How do you find the exact solutions to the system $y = 2 x + 1$ $y=2x+1$ and $2 x^{2} + y^{2} = 11$ $2x^2+y^2=11$ ?

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Answer 1 · 2016-11-02T02:58:44

We can directly substitute equation $1$ $1$ into equation $2$ $2$ .

$y = 2 x + 1 \to 2 x^{2} + {(2 x + 1)}^{2} = 11$ $y = 2x + 1 -> 2x^2 + (2x + 1)^2 = 11$

$2 x^{2} + 4 x^{2} + 4 x + 1 = 11$ $2x^2 + 4x^2 + 4x + 1 = 11$

$6 x^{2} + 4 x - 10 = 0$ $6x^2 + 4x - 10 = 0$

$2 (3 x^{2} + 2 x - 5) = 0$ $2(3x^2 + 2x - 5) = 0$

$3 x^{2} + 2 x - 5 = 0$ $3x^2 + 2x - 5 = 0$

$3 x^{2} - 3 x + 5 x - 5 = 0$ $3x^2 -3x + 5x - 5 = 0$

$3 x (x - 1) + 5 (x - 1) = 0$ $3x(x - 1) + 5(x - 1) =0$

$(3 x + 5) (x - 1) = 0$ $(3x + 5)(x - 1) = 0$

$x = - \frac{5}{3} and 1$ $x = -5/3 and 1$

Case 1: $x = - \frac{5}{3}$ $x = -5/3$

$y = 2 x + 1$ $y = 2x + 1$

$y = 2 (- \frac{5}{3}) + 1$ $y = 2(-5/3) + 1$

$y = - \frac{10}{3} + 1$ $y= -10/3 + 1$

$y = - \frac{7}{3}$ $y = -7/3$

Case 2: $x = 1$ $x = 1$

$y = 2 x + 1$ $y = 2x + 1$

$y = 2 (1) + 1$ $y = 2(1) + 1$

$y = 3$ $y = 3$

Hence, the solution sets are ${1, 3}$ ${1, 3}$ and ${- \frac{5}{3}, - \frac{7}{3}}$ ${-5/3, -7/3}$ .

Hopefully this helps!

How do you find the exact solutions to the system y=2x+1y=2x+1 and 2x2+y2=112x^2+y^2=11?