Can you manipulate?

Relevant wiki: System of Linear Equations (Simultaneous Equations)

The given equations imply $\begin{cases} E_1:\ a + 2d = b + 2c \\ E_2:\ a + 3d = 3b + c \end{cases}$ We want to multiply both equations by a factor and subtract them so that the left side becomes $a - d$ . It is not difficult to see that factors 4 and 3 do the trick, so that $\begin{array}{c} 4a + 8d = 4b + 8c \\ 3a + 9d = 9b + 3c \\ \hline a - d = -5b + 5c.\end{array}$ Clearly, $a - d = -5(b - c),\ \ \ \therefore\ \ \ \ \frac{a-d}{b-c} = \boxed{-5}.$

Relevant wiki: Componendo and Dividendo

$\begin{aligned} \dfrac{a-b}{c-d} = 2 & \implies \dfrac{(a-b)+(c-d)}{(a-b)-(c-d)} = \dfrac{2+1}{2-1} & \small {\color{#3D99F6} \text{by componendo and dividendo}} \\ & \implies \dfrac{(a-d)-(b-c)}{(a+d)-(b+c)} = 3 \qquad {\color{#3D99F6} (1)} \\ \dfrac{a-c}{b-d} = 3 & \implies \dfrac{(a-c)+(b-d)}{(a-c)-(b-d)} = \dfrac{3+1}{3-1} \\ & \implies \dfrac{(a-d)+(b-c)}{(a+d)-(b+c)} = 2 \qquad {\color{#3D99F6} (2)} \end{aligned}$

${\color{#3D99F6} (1)} \div {\color{#3D99F6} (2)}$ gives

$\begin{aligned} \dfrac{(a-d)-(b-c)}{(a-d)+(b-c)} = \dfrac 32 & \implies \dfrac{\{(a-d)-(b-c)\}+\{(a-d)+(b-c)\}}{\{(a-d)-(b-c)\}-\{(a-d)+(b-c)\}} = 5 & \small {\color{#3D99F6} \text{applying componendo and dividendo again}}\\ & \implies - \dfrac{a-d}{b-c} = 5 \\ & \implies \dfrac{a-d}{b-c} = \boxed{-5} \end{aligned}$

I decided to use a linear algebra approach:

Let the above two equations be rewritten as the system:

$a - b - 2c + 2d = 0$

$a - 3b - c + 3d = 0$

which after performing row reduction yields $c$ and $d$ as free variables (since we have a 2x4 system). The full solution set is:

$a = \frac{5p}{2} - \frac{3q}{2}, b = \frac{p}{2} + \frac{q}{2}, c = p, d = q$ ; $p,q \in \mathbb{R}$ .

We now compute: $\frac{a-d}{b-c} = \frac{(\frac{5p}{2} - \frac{3q}{2}) - q}{(\frac{p}{2} + \frac{q}{2}) - p} = \frac{\frac{5p}{2} - \frac{5q}{2}}{\frac{-p}{2} + \frac{q}{2}} = \frac{\frac{5}{2} \cdot (p-q)}{-\frac{1}{2} \cdot (p-q)} = \boxed{-5}.$

a-b = 2(c-d) => a + 2d = b + 2c

a-c = 3(b-d) => a + 3d = 3b + c.

Subtracting these 2 equations give d = 2b - c.

Substitution gives a = - 3b - 6c.

Putting a and d in the given equation gives: a-d=-5b-5c =-5(b-c). Dividing by b-c gives the answer - 5.

IM NOT THAT GOOD AT MATH SO USED TRIAL AND ERROR

A=6 B=2 C=3 D=1

Thus,

(A-D)÷(B-C)=(6-1)÷(2-3)

= -5

P.S IM still Grade 2 😂

Since we have nothing to do with d, Let d=0, then a-c=3b; and a-b=2c; then subtracting both we get c=2b, resubstitute this in a-c=3b we get a=5b, hence write all in terms of b in question we get 5b/-b = -5

How does a 15-year old get to set up these?

Anywhere here is my brute force solution:

Re-write the two equations and eliminate a by addition to give d = 2b -c

Now eliminate a by subtraction to give d = 4b - c

From these, b = 0 and c = - d

So the required quotient is -5d/5 = -5

I am pleased with myself, rather in the way that a man who digs a ditch with a shovel is when he sees a JCB do the job more elegantly and quickly.

FOS

I was most interested to know if the values of 2 and 3 had to be 'specially chosen' to make the requested value of $\frac{a-d}{b-c}$ come out to a constant like -5 independent of a, b, c, d.

So I just redid the algebra for the general case of $\frac{a-b}{c-d}=X$ and $\frac{a-c}{b-d}=Y$ .

Following linear algebra steps similar to those used by Tom Engelsman, Arjen Vreugdenhil, and others, the result is $\frac{a-d}{b-c}=\frac{XY-1}{X-Y}$ .

(I may upload the steps later.) So the dependence on the particular values of a, b, c, d always drops out. I can't say I really understand why, though, or how the repeating structure of the three fractions leads to this.

input in WolframAlpha solve (a-b)/(c-d)=2, (a-c)/(b-d)=3, m=(a-d)/(b-c) result: m=-5

I tried numbers. I set $d = 1$ .

The easiest way to solve this is to try to set the first numerator, $c - d$ , to 1. So I set $c = 2$ .

To pick different numbers, I chose $b = 3$ , which meant $a = 5$ .

Well, $\frac {5 - 2}{3 - 1} = \frac {3}{2}$ , so, that doesn't work.

So I set $b = 4$ and $a = 6$ .

Since $\frac {6 - 2}{4 - 1} = \frac {4}{3}$ , that doesn't work either. But I figured let me keep the $6$ since it equals $2 \times 3$ .

So then I tired setting $c = 3$ , which meant I had to set $b = 2$ , which worked because $\frac {6 - 3}{2 - 1} = 3$ .

Now plug in $\frac {6 - 1}{2 - 3} = -5$ .

By componendo and dividendo on the given information, we get (b-d)(c-d)=1/2 Also, If (a-d)/(b-c)=k, (4/3)[(b-d)/(c-d)]=(k+1)/(k-1) Putting the value of (b-d)/(c-d)=1/2 we get k=-5