Find the value of a and b for which the given system of equations has an infinite number of solutions :
4x + 6y = 14 ; (a + b + 1) x + (a + 2b + 2)y = 4 (a + b) + 1

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Solution

For infinite solutions

$\frac{a_{1}}{a_{2}}$ = $\frac{b_{1}}{b_{2}}$ = $\frac{c_{1}}{c_{2}}$

$\frac{4}{(a + b + 1)}$ = $\frac{6}{(a + 2 b + 2)}$ = $\frac{14}{(4 a + 4 b + 1)}$

3a + 3b + 3 = 2a + 4b + 4 &

7a + 7b + 7 = 8a + 8b + 2

a-b = 1 & a+b =5

a=3, b=2

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Similar questions

2 x + 3 y = 7, (a + b + 1) x + (a + 2 b + 2) y = 4 (a + b) + 1 .

2 x - 3 y = 7 (a + b) x - (a + b - 3) y = 4 a + b

2 x - 3 y = 7

(a + b) x - (a + b - 3) y = 4 a + b

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