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(C) The given equation is $(a+2b)x + (a-b)y + (a+5b) = 0$. Rearranging the terms based on $a$ and $b$: $a(x + y + 1) + b(2x - y + 5) = 0$. For this to

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$(-2, 1)$

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(C) The given equation is $(a+2b)x + (a-b)y + (a+5b) = 0$. Rearranging the terms based on $a$ and $b$: $a(x + y + 1) + b(2x - y + 5) = 0$. For this to hold for all values of $a$ and $b$,the coefficients must be zero: $x + y + 1 = 0$ (Equation $1$) $2x - y + 5 = 0$ (Equation $2$) Adding Equation $1$ and Equation $2$: $(x + y + 1) + (2x - y + 5) = 0$ $3x + 6 = 0 \implies x = -2$. Substituting $x = -2$ into Equation $1$: $-2 + y + 1 = 0 \implies y = 1$. Thus,the fixed point is $(-2, 1)$.

For all values of $a$ and $b$,the line $(a+2b)x + (a-b)y + (a+5b) = 0$ passes through a fixed point. Find that point.

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