Southbank International School Weblogs

Systems of Equation
How to solve an equation with 3 variables:
$2x+3y+3z=8$
$1x+1y+1z=3$
$5x-3y+1z=3$
To solve this equation two of the three variables needs to be canceled out. Multiply the second equation by two than subtract it from equation one. This gives us $1y+1z=2$ Multiple the second equation by five than subtract equation 3 from equation 2. This gives us $8y+4z=12$ We now have 2 equation with two variables:
$1y+1z=2$
$8y+4z=12$
Multiple $1y+1z=2$ by -8. This gives us $-8y-8z=-16$ Next, add the two equations together. This gives us $-4z=-4$ which can be solved as a linear equation.
$-4z=-4$
$z=1$
This can be plugged in to the previous equations.
$1y+1z=2$
$1y+1=2[tex] <br>[tex]1y=1$
They can then be susituted into the original equation.
$2x+3y+3z=8$
$2x+3+3=8$
$2x=2$
$x=1$
$x=1 y=1 z=1$

Completeing the Square
Completeing the square can be used to solve quadratic equation. For example: [tex}ax^2+3bx+6x=0[/tex] Here completing the square can be used to solve for x.
For Example:
$3x^2+10x+3=0$ Here is is easiest to divide the equation by the coefficient of x.
$x^2+\frac{10}{3}+1+G=0+G$ where G is the number needed to complete the square.
$(x+\frac{10}{3})^2+1=0+G$
$(x+\frac{5}{3})^2+1=\frac{25}{9}$ Now it can be solved like a linear equation.
$(x+\frac{5}{3})^2=\frac{16}{9}$
$x+\frac{5}{3}=\sqrt{\frac{16}{9}}$
$x+\frac{5}{3}=\frac+-{4}{3}$
$x=+\frac{-5}{3}\frac+-{4}{3}$
$x=-3 or \frac{-1}{3}$

0 Comments

0 Comments

Post a comment

About

Google Translate

Category Archives

Monthly Archives

Other