# Thread: Solving an equation with fractions

1. ## Solving an equation with fractions

Equation

-[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

x/y = 1/2

2. Originally Posted by parkparky

Equation

-[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

x/y = 1/2

$\displaystyle -(\frac {3}{(\frac {1}{2} + \frac{x}{y})}) + \frac {9y}{2x} -6 = 0$

$\displaystyle -(3 \cdot \frac {1}{\frac{1}{2} + \frac{x}{y}}) + \frac {9y}{2x} -6 = 0$

$\displaystyle -3 \cdot \frac {-1}{\frac{1}{2} + \frac{x}{y}} + \frac {9y}{2x} -6 = 0$

$\displaystyle -3 \cdot \frac {-1}{2x(\frac{1}{2} + \frac{x}{y})} + \frac {9y}{(\frac{1}{2} + \frac{x}{y})2x} -6 = 0$

$\displaystyle -3 \cdot \frac {-1}{x + \frac{2x^2}{y}} + \frac {9y}{x + \frac{2x^2}{y}} -6 = 0$

$\displaystyle -3 \cdot \frac {-1 + 9y}{x + \frac{2x^2}{y}} -6 = 0$

$\displaystyle \frac {3 - 27y}{x + \frac{2x^2}{y}} -6 = 0$

$\displaystyle 3 - 27y = 6(x+ \frac{2x^2}{y})$

$\displaystyle 3 - 27y = 6x+ \frac{12x^2}{y}$

$\displaystyle 3(1 - 9y) = 6x(1 + \frac{2x}{y})$

I simplified it the most I could, but I can't seem to find any possible way to finish this. Let's see what the other people can find...

3. ## Substitute

Greetings.

We have the equation: -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

Rewritten as: 9y/2x  [3/(1/2 + x/y)]  6 = 0

Let x/y = K, hence y/x = 1/K

So the equation becomes: 9/2K  [3/(1/2 + K)]  6 = 0

Multiply the equation by 2K: 9  [6K/(1/2 + K)]  12K = 0

Multiply the equation by (1/2 + K): 4.5 +9K  6K  6K  12K^2 = 0

Multiply the equation by 2: 9 + 18K  12K  12K  24K^2 = 0

Rearrange and simplify: 8K^2 + 2K 3 = 0

4. Originally Posted by I-Think
Greetings.

We have the equation: -[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

Rewritten as: 9y/2x  [3/(1/2 + x/y)]  6 = 0

Let x/y = K, hence y/x = 1/K

So the equation becomes: 9/2K  [3/(1/2 + K)]  6 = 0

Multiply the equation by 2K: 9  [6K/(1/2 + K)]  12K = 0

Multiply the equation by (1/2 + K): 4.5 +9K  6K  6K  12K^2 = 0

Multiply the equation by 2: 9 + 18K  12K  12K  24K^2 = 0

Rearrange and simplify: 8K^2 + 2K 3 = 0

Thank you so much. Really helpful. Do you know what topic this kind of question would go under so I can try more exercises like it?

Thanks xxx

5. Since the problem involves two variables in one equation, you can't solve for either x or y separately. You can solve for y/x or x/y since x and y always occur together. And I would recommend replacing "y/x" by a single variable: say a= y/x.
-[3/(1/2 + x/y)] + (9y/2x) - 6 = 0

Then $\displaystyle -\frac{3}{1/2+ 1/a}+ (9/2)a- 6= 0$.
Multiply numerator and denominator of that first fraction by 2a to simplify it: $\displaystyle -\frac{6a}{a+ 2}- (9/2)a- 6= 0$

Multiply the entire equation by 2(a+2) to get rid of all the denominators: $\displaystyle 12a- 9a(a+2)- 12(a+2)= 0$
$\displaystyle 12a- 9a^2- 18a- 12a- 24= -9a^2- 18a- 24= 0$
That's a quadratic equation which could be solved by the quadratic formula or completing the square.

6. ## Technique: Substitution

Allo. This topic falls under (Pre-Algebra + Algebra)?...

Okay, I'm not totally sure what topic this falls under, but to solve questions such as these, we usually substitute another variable to make the question easier to solve.

In your question, we let x/y=K

Another example, :x^4  3x^2 + 2 = 0
We let x^2=Z
So the equation becomes :Z^2 - 3Z + 2 = 0
Solve for Z and we then determine x.

Final point, when you see a variable as part of the denominator of one of the terms in the question, multiply both sides of the equation to get rid of it.

Eg. 6f + 3/f - 2 = 0

Multiply by f: 6f^2 + 3 - 2f = 0, Solve as normal

Eg 2. 5/(1+z) -3 + 11z = 0

Multiply by (1+z): 5- 3-3z +11z+11z^2 = 0
11z^2 +8z +2 = 0, Solve as normal

That's techniques I believe are necessary to solve questions of this *mysterious* topic.