If the system of linear equations 2x - 3y = g | vedclass

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(C) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} =

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$58$

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(C) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Given equations are $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$.Applying the condition: $\frac{\alpha}{2} = \frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$.From $\frac{\alpha}{2} = \frac{5}{-3}$,we get $\alpha = -\frac{10}{3}$.Multiplying by $9$,we get $9\alpha = -30$.From $\frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$,we get $5(\gamma + 5) = -3(\beta + 1)$.$5\gamma + 25 = -3\beta - 3$.$3\beta + 5\gamma = -28$.Now,we need to find $|9\alpha + 3\beta + 5\gamma|$.Substituting the values: $|-30 + (-28)| = |-58| = 58$.

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

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