An algebra problem by satyajit panda

Algebra Level 3

For which of the following pairs of values of a a and b b will the system:

{ 2 x − 3 y = 11 ( a + b ) x − ( a + b − 3 ) y = 4 a + b \begin{cases} 2x - 3y = 11 \\ (a+b) x - (a+b-3) y = 4a + b \\ \end{cases}

have an infinite number of solutions?

( a , b ) = ( − 9 , 3 ) (a,b) = (-9,3) ( a , b ) = ( 9 , − 3 ) (a,b) = (9,-3) ( a , b ) = ( 9 , 3 ) (a,b) = (9,3) ( a , b ) = ( − 9 , − 3 ) (a,b) = (-9,-3)

Satyajit Panda by satyajit panda , 20, India

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2 solutions

Viki Zeta
Sep 29, 2016

For 2 linear equations in 2 variable to have infinite solution, we have the following consistency v a 1 a 2 = b 1 b 2 = c 1 c 2 2 ( a + b ) = − 3 − ( a + b − 3 ) = − 11 − ( 4 a + b ) 2 a + b = 3 a + b − 3 = 11 4 a + b Equating first 2 and last 2 we get 2 different equations 2 a + b = 3 a + b − 3 ; and; 3 a + b − 3 = 11 4 a + b 2 ( a + b − 3 ) = 3 ( a + b ) ; and; 3 ( 4 a + b ) = 11 ( a + b − 3 ) 2 a + 2 b − 6 = 3 a + 3 b ; and; 12 a + 3 b = 11 a + 11 b − 33 a + b = − 6 ; and; − a + 8 b = 33 ⟹ 9 b = 27 b = 3 a + b = − 6 a = − b − 6 = − 3 − 6 = − 9 a = − 9 ∴ ( a , b ) = ( − 9 , 3 ) \text{For 2 linear equations in 2 variable to have infinite solution, we have the following consistency} \\v \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \\ \dfrac{2}{(a+b)} = \dfrac{-3}{-(a+b-3)} = \dfrac{-11}{-(4a+b)} \\ \dfrac{2}{a+b} = \dfrac{3}{a+b-3} = \dfrac{11}{4a+b} \\ \text{Equating first 2 and last 2 we get 2 different equations}\\ \dfrac{2}{a+b} = \dfrac{3}{a+b-3} \text{; and; } \dfrac{3}{a+b-3} = \dfrac{11}{4a+b} \\ 2(a+b-3) = 3(a+b) \text{; and; } 3(4a+b) = 11(a+b-3) \\ 2a+2b-6 = 3a+3b \text{; and; } 12a + 3b = 11a + 11b - 33 \\ a+b = -6 \text{; and; } -a + 8b = 33 \\ \implies 9b = 27 \\ \boxed{b = 3} \\ a+b = -6 \\ a = -b - 6 = -3 - 6 = -9 \\ \boxed{a=-9} \\ \boxed{\therefore (a, b) = (-9, 3)}

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Roger Erisman
Sep 30, 2016

To have infinite solutions, the second equation must have coefficients that are a multiple of the first.

Let that multiple be "n".

{1} Therefore, a + b = 2 * n

{2} a + b - 3 = 3 * n

{3} 4a + b = 11 * n

Subtracting first two equations:

yields n = - 3

Substituting n = -3 into 1st and 3rd equation:

{1} a + b = -6

{3} 4a + b = -33

Subtracting gives 3a = -27 so a = -9

Substitute into first equation means b = 3

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