I have the following equation to solve. I have the answer, but don't know how to get to that answer. Please help:
Equation
-[3/(1/2 + x/y)] + (9y/2x) - 6 = 0
The answer is given as:
x/y = 1/2
Please help, and write out each step to getting the answer if possible. I know it's probably simple, but I'm very bad at maths.
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
Solve the quadratic equation and youll obtain your solution.
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 .
Multiply numerator and denominator of that first fraction by 2a to simplify it:
Multiply the entire equation by 2(a+2) to get rid of all the denominators:
That's a quadratic equation which could be solved by the quadratic formula or completing the square.
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.
Farewell and adieu. Church beckons.