Question

Find the values of $\alpha and\beta$ for which the following system of linear equations has infinite number of solution

$2x+3y=7;2\alpha x+(\alpha +\beta )y=28$

Answer 1

Step 1:If the pair of equations has infinite number of solutions

then,

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

Given,

$$\begin{gathered} 2x+3y=7 \\ \Rightarrow 2x+3y-7=0.................................\left(1\right) \\ 2\alpha x+(\alpha +\beta )y=28 \\ \Rightarrow 2\alpha x+(\alpha +\beta )y-28=0.............\left(2\right) \end{gathered}$$

Equation (1) and (2) are in standard form

$$\begin{gathered} herea1=2b1=3c1=-7 \\ a2=2\alpha b2=\alpha +\beta c2=-28 \end{gathered}$$

Step 2:Substituting the values in the condition $\frac{{\mathbf{a}}_{\mathbf{1}}}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{b}}_{\mathbf{1}}}{{\mathbf{b}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{{\mathbf{c}}_{\mathbf{2}}}$ by cross multiplication method

Therefore,

$$\begin{gathered} \frac{2}{2\alpha }=\frac{3}{\alpha +\beta }=\frac{-7}{-28} \\ \Rightarrow \frac{1}{\alpha }=\frac{3}{\alpha +\beta }=\frac{1}{4} \\ Taking\frac{1}{\alpha }=\frac{1}{4} \\ \Rightarrow \alpha =4 \\ Taking\frac{3}{\alpha +\beta }=\frac{1}{4} \\ \Rightarrow \frac{3}{4+\beta }=\frac{1}{4} \\ \Rightarrow 12=4+\beta \\ \Rightarrow \beta =8 \end{gathered}$$

Hence the value of $\alpha =4and\beta =8$