How do you solve the system $5 x^{2} + 3 y^{2} = 17$ $5x^2+3y^2=17$ and $- x + y = - 1$ $-x+y=-1$ ?

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Answer 1 · 2016-08-19T15:02:06

Solution set:
${\frac{7}{4}, \frac{3}{4}}, {- 1, - 2}$ ${7/4, 3/4}, {-1, -2}$

$y = - 1 + x$ $y = -1 + x$

$\Rightarrow 5 x^{2} + 3 {(- 1 + x)}^{2} = 17$ $=>5x^2 + 3(-1 + x)^2 = 17$

$\Rightarrow 5 x^{2} + 3 (1 - 2 x + x^{2}) = 17$ $=>5x^2 + 3(1 - 2x + x^2) = 17$

$\Rightarrow 5 x^{2} + 3 - 6 x + 3 x^{2} = 17$ $=>5x^2 + 3 - 6x + 3x^2 = 17$

$\Rightarrow 8 x^{2} - 6 x - 14 = 0$ $=>8x^2 - 6x - 14 = 0$

$\Rightarrow 2 (4 x^{2} - 3 x - 7) = 0$ $=>2(4x^2 - 3x - 7) = 0$

$\Rightarrow 4 x^{2} + 4 x - 7 x - 7 = 0$ $=>4x^2 +4x - 7x - 7 = 0$

$\Rightarrow 4 x (x + 1) - 7 (x + 1) = 0$ $=>4x(x +1) - 7(x + 1) = 0$

$\Rightarrow (4 x - 7) (x + 1) = 0$ $=>(4x - 7)(x + 1) = 0$

$\Rightarrow x = \frac{7}{4} and - 1$ $=>x = 7/4 and -1$

Now, solving for y:

$y = - 1 + \frac{7}{4} and y = - 1 - 1$ $y = -1 + 7/4" and " y = -1 - 1$

$y = \frac{3}{4} and y = - 2$ $y = 3/4" and "y = -2$

Hence, the solution set is ${\frac{7}{4}, \frac{3}{4}} and {- 1, - 2}$ ${7/4, 3/4} and {-1, -2}$ .

Hopefully this helps!

How do you solve the system 5x2+3y2=175x^2+3y^2=17 and −x+y=−1-x+y=-1?