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To find the value of `(3x+5y)/(2x+5y)`
`x^2-3xy-10y^2=0` becomes `(x-5y)(x+2y)=0`
`therefore x-5y=0 and x+2y=0`
`therefore x=5y` or `x=-2y` but, the conditions where `xgt0 and ygt0` will exclude `x=-2y` which will render a result of -1.
`therefore x=5y` Substitute;
`therefore (3x+5y)/(2x+5y)` becomes `(3x+x)/(2x+x)` which simplifies to:
Therefore `(3x+5y)/(2x+5y) = 4/3`
We can start focus on constraint `x^2 - 3xy - 10y^2 = 0` and we can divide the constraint by `y^2` :
`(x^2)/(y^2) - (3xy)/(y^2) - (10y^2)/(y^2) = 0`
Some simplifications are required:
`(x^2)/(y^2) - 3(x/y) - 10 = 0`
We can transform the equation into a quadratic just by changing the fraction x/y into unknown v:
`v^2 - 3v - 10 = 0`
Solving the quadratic becomes the easy part because you have many methods to do it. I will select the quadratic formula that I will present you now:
`v1 = (-b + sqrt(b^2 - 4ac))/(2a)` `v2 = (-b + sqrt(b^2 - 4ac))/(2a) `
You may be wondering what are a,b,c in formula. Well, they are coefficients of powers `v^2,v,v^0` .
So, a is the leading coefficient, a = 1.
The coefficient of v is b = -3 and constant is c = -10.
Just change the letters a,b,c with the new values in formula:
`v1 = (-(-3) + sqrt((-3)^2 - 4(-10)))/2`
`v2 = (-(-3)- sqrt((-3)^2 - 4(-10)))/2`
`v1 = 5`
`v2 = -2`
Do not forget what v is:
`v = x/y`
So, the fraction `x/y` has two values, -2 and 5.
Oops! You cannot accept -2 because the fraction of two positives must be a positive, too.
Now it's time to calculate the required fraction (3x+5y)/(2x+5y). Why don't you make the following trick, so factor numerator and denominator by y:
What do you notice? Exactly, the presence of fraction `x/y` . All you need to do is to put 5 instead of `x/y` .
`(3(x/y)+5)/(2(x/y)+5) = (3*5 + 5)/(2*5 + 5) = 20/15 = 4/3`
You answer is `(3x+5y)/(2x+5y) = 4/3` .
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