Sovle: (y-7)/(y^2+5y) - (y-1)/(y^2-25).?

Question

This is one fraction minus another fraction to solve for "y".

Answer 1

((y - 7)/(y^2 + 5y)) - ((y - 1)/(y^2 - 25))
((y - 7)/(y(y + 5))) - ((y - 1)/((y - 5)(y + 5)))

Multiply everything by y(y + 5)(y - 5)

(((y - 7)(y - 5)) - y(y - 1))/(y(y + 5)(y - 5))

((y^2 - 5y - 7y + 35) - y^2 + y)/(y(y + 5)(y - 5))

(y^2 - 12y + 35 - y^2 + y)/(y(y + 5)(y - 5))

(-11y + 35)/(y^3 - 25y)

Answer 2

hey, I take it that this all equals zero?

If thats the case then this is what you do;

you have two fractions and you can make this a whole lot easier if you make it one big fraction - to do this the they must have the same denominator - so multiple the first fraction by (y^2-25)/(y^2-25) which is basically multiplying it by 1. Multiply the secont fraction by (y^2 +5y)/(y^2+5y) which does the same thing. Now both fractions have the same denominator - which is (y^2 +5y)(y^2 -25) and the whole equation looks like;

[(y-7)(y^2-25) - (y-1)(y^2+5y)]/[(y^2+5y)(y^2-25)] = 0

now the only way this equation will equal zero is if the numerator equals zero, you dont need to worry about whats below the fraction sign, so the equation becomes;

(y-7)(y^2-25) - (y-1)(y^2+5y) = 0

multiply everything out and you get;

-13y^2 -30y +175, but im kinda drunk and so you should check this for yourself

you then have a quadratic equation where a=-13, b=-30 and c=175

using the quadratic formula;

y= [±b-√(b^2-4ac)]/2a

which when you plug in the numbers you get 2 roots;

y=-2.82, -0.51

you should check the sums but thats the way you solve the equation

hope this helps

Answer 3

(y-7)/(y^2+5y) - (y-1)/(y^2-25)

(y-7)/(y^2 x y^2 + y^2 x -25 + 5y x y^2 + 5y x -25) - (y-7)/(y^2 x y^2 + y^2 x -25 + 5y x y^2 + 5y x -25)

(y-7)/(y^4 + -25y^2 + 5y^3 + -125y) - (y-1)/(y^4 + -25y^2 + 5y^3 + -125y)

(y-6)/(y^4 + 5y^3 -25y^2 - 125y)

Answer 4

i got SYNTAX Error in my calculator bud