Answer 2.3:2
From Förberedande kurs i matematik 1
(Difference between revisions)
(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 ...) |
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| Line 5: | Line 5: | ||
|width="33%" | <math>\left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr }\right.</math> | |width="33%" | <math>\left\{ \eqalign{ y_1 &= -5 \cr y_2 &= 3\cr }\right.</math> | ||
|c) | |c) | ||
| - | |width="33%" | | + | |width="33%" | no (real) solution |
|- | |- | ||
|d) | |d) | ||
Revision as of 10:27, 8 September 2008
| 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) | no (real) solution |
| 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. |
