## Hollywood_chrissy Group Title Can somebody please explain th3 attached to me. is there a proof or something??? one year ago one year ago

1. Hollywood_chrissy Group Title

2. koalamon Group Title

just think about it, it will come to you.

3. Hollywood_chrissy Group Title

4. asnaseer Group Title

hint: in this equation:$\frac{a+b}{a-b}=\frac{c+d}{c-d}$divide the numerator and denominator of the left-hand-side by 'b' and divide the numerator and denominator of the right-hand-side by 'd'.

5. Hollywood_chrissy Group Title

@asnaseer, even after your hint, i am still running into difficulty

6. DLS Group Title

Isn't this Componendo & Dividendo? ;)

7. Hollywood_chrissy Group Title

what is componedo and dividendo???

8. DLS Group Title

It is a theorem in which the denominator is added to the numerator & numerator is added with negative denominator e. g. if $\frac{x}{y}=\frac{a}{b}$ then by componendo & dividendo we have $\frac{( x + y)}{( x- y ) }$= $\frac{( a +b )}{( a - b )}$ (Proof) $\frac{3}{2}$ =$\frac{ 6}{4}$ then by componendo & dividendo $\frac{3+ 2}{3-2}$ = $\frac{6 +4}{6- 4}$ OR 5/ 1 = 10/ 2 that is true

9. Hollywood_chrissy Group Title

@DLS do you understand asnaseer's hint???

10. DLS Group Title

I think the theorem is better, he just said if u do the same thing/operation with a digit on one side,u do that on other too so overall effect nullifies

11. Hollywood_chrissy Group Title

you helped me greatly thanks

12. asnaseer Group Title

This is what I meant - starting with this:$\frac{a+b}{a-b}=\frac{c+d}{c-d}$lets first divide the numerator and denominator of the left-hand-side by 'b' to give:$\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{c+d}{c-d}$then lets divide the numerator and denominator of the right-hand-side by 'd' to give:$\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{\frac{c}{d}+1}{\frac{c}{d}-1}$now note that you are given that:$\frac{a}{b}=\frac{c}{d}$so the equality of both sides follows from this relation.

13. asnaseer Group Title

I don't think you can just assume the "Componendo and dividendo" theorem directly as this question is effectively asking you to prove it.