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it has to be x=
I just started it
Substitute y=15-4x in first equation, you will get the value of x
Cool people use matrices. Be a cool person. Use matrices.
oh i apologize.
i'm taking a pretest. So i have never seen this stuff before. so if you'll explaine it or show your work id appraciate it! (:
You could solve it by saying \(y=15-4x\), then sticking that into \(6x+3\left(15-4x\right)=33\), and solving for \(x\) which should make \(y\) easy to find.
So can you work it out for me? :D
No, WolframAlpha has you covered on that. http://www.wolframalpha.com/ I'm not going to do a machine's job! http://www.youtube.com/watch?v=gAYL5H46QnQ
heres the answer for the first one: http://www.wolframalpha.com/input/?i=6x++%2B++3y++%3D++33
heres the answer for the second one: http://www.wolframalpha.com/input/?i=4x++%2B+++y++%3D++15 hope this helps !(:
Hey @Kyrie - I can help you with this
So, here you have two systems of equations, and you want to find a way to make at least one of the equations have only ONE variable - does that make sense? So, in this example, we want to get rid of the y in one of the equations
y=11-2x...... thats what the machine told me. but Idk what the real answer is i'll post the multiple choice kk?
So... we need to have something that equals y in our equation instead, understand? In this case, we have two sets of equations, but I'm going to use the second equation to find something that equals Y. \[4x + y = 15\] subtract 4x from each side of the equation (everything you do in algebra has to be done to BOTH sides of the equation to make it equal). This gives us: \[y = 15-4x\] Now, we have something that equals Y. So, going back to our first equation, let's put in our new 'Y': \[6x + 3(15-4x) = 33\]
A. x = 2, y = 7 B. x = -13, y = 7 C.x=-2/3 D.x = 5, y = 1
Well, do you understand what I've done so far? This is the substitution method btw
Oh i done C wrong :(
x=2, y=7 Right?
Nice work - you got it!