Svar 2.3:2
Förberedande kurs i matematik 1
(Skillnad mellan versioner)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <math>\left\{ \eqalign{ x_1 &= 1 \cr x_2 &= 3\cr }\right.</math> |b) |width="33%" | <math>\left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr ...) |
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="33%" | <math>\left\{ \eqalign{ x_1 &= 1 \cr x_2 &= 3\cr }\right.</math> |b) |width="33%" | <math>\left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr ...) |
Nuvarande version
| a) | \displaystyle \left\{ \eqalign{ x_1 &= 1 \cr x_2 &= 3\cr }\right. | b) | \displaystyle \left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr }\right. | c) | saknar (reella) lösning |
| d) | \displaystyle \left\{ \eqalign{ x_1 &= \textstyle\frac{1}{2}\cr x_2 &= \textstyle\frac{13}{2}\cr }\right. | e) | \displaystyle \left\{ \eqalign{ x_1 &= -1 \cr x_2 &= \textstyle\frac{3}{5}\cr }\right. | f) | \displaystyle \left\{ \eqalign{ x_1 &= \textstyle\frac{4}{3}\cr x_2 &= 2\cr }\right. |
