Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 169669: I am a math tutor on this site, can someone please help with a method of systems of equations?
I need to know the name of this method of solving
Ex.
.
x + 4y = 20
.
2x - y = 40
.
First solve for a variable, we will solve for "x" in each equation
.
First equation, x + 4y = 20
.
x + 4y = 20 , move "4y" to the right side
.
x + 4y = 20 = x + 4y - 4y = 20 - 4y = x = 20 - 4y or x = (-4y) + 20
.
(-4y) + 20 is our first answer, keeping that answer in mind we solve for "x" in the other equation
.
Second equation, 2x - y = 40
.
2x - y = 40 , move (-y) to the right side
.
2x - y = 40 = 2x - y + y = 40 + y = 2x = 40 + y or 2x = y + 40 , now divide each side by "2"
.
2x = y + 40 = 2x/2 = (y + 40)/2 = x = (y + 40)/2
.
Second answer is (y + 40)/2 , then I would say, since these two answers equal "x" both answers will equal each other, then I would put them in an equation
.
(-4y) + 20 = (y + 40)/2 , then I would solve for "y"
.
((-4y) + 20)/1 = (y + 40)/2 , then I would show the student the cross multiplication
.
We come up with -8y + 40 = y + 40 , then just solve, in which we would get y = 0
.
Then I would replace "y", and go through the math, and find that x = 20
.
What would you call this method ( when you solve for a variable in both equations, and put your answers together in an equation and solve )
.
Solution set = (20,0)
.
Please don't answer if you don't know the name of this method
.
Thanks ahead of time, Electrified_Levi
: I am a math tutor on this site, can someone please help with a method of systems of equations?
I need to know the name of this method of solving
Ex.
.
x + 4y = 20
.
2x - y = 40
.
First solve for a variable, we will solve for "x" in each equation
.
First equation, x + 4y = 20
.
x + 4y = 20 , move "4y" to the right side
.
x + 4y = 20 = x + 4y - 4y = 20 - 4y = x = 20 - 4y or x = (-4y) + 20
.
(-4y) + 20 is our first answer, keeping that answer in mind we solve for "x" in the other equation
.
Second equation, 2x - y = 40
.
2x - y = 40 , move (-y) to the right side
.
2x - y = 40 = 2x - y + y = 40 + y = 2x = 40 + y or 2x = y + 40 , now divide each side by "2"
.
2x = y + 40 = 2x/2 = (y + 40)/2 = x = (y + 40)/2
.
Second answer is (y + 40)/2 , then I would say, since these two answers equal "x" both answers will equal each other, then I would put them in an equation
.
(-4y) + 20 = (y + 40)/2 , then I would solve for "y"
.
((-4y) + 20)/1 = (y + 40)/2 , then I would show the student the cross multiplication
.
We come up with -8y + 40 = y + 40 , then just solve, in which we would get y = 0
.
Then I would replace "y", and go through the math, and find that x = 20
.
What would you call this method ( when you solve for a variable in both equations, and put your answers together in an equation and solve )
.
Solution set = (20,0)
.
Please don't answer if you don't know the name of this method
.
Thanks ahead of time, Electrified_Levi

Answer by stanbon(19728) About Me  (Show Source):
You can put this solution on YOUR website!
I think "substitution" is the proper label for the method.
Cheers,
Stan H.