# Thread: Basic Algebra Problem - Is there a better way of solving this?

1. ## Basic Algebra Problem - Is there a better way of solving this?

Hey guys,

I've posted a basic Algebra problem below that I've solved however I'm curious if there is a better way to solve it. While my solution works it is a 2 part solution so for every answer I get from the first equation I then need to do a second calculation to see if it's correct.

Since I was doing this by hand (I was away from my computer at the time) it was a rather time consuming process of trial and error and I'm just curious if there is a better way of doing this.

The problem is:

The results of an election for a sporting club is as follows... "Fifty votes ahead of her closest competition, Amanda will be the new sports captain, obtaining 35% of all votes. Belinda came in second place, with Cameron not far behind her obtaining 134 more votes than Daniel. Daniel obtained 12% of all votes."

If these 4 people were the only candidates, find the number of votes each candidate obtained.
The only hint was to let x = the total number of votes.

So to solve this I simply created an equation where x = A + B + C + D where A = Amanda's votes, B = Belinda's votes, C = Cameron's votes and D = Daniel's votes.

Then I chose D as the main variable and created an equation to link all 4 candidates votes back to Daniel's. Using the above information the initial equation I came up with was:

${\rm{Total}}\,{\rm{Votes (x) = }}A + B + C + D\\$
${\rm{Total}}\,{\rm{Votes (x) = }}(D*\frac{{35}}{{12}}) + (D*\frac{{35}}{{12}} - 50) + (D + 134) + D$

I then simplified this to:

${{\rm{x = }}D*\frac{{35}}{{12}} + D*\frac{{35}}{{12}} - 50 + D + 134 + D}$

${{\rm{x = }}D(\frac{{35}}{{12}} + \frac{{35}}{{12}}) - 50 + 134 + D + D}$

${{\rm{x = }}\frac{{70}}{{12}}D + 84 + 2D}$

${x = (5\frac{{10}}{{12}}D + 2D) + 84}$

${x = 7\frac{5}{6}D + 84}$

So now for any value of D I can get the total number of votes that would go along with it. However the main rule that needs to be satisfied is $\frac{d}{x} = 12\%$.

So basically what I ended up doing was trying different values of D with the above equation to get the total number of votes and then doing a second calculation to divide D by x to see if it equaled 12%.

Eventually after 11 attempts I came up with:

D = 168 so...

$x = 7\frac{5}{6}*168 + 84\\$
$x = 1400$

and then confirmed it by:

$\frac{D}{x} = \frac{{168}}{{1400}} = 0.12{\mkern 1mu} \,{\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} \,\frac{A}{x} = \frac{{(168*\frac{{35}}{{12}}\,)}}{{1400}} = 0.35{\mkern 1mu} \,(also{\mkern 1mu} \,matches)$

which gives the final answer of:

Amanda's Votes: $168*\frac{{35}}{{12}} = 490$

Belinda's Votes: $168*\frac{{35}}{{12}} - 50 = 440$
Cameron's Votes: $168 + 134 = 302$
Daniel's Votes: $168$

So finally I managed to get the correct answer. However not having access to a computer meant it was a rather slow process of trial and error with 2 calculations per attempt.

As a result I was just wondering if any of you guys can see a better way of doing this??? Is there anyway to express this as one equation so by solving that one equation you could tell if your answer was correct instead of having to do the 2nd step I did by checking if $\frac{D}{x} = 0.12$??? Any advice would be greatly appreciated.

2. ## Re: Basic Algebra Problem - Is there a better way of solving this?

There is a shorter way, I think:

Let $x$ = total votes.

Then, from the question, we can rewrite each person's vote count in terms of $x$

Amanda has $0.35x$

Belinda has $0.35x-50$

Daniel has $0.12x$

Cameron has $0.12x+134$

$0.35x+0.35x+0.12x+0.12x-50+134=x$

$0.94x+84=x$

$0.06x=84$

$x=1400$

Once you know this, it should be simple to finish.

3. ## Re: Basic Algebra Problem - Is there a better way of solving this?

V = A + A-50 + C + C-134
V = 2A + 2C - 184

A = .35(V) ..... C = (3A + 644) / 7 [1]

C-134 = .12(V) ..... C = (6A + 2798) / 19 [2]

[1][2]: (3A + 644) / 7 = (6A + 2798) / 19
Solve: A = 490
Wrapup...

4. ## Re: Basic Algebra Problem - Is there a better way of solving this?

Hey Quacky and Wilmer... I just want to say thank you both very much for your help and I see where I was being in-efficient now.

Quacky I love your solution and realize now I could of made it easier by linking the total votes (x) = Ax, Bx, Cx, Dx like you did rather then to x = D which made them un-related hence my second calculation I had to do.

Wilmer your solution was more along the lines of what I was initially trying to do and I got up to the part where V = 2A + 2C - 184 (although I did V = 2A + 2D + 84). However I became stuck after that and so eventually went about it a different way. Seeing your solution I now know why I was stuck and it's simply because I haven't learnt enough algebra yet to understand what you're doing in [1] & [2] altho I get the final step when you solve [1] = [2] to get A = 490 (but not too sure how to solve that either). This could be fun tho. You've now given me something more to play with so I'm going to bookmark this page and return to it once I've learnt more algebra to see if I can figure out the remaining steps on my own and I'll post again if I get stuck.

Thanks again to both of you because you've both been a huge help