<s>[INST] Calculating the value of $\frac{a-d}{b-c}$ 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.
 [/INST] 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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