Solve the system of equations
$y = x^{2} + 4 x - 1$ , $- 3 x + y = 1$

(1, 4)

and

(- 2, 5)

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(1, - 4)

and

(2, 11)

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(1, 4)

and

(- 2, - 5)

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(- 2, 7)

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(0, 1)

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Solution

The correct option is C $(1, 4)$ and $(- 2, - 5)$
Given equations are $y = x^{2} + 4 x - 1$ and $- 3 x + y = 1$
From second equation, $y = 1 + 3 x$
$\Rightarrow 1 + 3 x = x^{2} + 4 x - 1$ (When substituted in first equation)
$\Rightarrow x^{2} + x - 2 = 0$
$\Rightarrow (x - 1) (x + 2) = 0$
$\Rightarrow x = + 1, - 2$
When $x = + 1, y = 1 + 3 \times (1) = 4$
When $x = - 2, y = 1 + 3 \times (- 2) = - 5$
Points of intersection are $(1, 4) (- 2, - 5)$ .

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