Chapter ISG, Problem 9.7.13P

To determine

To solve system of equation graphically.

Answer to Problem 9.7.13P

Then point of intersect are $(-1,3)$ and $(1,-3)$

Explanation of Solution

Given information:

  $x^2+y^2=15$

  $-2x+y=0$

Calculation:

  $x^2+y^2=15$

  $-2x+y=0$

Rewrite

  $y=2x$

Rewrite

  $x^2+y^2={(\sqrt{15})}^2\mathrm{..............}(1)$

This is equation of circle whose center is at $(0,0)$

  $y=2x\mathrm{...................}(2)$

Putting value of y from equation (2) in (1)

  $x^2+{(2x)}^2=15$

On solving,

  $x^2+4x^2=15$

On solving,

  $5x^2=15$

On solving,

  $x^2=3$

On solving,

  $x=\sqrt{3}$

On solving,

  $x=-\sqrt{3}$

  $x=\sqrt{3}$

On solving,

  $x=-1.73$

  $x=1.73$

If $x=-1.73$

  $y=2(-1.73)$

On solving,

  $y=-3.46$

If $x=1.73$

  $y=2(1.73)$

On solving,

  $y=3.46$

Then point of intersect are $(-1,3)$ and $(1,-3)$