Question

find the value of 'a' linear equation has no solution x+ay=2 and 2x-5y=3​

Answer 1

$\huge{\underline{\bf\red{Questi {\mathbb{O}} n} :}}$

$\sf Fine \: the \: value \: of \: 'a' \: for \: which \\ \sf the \: system \: of \: linear \: equations \\ x + ay = 2 \: { \sf{and}} \: 2x - 5y = 3 \\ \sf has \: no \: solution.$

$\huge{\underline{\: \bf\green{Answ {\mathbb{E}} r}\ \: :}}$

• CASE ( i ) :-

$\boxed{\qquad \qquad a = - \frac{5}{2} \qquad \qquad }$

• CASE ( ii ) :-

$\boxed{a = \bigg\{ x :x \in {\mathbb{R}},x \neq - \frac{10}{3} \bigg\}}$

$\huge{\underline{\bf\blue{Soluti {\mathbb{O}} n}\ :}}$

The given system of linear equations is of the form :—

• $a_1 x + b_1 y + c_1 = 0$
• $a_2 x + b_2 y + c_2 = 0$

where,

• $a_1 = 1, \ b_1 = a, \ c_1 = -2$
• $a_2 = 2, \ b_2 = -5, \ c_2 = -3$

Now,

$\because$ the given system of linear equations has no solution.

$\therefore$ $\boxed{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} }$

CASE ( i ) :

$\begin{aligned} \\ \qquad \ \frac{a_{1}}{a_{2}} & = \frac{b_{1}}{b_{2}} \\ \\ \implies\frac{1}{2} & = \frac{a}{ - 5} \\ \\ \implies a & = - \frac{5}{2} & & & & & & & \end{aligned}$

CASE ( ii ) :

$\begin{aligned} \\ \qquad \ \frac{b_{1}}{b_{2}} & \neq \frac{c_{1}}{c_{2}} \\ \\ \implies\frac{a}{ - 5} & \neq \frac{ - 2}{ - 3} \\ \\ \implies \quad a & \neq \frac{ - 10}{3} & & & & & & & \end{aligned}$

$\red{\rule{5.5cm}{0.02cm}}$

$\purple{\rule{7.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$

Answer 2

Answer:

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