Chapter ISG, Problem 9.7.11P

To determine

To find solution of system of equation algebraically.

Answer to Problem 9.7.11P

Then points of intersection are $(0.61,1.83)$ and $(-1.61,-4.83)$

Explanation of Solution

Given information:

  $y=x^2+4x-1$

  $y=3x$

Calculation:

  $y=x^2+4x-\mathrm{1.............}(1)$

  $y=3x\mathrm{....................}(2)$

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

  $3x=x^2+4x-1$

On solving,

  $0=x^2+4x-1-3x$

On solving,

  $0=x^2+x-1$

On solving,

  $x^2+x-1=0$

On solving,

  $x=\frac{-(1)\pm \sqrt{{(1)}^2-4(1)(-1)}}{2(1)}$

On solving,

  $x=\frac{-1\pm \sqrt{1+4}}{2}$

On solving,

  $x=\frac{-1\pm \sqrt{5}}{2}$

On solving,

  $x=\frac{-1+\sqrt{5}}{2}$

  $x=\frac{-1-\sqrt{5}}{2}$

On solving,

  $x\approx -1.61$

  $x\approx 0.61$

If $x\approx -1.61$

  $y=3(-1.61)$

On solving,

  $y\approx -4.83$

If $x\approx 0.61$

  $y=3(0.61)$

On solving,

  $y\approx 1.83$

Then points of intersection are $(0.61,1.83)$ and $(-1.61,-4.83)$