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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.

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

$$c-b=2c-3b+d$$

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

$$b-c=d-b$$

Then

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

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

$$2=\frac12\frac{a-d}{c-b}+\frac12\frac{d-b}{b-d}=\frac12\frac{a-d}{c-b}-\frac12$$

Therefore,

$$\frac{a-d}{c-b}=5$$

Or

$$\frac{a-d}{b-c}=-5$$


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

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

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

By adding the first row to twice the second,

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

Then

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

You then plug these values in your fraction,

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

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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., $k=-5$.

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  • $\begingroup$ 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
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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 $$\frac{a-d}{b-c}=-5$$

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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 :

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

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

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