Question 787911
<font face="Times New Roman" size="+2">


There are several methods:


Substitution:  Solve one of the equations for one of the variables in terms of the other.  In your case, you might choose to solve your first equation for n in terms of an expression in d, as in *[tex \LARGE n\ =\ 13\ -\ d].  Then substituting the expression part of this new equation for the solved for variable in the OTHER equation.  Thus:  *[tex \LARGE 5(13\ -\ d)\ +\ 10d\ =\ 95].  That gives you a single variable equation that you can solve by ordinary algebraic means.


Elimination:  Multiply one (or both, if necessary) of your equations by a constant so that the coefficients on one of the varibles will be additive inverses in the two equations.  For your example, multiply the first equation by -5 giving you *[tex \LARGE -5n\ -\ 5d\ =\ -65].  Next, add the two equations term-by-term so that one of the variables is eliminated.  *[tex \LARGE -5n\ +\ 5n\ -\ 5d\ +\ 10d\ =\ -65\ +\ 95] which simplifies to *[tex \LARGE 5d\ =\ 30]


There are still other methods, Gauss-Jordan row reduction and Cramer's Rule, but we'll save all that for later, shall we?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>