`
Wiskundeleraar
©2012 wiskundeleraar.nl
twee manieren op een stelsel op te lossen
$
\eqalign{
& \left\{ \matrix{
x - 2y = - 1 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
x = 2y - 1 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
x = 2y - 1 \cr
2\left( {2y - 1} \right) + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
x = 2y - 1 \cr
4y - 2 + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
x = 2y - 1 \cr
7y = 14 \cr} \right. \cr
& \left\{ \matrix{
x = 2y - 1 \cr
y = 2 \cr} \right. \cr
& \left\{ \matrix{
x = 3 \cr
y = 2 \cr} \right. \cr}
$
$
\eqalign{
& \left\{ \matrix{
x - 2y = - 1 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
2x - 4y = - 2 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
- 7y = - 14 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
y = 2 \cr
2x + 3y = 12 \cr} \right. \cr
& \left\{ \matrix{
y = 2 \cr
2x + 3 \cdot 2 = 12 \cr} \right. \cr
& \left\{ \matrix{
y = 2 \cr
2x = 6 \cr} \right. \cr
& \left\{ \matrix{
y = 2 \cr
x = 3 \cr} \right. \cr}
$
Login
View