Question

The instructions say: "Prove the system has infinitely many solutions. Assume that D=0, Dx=0, Dy=0 and solve system of equations using the addition method."
It gave me this system of equations.
ax+by=e
and cx+dy=f
If I insert the given solutions [x=0, y=0] then it's 0+0=0. Does that prove infinitely many solutions, since it's a true statement? Or am I doing it wrong? PLEASE HELP.

Answer 1

what is the D,Dx,Dy? derivatives?

Answer 2

@dan815 I'm sorry idk what they are called. But they are how you find solutions using determinants. For example x=Dx/D, y=Dy/D
Here is a picture of the equation on paper. When I type it out, it makes no sense.
http://i63.tinypic.com/2dchc3t.jpg

Answer 3

I think that we can write this:
\[\Large \begin{gathered}
D = \det \left( {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right),\quad {D_x} = \det \left( {\begin{array}{*{20}{c}}
e&b \\
f&d
\end{array}} \right), \hfill \\
\hfill \\
{D_y} = \det \left( {\begin{array}{*{20}{c}}
a&e \\
b&f
\end{array}} \right) \hfill \\
\end{gathered} \]
so from the text of the problem we get the subsequent equations:
\[\Large \begin{gathered}
D = 0, \Rightarrow ad - bc = 0 \Rightarrow ad = bc\quad \left( 1 \right) \hfill \\
\hfill \\
{D_x} = 0 \Rightarrow ed - bf = 0 \Rightarrow ed = bf\quad \left( 2 \right) \hfill \\
\hfill \\
{D_y} = 0 \Rightarrow af - ec = 0 \Rightarrow af = ec\quad \left( 3 \right) \hfill \\
\end{gathered} \]:

Answer 4

now If I divide equation \(2\) side by side, by equation \(3\), the I get:
\[\Large \frac{e}{f}\frac{d}{a} = \frac{b}{c}\frac{f}{e}\]
or, after a simplification:
\[\Large {\left( {\frac{e}{f}} \right)^2} = \frac{b}{c}\frac{a}{d}\]
now using the first equation, we can write:
\[\Large \frac{a}{c} = \frac{b}{d}\]
and substituting this condition, we get:
\[\Large {\left( {\frac{e}{f}} \right)^2} = {\left( {\frac{a}{c}} \right)^2} = {\left( {\frac{b}{d}} \right)^2}\]
namely:
\[\Large \frac{e}{f} = \frac{a}{c} = \frac{b}{d}\]
so the 2 equation of the algebraic system are proportional each to other, and they are equivalent, in other words we have only one equation, which means infinite solutions

Answer 5

@Michele_Laino Thank you. You solved this using Cramer's Rule, right? Is this also considered the Addition Method? I thought about solving it how you did, but I didn't know if it was still considered Addition Method.

Answer 6

correct! I used the Cramer's rule, which is different from the addition method

Answer 7

nevertheless, when I saw the quantities \(D,D_x,D_y\) I supposed that we are using the Cramer's rule

Answer 8

@Michele_Laino That's where my confusion lies. I don't understand how to do use the Addition Method, given the system of equations.
\[D,D _x, D_y \] all equal zero. Meaning x and y= 0, right?
Since they are all zero when I plug 0 into the system of equations for x and y.
I get 0+0=0 Would that be the answer, using this method?

Answer 9

not exactly, if we write:
\(D=D_x=D_y=0\) then we can not infer any statement on the values of \(x,y\)

Answer 10

from the text of your problem, I understand that such problem asks to you 2 steps:
1) the proof that we have infinite solutions
2) to solve the system using the addition method

Answer 11

and, with my reasoning above, I showed the first step

Answer 12

Oh! So above (the part using cramer's Rule) is how I solve the first part?

Answer 13

yes! I think so!

Answer 14

since the subsequent formulas:
\[\Large x = \frac{{{D_x}}}{D},\quad y = \frac{{{D_y}}}{D}\]
are the expression of Cramer's rule

Answer 15

I think I get it now. Thank you so much @Michele_Laino

Answer 16

:)