y \approx 165.\)

Let's check.

\(\displaystyle 16 * 202 + 24 * 165 = 3232 + 3960 = 7192 \approx 7200.\)

\(\displaystyle \dfrac{202}{202 + 165} = \dfrac{202}{367} \approx 55\%\)

The answers are approximate because only integer answers are meaningful.

I was using \(\displaystyle y = \dfrac{.45x}{.5}\).

I got lost at the part where you put everything over 11. I did \(\displaystyle 16(X) + (\dfrac{24}{1} * \dfrac{.45x}{.55}) = 7,200 \implies 16(X) + \dfrac{10.8x}{.55} =7,200\).

I haven't had a math class in some time. Can you describe to me how the 16(x) also got multiplied by \(\displaystyle \dfrac{11}{11}\)?

 

[JeffM]

I was using \(\displaystyle y = \dfrac{.45x}{.5}\).

I got lost at the part where you put everything over 11. I did \(\displaystyle 16(X) + (\dfrac{24}{1} * \dfrac{.45x}{.55}) = 7,200 \implies 16(X) + \dfrac{10.8x}{.55} =7,200\).

I haven't had a math class in some time. Can you describe to me how the 16(x) also got multiplied by \(\displaystyle \dfrac{11}{11}\)?

When adding fractions, they must have a common denominator. That goes way back to early arithmetic.

16x isn't even a fraction but multiplying a number by 1 does not change it. Now I want a denominator of 11, and 11/11 = 1 so I multiply 16x by 11/11 to get a fraction equivalent to 16x with a denominator of 11. Does this help?

EDIT: I simplified \(\displaystyle \dfrac{55}{45} = \dfrac{5 * 11}{5 * 9} = \dfrac{11}{9}.\) I find working with 9 and 11 easier than working with 45 and 55.