How do you solve $x^{2} + y^{2} = 25, 4 x + 3 y = 0$ $x^2+y^2=25, 4x+3y=0$ ?

Question

How do you solve $x^{2} + y^{2} = 25, 4 x + 3 y = 0$ $x^2+y^2=25, 4x+3y=0$ ?

1 Answer

Impact of this question

7525 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T21:50:42

$(- 3, 4) and (3, - 4)$ $(-3, 4) and (3, -4)$

Explanation:

$x^{2} + y^{2} = 25$ $x^2+y^2=25$
$4 x + 3 y = 0$ $4x+3y=0$ => solve for y in terms of x:
$3 y = - 4 x$ $3y=-4x$
$y = \frac{- 4 x}{3}$ $y=(-4x)/3$ => substitute in the first equation:
$x^{2} + {[\frac{- 4 x}{3}]}^{2} = 25$ $x^2+[(-4x)/3]^2=25$
$x^{2} + \frac{16 x^{2}}{9} = 25$ $x^2 +(16x^2)/9=25$ => multiply all terms by 9:
$9 x^{2} + 16 x^{2} = 25 \cdot 9$ $9x^2+16x^2=25*9$
$25 x^{2} = 25 \cdot 9$ $25x^2=25*9$
$x^{2} = 9$ $x^2=9$ => solve for x:
$x = \pm 3$ $x=+-3$
$x = - 3, y = 4$ $x=-3, y=4$
$x = 3, y = - 4$ $x=3, y=-4$
What we have is a straight line(2nd equation) that intercepts a circle(1st equation) in the above points(the solutions)

Science

Math

Humanities

... and beyond

How do you solve $x^{2} + y^{2} = 25, 4 x + 3 y = 0$ $x^2+y^2=25, 4x+3y=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve x^2+y^2=25, 4x+3y=0x2+y2=25,4x+3y=0x^2+y^2=25, 4x+3y=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $x^{2} + y^{2} = 25, 4 x + 3 y = 0$ $x^2+y^2=25, 4x+3y=0$ ?