Question

The equation
$\frac{x}{a} + \frac{y}{b} = 1 \: and \: \frac{x}{b} + \frac{y}{a} = 1$
are inconsistent if __________.


✍ANSWER with Full Solutions.

Answer 1

$\huge{\pink{\underline{\ulcorner{\star\: Solution\: \star}\urcorner}}}$

$\frac{x}{a} + \frac{y}{b} =1\\ \frac{x}{b}+\frac{y}{a}=1\\\\ \frac{xb+ya}{ab} =1 \\\\ \frac{xa+yb}{ab} =1\\ \implies bx + ya = ab ....eq(1) \\\implies ax + by = ab ....eq(2)$

$a_1 = b ,\: a_2 = a \\ b_1 = a ,\: b_2 = b \\ c_1 = ab ,\: c_2 = ab$

$\huge{\bold{\frac{a_1}{a_2}= \frac{b_1}{b_2} = \frac{c_1}{c_2}}}$

$\frac{b}{a}= \frac{a}{b} = \frac{\cancel{ab}}{\cancel{ab}} \\\\ \frac{b}{a}=\frac{a}{b}=\frac{1}{1}$

$\frac{b}{a}\:\cancel{=}\:\frac{a}{b}\:\cancel{=}\:\frac{1}{1}$

It means the system is inconsistent.

This equation have a unique solution

Answer 2

Hello!

$\star \: \dfrac{\sf x}{\sf a} + \dfrac{\sf y}{\sf b} = 1 \\ \implies \dfrac{\sf xb+ya}{\sf ab} = 1 \\ \implies \boxed{\red{\sf bx + ya = \sf ab}}$

$\star \: \dfrac{\sf x}{\sf b} + \dfrac{\sf y}{\sf a} = 1 \\ \implies \dfrac{\sf xa+yb}{\sf ab} = 1 \\ \implies \boxed{\red{\sf ax + by = \sf ab}}$

Then,

$\dfrac{\sf a_{1}}{\sf a_{2}} = \dfrac{\sf b_{1}}{\sf b_{2}} = \dfrac{\sf c_{1}}{\sf c_{2}} \\ \\ \implies \dfrac{\sf b}{\sf a} = \dfrac{\sf a}{\sf b} = \dfrac{\cancel{\sf ab}}{\cancel{\sf ab}}$

$\implies \dfrac{\sf b}{\sf a} = \dfrac{\sf a}{\sf b} = \dfrac{1}{1} \\ \\ \implies \dfrac{\sf b}{\sf a} \cancel{=} \dfrac{\sf a}{\sf b} \cancel{=} \dfrac{\cancel{\sf ab}}{\cancel{\sf ab}}$

Hence,
It has $\bf{Unique\: Solution!}$

Cheers!