Part 1: Decide whether you would use the graphing, substitution, or elimination method to solve the following system of equations. Explain, in complete sentences, why you chose that method. Part 2: Solve the following system of equations and show all of your work. 4x + 3y = -1 3x + y = 3
ok the last time, I used the elimination method because we would have ended up with fractions when we tried to solve for any variable, can you solve for a variable in either one of these equations? hint: the second one....solve for y. then tell me what you got.
Use elimination.
shouldn't it be a fraction?
y=3-3x
wait I got two
y=3-3x
using subsitution.
so its pretty easy to now pluf in y=3-3x into the first equation and then solve for x.
pluf=plug
if you have a fraction, it will be the last step and it will not have variables...
you want to use substitution, in algegebra they taught us that if any time you have a variable by its self, in this case the y use substitution.
good idea^^
true. :)
no fractions:)
i ment algebra
I think elimination is much faster, and with a little practice, you can do it in your head pretty fast, but....to each their own. Just know why you are picking one over the other.
so y equals 13x plus 3
so 4x+3(3-3x)=-1 4x+9-9=-1 4x=-1 x=-4
substitution might not be the fastest way, but i just took algebra last year and they said if thers a variable by its self use substition, i would put that since thats what they "want"
Sometimes Substitution is pretty fast....annoying summer hw. :(
4x+3(13x+3)
I would personally use whatever way you are most comfortable with, or whichever way has the least possibility of error. Just my 2 cents worth.
im confused
distribute to find x, then plug it
4x + 3y = -1 3x + y = 3 one method y=3-3x plug into first equation 4x+3(3-3x)=-1 4x+9-9x=-1 -5x=-10 x=2 pluf this into either to solve for y y=3-3*2 y=-3 so(2,-3) second method 4x + 3y = -1 3x + y = 3 multiply second equation by -3 4x + 3y = -1 -9x - 3y = -9 add them -5x=-10 x=2 plug in solve for y y=-3 third method graph them, this is lame and takes time.....these are linear so to easy to solve.
do u understand now?
@jaderbrown
@jaderbrown
first method = substitution second method = elimination I think second is way easy, but you do it both ways and decide.
yes thank you!
most people would use substitution here...
I think:)
Substitution is probably the easiest and most straight forward method for this particular problem. It is glaringly obvious by the lone y variable in the second equation.
and actually I misspoke, these are not linear equations. They are called affine. Linear must go through the origin.
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