Question

for which values of a and b does the following pair of linear equation have an infinity no. of solutions 2x+3y=5 (a+b)x+ (a-b)y = ( 2a+b-1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of lines are

$\rm \implies\:2x + 3y = 5 \:$

and

$\rm :\longmapsto\:(a + b)x + (a - b)y = 2a + b - 1$

It is given that, System of equations have infinitely many solutions.

We know that,

$\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

Then

Two lines have infinitely many solutions, iff

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \: }}}$

Here,

• • a₁ = 2

• • a₂ = a + b

• • b₁ = 3

• • b₂ = a - b

• • c₁ = 5

• • c₂ = 2a + b - 1

So, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{2}{a + b} = \dfrac{3}{a - b} = \dfrac{5}{2a + b - 1}$

Taking first and second member, we get

$\rm :\longmapsto\:\dfrac{2}{a + b} = \dfrac{3}{a - b}$

$\rm :\longmapsto\:3(a + b) = 2(a - b)$

$\rm :\longmapsto\:3a + 3b= 2a - 2b$

$\rm :\longmapsto\:3a - 2a= - 3b - 2b$

$\rm \implies\:\boxed{ \tt{ \: a \: = \: - \: 5b \: }} - - - - (1)$

Now, Taking second and third member, we get

$\rm :\longmapsto\: \dfrac{3}{a - b} = \dfrac{5}{2a + b - 1}$

$\rm :\longmapsto\:3(2a + b - 1) = 5(a - b)$

$\rm :\longmapsto\:6a + 5b - 5= 5a - 5b$

$\rm :\longmapsto\:6a - 5a = 5 - 5b - 5b$

$\rm :\longmapsto\:a = 5 - 10b$

On substituting the value of a, from equation (1), we get

$\rm :\longmapsto\: - 5b = 5 - 10b$

$\rm :\longmapsto\: - 5b + 10b = 5$

$\rm :\longmapsto\: 5b = 5$

$\bf\implies \:\boxed{ \tt{ \: b \: = \: 1 \: }}$

On substituting the value of b, in equation (1), we get

$\rm \implies\:\boxed{ \tt{ \: a \: = \: - \: 5 \: }}$

Hence,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{a \: = \: - \: 5} \\ \\ &\sf{b \: = \: 1} \end{cases}\end{gathered}\end{gathered}}$

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More to know :-

$\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then

1. Two lines are parallel iff

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \tt \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \: }}}$

2. Two lines have unique solution iff

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \tt \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \: }}}$