For all values of a and b,the line (a+2b)x + | vedclass

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

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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 by grouping $a$ and $b$: $a(x + y + 1) + b(2x - y + 5) = 0$. For this equation to hold for all values of $a$ and $b$,the coefficients of $a$ and $b$ must be zero independently: $x + y + 1 = 0$  --- $(1)$ $2x - y + 5 = 0$ --- $(2)$ Adding equations $(1)$ and $(2)$: $(x + y + 1) + (2x - y + 5) = 0$ $3x + 6 = 0$ $3x = -6$ $x = -2$. Substituting $x = -2$ into equation $(1)$: $-2 + y + 1 = 0$ $y - 1 = 0$ $y = 1$. Thus,the line passes through the fixed point $(-2, 1)$.

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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