If $\frac{a-b}{c-d}=2$ and $\frac{a-c}{b-d} = 3$ then determine the value of: $$\frac{a-d}{b-c}$$ Where $a,b,c,d$ are real numbers.

Can someone please help me with this and give me a hint? I tried substitutions and solving them simultaneously but I couldn't determine this value. Please help.

  • $\begingroup$ HINT : Try manipulating these equations with componendo and dividendo, u might come up with something useful $\endgroup$ – Shobhit Nov 14 '14 at 12:13

You have $\displaystyle{\frac{a-b}{c-d}}=2$ and $\displaystyle{\frac{a-c}{b-d}}=3$, hence

$$a-b = 2c-2d\\a-c = 3b-3d$$

By subtracting,


Or, adding $2b-2c$ to both terms,




In denominator, $c-b$ and $b-d$ are equal, so






Alternate solution, write the following as a system of equations where $c$ and $d$ are the unknowns:

$$a-b = 2c-2d\\a-c = 3b-3d$$


By adding the first row to twice the second,

$$-2d+6d=a-b+6b-2a$$ $$d=\frac{5b-a}{4}$$


$$2c=a-b+2d=\frac{2a-2b+5b-a}{2}$$ $$c=\frac{3b+a}{4}$$

You then plug these values in your fraction,



This might not be the most mathletic solution, but here it is anyway.

Note that for the given information to make sense, we must have $c\neq d$ and $b\neq d$. From these two observations, we also see that $b\neq c$ since otherwise the two given equations are equal, i.e., $2=3$ which is nonsense. Let $k=\frac{a-d}{b-c}$.

So we can then form a system of three equations in four unknowns: $$a-b-2c+2d=0, \\ a-3b-c+3d=0,\\ a-kb+kc-d=0.$$

So the values of $a,b,c,d$ lie in the nullspace of the matrix $$\begin{bmatrix} 1 & -1 & -2 & 2\\ 1 & -3 & -1 & 3\\1 & -k & k & -1\end{bmatrix}.$$

We can apply the usual row reduction procedure to see that the nullspace of that matrix is equal to the nullspace of $$\begin{bmatrix} 1 & -1 & -2 & 2\\0 & 1 & -\frac{1}{2} & -\frac{1}{2} \\0 & 0 & \frac{k}{2}+\frac{5}{2} & -\frac{k}{2}-\frac{5}{2}\end{bmatrix}.$$

Now if $\frac{k}{2}+\frac{5}{2}\neq 0$, then we can continue the row reduction to obtain $$\begin{bmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1\\0 & 0 & 1 & -1\end{bmatrix}.$$ But this means that $a=b=c=d$, which we know is not the case. Thus $\frac{k}{2}+\frac{5}{2}=0$ must be true, i.e., $k=-5$.

  • $\begingroup$ nicely done, easy explanation (+1) $\endgroup$ – Shobhit Nov 14 '14 at 13:09
  • $\begingroup$ Thanks, although it's unknown if the OP has done row reduction yet, so it may not be easy to them! Jean-Claude's answer is probably the "right" one. $\endgroup$ – Casteels Nov 14 '14 at 13:10
  • $\begingroup$ well i dont know about that, i m just admiring your answer. $\endgroup$ – Shobhit Nov 14 '14 at 13:12

Cross-multiplying the two equations results in $$a-b=2(c-d)\text{ (1)}\\a-c=3(b-d)\text{ (2)}$$ Eliminating $a$ by subtracting $(1)$ from $(2)$ results in $$b-c=3b-d-2c\text{ (3)}$$ Eliminating $d$ by subtracting $2\times(1)$ from $3\times(2)$ results in $$6(b-c)=-2c+3b-a\text{ (4)}$$ Finally subtracting $(3)$ from $(4)$ results in $$5(b-c)=d-a$$ leading to $$\frac{a-d}{b-c}=-5$$


componendo dividendo = CD

$\frac{a-b}{c-d}=2$ $....(1)$

$\frac{a-c}{b-d}=3$ $.....(2)$

Using CD in eqn (1) to get:

$\frac{a+c-(b+d)}{a+d-(b+c)}=2$ $....(3)$

Again using CD in eqn (2) we get :

$\frac{a+b-(c+d)}{a+d-(b+c)}=3$ $....(4)$

Dividing (3) by (4) we get :


Now use CD in this eqn to come up with an answer.

  • 1
    $\begingroup$ @JohnZHANG thanks for the edit $\endgroup$ – Shobhit Nov 14 '14 at 12:26

plugging $c=\frac{a+3b}{4}$ and $d=-\frac{a}{4}+\frac{5b}{4}$ in the given term we get $-5$ as the given result.


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