Section 42
(26) 2 R_{1} > R_{1}
(28) 2/3R_{2} + R_{1} > R_{1}
(30) 4R_{1} + R_{2} > R_{2}
(38) Solve the system:
2x_{1}

+

x_{2}

=

0

x_{1}



2x_{2}

=

5


R_{1} <>R_{2}


2R_{1}+R_{2}>R_{2}


1/5R_{2}>R_{2}


2R_{2}+R_{1}>R_{1}


so the
solution is 

(40) Solve the system:
2x_{1}



3x_{2}

=

2

4x_{1}

+

6x_{2}

=

7


2R_{1}+R_{2}>R_{2}


(46) Solve the system:
6x_{1}

+

2x_{2}

=

4

3x_{1}



1x_{2}

=

2

First we put this into matrix form, below. We could start with multiplying the first row by 1/6 because that would give a 1 in the upper left. However, we would end up with fractions for the other entries in the frist row. Thus, let's start with getting a 0 under the 6.

1/2R_{1}+R_{2}>R_{2}


1/6R_{1}>R_{1}


This corresponds
to the system 

Since any pair solves the second equation, the solutions to the system will be those pairs which solve the first. Thus the solution is
x_{1}

=

1/3t



2/3

x_{2}

=

t
