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Algebra 19 Online
OpenStudy (anonymous):

to determine if an ordered pair is a solution to a system of linear equations in two variables

OpenStudy (anonymous):

The simplest solution I can think of is just to place it in both equations and see if it fits: In the ordered pair you have both let's say (x, y), so you can substitute it with the variables in those equations and see if it is a real equality for both

OpenStudy (anonymous):

do you want an example or is that clear enough? o.o

OpenStudy (anonymous):

K i'm back. so let's see.. an example I guess =]

OpenStudy (anonymous):

let's take just 2 linear functions I just made up: \[ x = y +6 \\ y = 3x + 2 \] Now just give me a random point (x,y)

OpenStudy (anonymous):

thats good enough. thanks.

OpenStudy (anonymous):

In the system –12c + 3d = 6 and –3c + 6d = –9, which expression could you use for substitution?

OpenStudy (anonymous):

Well, you asked to determine if an ordered pair is the solution of the system, so I gave the simplest way doing it. If you don't have any ordered pair, and you want to find the solution that's a different case. Is that what you need?

OpenStudy (anonymous):

this is a different question.

OpenStudy (anonymous):

\[ \left \{ \begin{array}{l l} -12c + 3d = 6 \;// +12c \\ -3c + 6d = -9 \;// +3c \end{array} \right. \\ \left \{ \begin{array}{l l} 3d = 6 + 12c \;// \div 3\\ 6d = -9 + 3c \;// \div 6 \end{array} \right. \\ \left \{ \begin{array}{l l} d = 2 + 4c \\ d = -\frac{3}{2} + \frac{c}{2} = \frac{c - 3}{2} \end{array} \right. \\ \] Now since we know what d is in both equations, we can check for which 'c' values this is equal: \[ 2 + 4c = \frac{c - 3}{2} \;// \cdot 2 \\ 4 + 8c = c - 3 \\ 7c = -7 \\ c = -1 \] Now finally since we know those 2 equations have the same spot at c=1, we want to find what is the d at that spot. we can use either one of the original equations: \[ d = 2 + 4c \\ d = 2 + 4 \cdot (-1) = 2- 4 = -2 \] so. c = -1, d = -2 last one, a check (which is what I meant above): \[ -12c + 3d = 6 \implies -12 \cdot (-1) + 3 \cdot (-2) = 6 \implies \\ \implies 12 - 6 = 6 \implies 6=6\\ \;\\ -3c + 6d = -9 \implies -3 \cdot (-1) + 6 \cdot (-2) = -9 \implies \\ \implies 3 - 12 = -9 \implies -9 = -9 \] So here, it's all good.

OpenStudy (anonymous):

thanks i gave you a medal since you got it right.

OpenStudy (anonymous):

I prefer a medal if you got my explanation, otherwise there is no really point heh.. I could just give you a solution, but I truly hope you would understand all those at the end and be able to solve them by yourself..

OpenStudy (anonymous):

i got your explanation. how old are you? and are you a boy or girl/

OpenStudy (anonymous):

What is the first step in solving the following system of equations by substitution? 9x-2=4y x=6-y

OpenStudy (anonymous):

sorry back. was in different question. reading

OpenStudy (anonymous):

Well.. I'm a 20 years old (old...) boy. And to the question first step would be to get in both equations the same variable isolated. I would go for the 'y' because it has smaller coefficient. Can do?

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