# simultaneous equation

• Jun 1st 2010, 02:46 AM
mastermin346
simultaneous equation
Solve the simultaneous equation:

$\displaystyle x-2y+3=0$

$\displaystyle x^2+xy-27=0$
• Jun 1st 2010, 02:49 AM
Sudharaka
Quote:

Originally Posted by mastermin346
Solve the simultaneous equation:

$\displaystyle x-2y+3=0$

$\displaystyle x^2+xy-27=0$

Dear mastermin346,

Subject "y" in the first equation and substitute it in the second equation. Then you would have a quadratic equation.

Hope this helps.
• Jun 1st 2010, 02:54 AM
Prove It
Quote:

Originally Posted by mastermin346
Solve the simultaneous equation:

$\displaystyle x-2y+3=0$

$\displaystyle x^2+xy-27=0$

From the first equation, we can see that $\displaystyle x = 2y - 3$.

Substituting into the second equation:

$\displaystyle (2y - 3)^2 + (2y - 3)y - 27 = 0$

$\displaystyle 4y^2 - 12y + 9 + 2y^2 - 3y - 27 = 0$

$\displaystyle 6y^2 - 15y - 18 = 0$

$\displaystyle 2y^2 - 5y - 6 = 0$

$\displaystyle y^2 - \frac{5}{2}y - 3 = 0$

$\displaystyle y^2 - \frac{5}{2}y + \left(-\frac{5}{4}\right)^2 - \left(-\frac{5}{4}\right)^2 - 3 = 0$

$\displaystyle \left(y - \frac{5}{4}\right)^2 - \frac{25}{16} - \frac{48}{16} = 0$

$\displaystyle \left(y - \frac{5}{4}\right)^2 - \frac{73}{16} = 0$

$\displaystyle \left(y - \frac{5}{4}\right)^2 = \frac{73}{16}$

$\displaystyle y - \frac{5}{4} = \pm \frac{\sqrt{73}}{4}$

$\displaystyle y = \frac{5 \pm \sqrt{73}}{4}$.

So substitute $\displaystyle y = \frac{5 - \sqrt{73}}{4}$ and $\displaystyle y = \frac{5 + \sqrt{73}}{4}$

into $\displaystyle x = 2y - 3$

to find the two values of $\displaystyle x$.