Question

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Answer 1

Question: Find the value of 'α' and 'β' for which the following linear equations have infinite number of solutions:

2x + 3y = 7

2αβ + (α + β) = 28

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We know that for a pair of linear equations to have infinite solutions, the condition is that,

$\boxed{\begin{minipage}{3.5 cm} \sf \longrightarrow \ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \end{minipage}\\}$

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The given equations are,

2x + 3y = 7

2αx + (α + b)y = 28

Hence,

a₁ $\longrightarrow$ 2

b₁ $\longrightarrow$ 3

c₁ $\longrightarrow$ - 7

a₂ $\longrightarrow$ 2α

b₂ $\longrightarrow$ (α + β)

c₂ $\longrightarrow$ -28

Now, we substitute these values in the condition,

$\sf \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\sf \dfrac{2}{2\alpha} = \dfrac{3}{\alpha + \beta} = \dfrac{-7}{-28}$

$\sf \dfrac{2}{2\alpha} = \dfrac{3}{\alpha + \beta} = \dfrac{7}{28}$

$\sf Now \ we \ equate \ \dfrac{2}{2\alpha} \ and \ \dfrac{7}{28}$

$\sf \dfrac{2}{2\alpha} = \dfrac{7}{28}$

$\sf 2 \times 28 = 7 \times 2\alpha$

$\sf 56 = 14\alpha$

$\sf \alpha = \dfrac{56}{14}$

$\sf \alpha = 4$

$\sf Now \ we \ equate \ \dfrac{3}{\alpha + \beta} \ \ and \ \ \dfrac{2}{2\alpha} \ and \ substitute \ the \ value \ of \ \alpha \ in \ it.$

$\sf \dfrac{2}{2\alpha} = \dfrac{3}{\alpha + \beta }$

$\sf 2(\alpha + \beta) = 3 \times 8$

Substituting the value of 'α' we get,

$\sf 2(4 + \beta) = 3 \times 8$

$\sf 8 + 2\beta = 24\\ \\ 2\beta = 24 - 8\\ \\ 2\beta = 16\\ \\ \beta = \dfrac{16}{2}\\ \\ \beta = 8$

Hence,

α $\longrightarrow$ 4

β $\longrightarrow$ 8