Question

31. For what value of a and b does the pair of equations 3x + 4y = 11 and x + 2ay = 3a + 4b + 1 have an infinite number of solutions.​

Answer 1

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Step-by-step explanation:

Answer 2

Answer:

$3x + 4y - 11 = 0 \\ x + 2ay - (3a + 4b + 1) = 0 \\ condition \: for \: infinite \: solutions \\ \frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} \\ \frac{3}{1} = \frac{4}{2a} = \frac{11}{3a + 4b + 1} \\ 3 = \frac{11}{3a + 4b + 1} \\ 9a + 12b + 3 = 11 \\ 9a + 12b = 8 \: \: \: \: \: eq1 \\ \frac{4}{2a } = \frac{11}{3a + 4b + 1} \\ 12a + 16b + 4 = 22a \\ 10a - 16b - 4 = 0 \: \: \: \: \: eq 2 \\ \\ by \: solving \: the \: equations \\ a = \frac{2}{3} \\ b = \frac{1}{6}$

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