Can we get correct answer when we add fractions in a wrong way Are there any two fractions $\frac{a}{b}$ and $\frac{c}{d}$ such that
$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$ with conditions $HCF(a,b)=1$ and $HCF(c,d)=1$ and $b \ne d$
I have just simplified the given equation and got $$ad^2=-b^2c$$ so one of $a$ and $c$ should be negative. so is there any possibility of such fractions?
 A: $ad^2=-b^2c$ 
$gcd(a,b)=1$ => $a|c\enspace$ and $\enspace gcd(c,d)=1$ => $c|a$ , $\enspace$ therefore $a=\pm c$
$gcd(a,b)=1$ => $b|d\enspace$ and $\enspace gcd(c,d)=1$ => $d|b$ , $\enspace$ therefore $b=\pm d$ 
$ad^2=-b^2c\enspace$ => $\enspace a=-c$ 
$b=-d$ is not possible (-> denominator $b+d$) and therefore $b=d$ which is not allowed here by the conditions. Therefore no solution.
A: You have shown that at least one of them should be negative.
On the other hand, it is easy to show $$\frac{a+c}{b+d} \le max(a/b, c/d) < a/b + c/d$$
if all are positive. 
A: The relation you derived $ad^2=-b^2c$ proves that one of $a,c$ must be negative, so there is no solution with all positives.
To show that the equality is not possible, even with negative numbers, note that it implies:


*

*$a \mid b^2 c$, which given $\gcd(a,b)=1$ implies $a \mid c$, and in particular $|a| \le |c|\;$;

*$d^2 \mid b^2 c$, which given $\gcd(d,c)=1$ implies $d^2 \mid b^2$, so $d^2 \le b^2$; since $b \ne d$ and $b \ne -d$ (otherwise the RHS denominator $b+d$ would vanish) the strict inequality follows $d^2 \lt b^2$.
Multiplying the two inequalities above gives $|a|\; d^2 \lt |c|\; b^2$ which contradicts the initial equality. Therefore, there are no solutions in negative integers, either.
A: With your equation, we can deduce that $b^2\mid |d| $ and that $d^2 \mid |b|$ (use unique prime factorization). Can you see why this is impossible (if $|b| \neq |d|$)?
