How do you solve the system $- 3 x^{2} + y^{2} = 9$ $-3x^2+y^2=9$ and $- 2 x + y = 0$ $-2x+y=0$ ?

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How do you solve the system $- 3 x^{2} + y^{2} = 9$ $-3x^2+y^2=9$ and $- 2 x + y = 0$ $-2x+y=0$ ?

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Answer 1 · 2018-03-27T07:29:15

See explanation.

Explanation:

From the second equation you can calculate variable $y$ $y$

$y = 2 x$ $y=2x$

If you substitute $2 x$ $2x$ for $y$ $y$ in the first equation you get an equation with one variable $x$ $x$ :

$- 3 x^{2} + {(2 x)}^{2} = 9$ $-3x^2+(2x)^2=9$

$- 3 x^{2} + 4 x^{2} = 9$ $-3x^2+4x^2=9$

$x^{2} = 9$ $x^2=9$

$x = - 3 \lor x = 3$ $x=-3vvx=3$

Now we can put the calculates values to find the corresponding values of $y$ $y$ :

Answer: The system has 2 solutions:

${(x=-3),(y=-6):}$

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How do you solve the system $- 3 x^{2} + y^{2} = 9$ $-3x^2+y^2=9$ and $- 2 x + y = 0$ $-2x+y=0$ ?

1 Answer

Explanation:

$y = 2 x$ $y=2x$

$- 3 x^{2} + {(2 x)}^{2} = 9$ $-3x^2+(2x)^2=9$

$- 3 x^{2} + 4 x^{2} = 9$ $-3x^2+4x^2=9$

$x^{2} = 9$ $x^2=9$

$x = - 3 \lor x = 3$ $x=-3vvx=3$

${(x=-3),(y=-6):}$

${(x=3),(y=6):}$

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How do you solve the system -3x^2+y^2=9−3x2+y2=9-3x^2+y^2=9 and -2x+y=0−2x+y=0-2x+y=0?

1 Answer

Explanation:

y=2xy=2xy=2x

-3x^2+(2x)^2=9−3x2+(2x)2=9-3x^2+(2x)^2=9

-3x^2+4x^2=9−3x2+4x2=9-3x^2+4x^2=9

x^2=9x2=9x^2=9

x=-3vvx=3x=−3∨x=3x=-3vvx=3

{(x=-3),(y=-6):}{(x=-3),(y=-6):}

{(x=3),(y=6):}{(x=3),(y=6):}

Related questions

Impact of this question

How do you solve the system $- 3 x^{2} + y^{2} = 9$ $-3x^2+y^2=9$ and $- 2 x + y = 0$ $-2x+y=0$ ?

$y = 2 x$ $y=2x$

$- 3 x^{2} + {(2 x)}^{2} = 9$ $-3x^2+(2x)^2=9$

$- 3 x^{2} + 4 x^{2} = 9$ $-3x^2+4x^2=9$

$x^{2} = 9$ $x^2=9$

$x = - 3 \lor x = 3$ $x=-3vvx=3$

${(x=-3),(y=-6):}$

${(x=3),(y=6):}$