Choose the value of the y determinant (Dy) in the following system. 2x + 3y = 4
Where did the second equation go?
Yeah thats what im going to ask too?
5x – y = 8 sorry this is it
Use multiplication of the second equation by 3 and then add the two equations. From there, it is easy.
yeah ^he got it..
thanks
Wait, wasn't the question the determinant for y? 2x + 3y = 4 5x - y = 8 Replace the coefficients of the y's with the values to the tight of the equal sign. 2x + 4 5x - y Remove the variables and set up a matrix. \[\left[\begin{matrix}2 & 4 \\ 5 & -1\end{matrix}\right]\]
You are welcome. Have a good day now.
@Calcmathlete, I'm getting tired. Maybe I goofed. I don't understand the use of a determinant in this context. Care to explain?
Well you would use Cramer's Rule. http://www.purplemath.com/modules/cramers.htm They showed a 3 x 3 matix, but it works for 2 x 2 too.
Irrelevant and unnecessary, in my opinion. However, from a pedagogical viewpoint, I guess the question is asking for that setup. Question: Are they asking for the determinant which evaluates to \(y\) or are they asking for the determinant of the system? I assume the former. What do you think?
I concur. For this situation it's easier to solve by elimination or even substitution, but I do find that in some situations, it's easier to work with determinants that to work with he first two processes.
Determinants are the systematic way of solving the equations. That's all. They are a more generalized solving technique. They're also very, very straightforward. There was an elegant theory behind all of this. However, I never learned this aspect, unfortunately. As such, I cannot quite appreciate the beauty of this technique.
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