For which values of a and b does the followin | vedclass

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(A) The given equations are:$2x + 3y - 7 = 0$$(a-b)x + (a+b)y - (3a + b - 2) = 0$For a system of linear equations $\frac{a_1}{a_2}x + \frac{b_1}{b_2}y

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$a=5, b=1$

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(A) The given equations are:$2x + 3y - 7 = 0$$(a-b)x + (a+b)y - (3a + b - 2) = 0$For a system of linear equations $\frac{a_1}{a_2}x + \frac{b_1}{b_2}y + \frac{c_1}{c_2} = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Here,$\frac{a_1}{a_2} = \frac{2}{a-b}$,$\frac{b_1}{b_2} = \frac{3}{a+b}$,and $\frac{c_1}{c_2} = \frac{-7}{-(3a+b-2)} = \frac{7}{3a+b-2}$.Equating the ratios:$1) \frac{2}{a-b} = \frac{3}{a+b} \implies 2a + 2b = 3a - 3b \implies a - 5b = 0 \implies a = 5b$ (Equation $1$)$2) \frac{2}{a-b} = \frac{7}{3a+b-2} \implies 6a + 2b - 4 = 7a - 7b \implies a - 9b = -4$ (Equation $2$)Substituting $a = 5b$ into Equation $2$:$5b - 9b = -4 \implies -4b = -4 \implies b = 1$.Now,substituting $b = 1$ into $a = 5b$:$a = 5(1) = 5$.Thus,the values are $a = 5$ and $b = 1$.

For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?
$2x + 3y = 7$
$(a-b)x + (a+b)y = 3a + b - 2$

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For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?$2x + 3y = 7$$(a-b)x + (a+b)y = 3a + b - 2$