Question 126789
Problem 2.
Eq. 1 {{{t+m=2.1}}}
Eq. 2 {{{2t+3m=5.15}}}

Multiply Eq. 1 by -2.  That gives you -2t in Eq. 1 and 2t in Eq. 2

Eq. 1 {{{-2t-2m=-4.2}}}
Eq. 2 {{{2t+3m=5.15}}}

Now add the two equations

{{{0t+m=.95}}}
{{{m=.95}}}

Now you can proceed either one of two ways.  Either substitute the .95 value you got for m back into Eq. 1 ({{{t+.95=2.1}}} => {{{t=1.15}}}), or you can start with the original 2 equations and multiply Eq. 1 by -3 so that the addition step will eliminate m.


Eq. 1 {{{-3t-3m=-6.3}}}
Eq. 2 {{{2t+3m=5.15}}}
========================
{{{-t+0m=-1.15}}}
{{{t=1.15}}}  Same result.


Problem 3.
Do this one the same way, except that you need to be a little more creative in finding the multipliers for the equations.

Eq. 1  {{{4o+5a=2.00}}}
Eq. 2  {{{3o+4a=1.56}}}


Multiply Eq. 1 by 3 and Eq. 2 by -4.  That gives you {{{12o}}} in Eq. 1 and {{{-12o}}} in Eq. 2.  Then proceed as in the first problem.


Problem 7.  There is no effective solution method for this system because you have 4 different variables and only 2 equations.