SOLUTION: 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 }}}

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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+%2B+4y+=+20+
.
+2x+-+y+=+40+
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First solve for a variable, we will solve for "x" in each equation
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First equation, +x+%2B+4y+=+20+
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+x+%2B+4y+=+20+, move "4y" to the right side
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+x+%2B+4y+=+20+ = +x+%2B+4y+-+4y+=+20+-+4y+ = +x+=+20+-+4y+ or +x+=+%28-4y%29+%2B+20+
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%28-4y%29+%2B+20+ is our first answer, keeping that answer in mind we solve for "x" in the other equation
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Second equation, +2x+-+y+=+40+
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+2x+-+y+=+40+, move (-y) to the right side
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+2x+-+y+=+40+ = +2x+-+y+%2B+y+=+40+%2B+y+ = +2x+=+40+%2B+y+ or +2x+=+y+%2B+40+, now divide each side by "2"
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+2x+=+y+%2B+40+ = +2x%2F2+=+%28y+%2B+40%29%2F2+ = +x+=+%28y+%2B+40%29%2F2+
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Second answer is +%28y+%2B+40%29%2F2+, then I would say, since these two answers equal "x" both answers will equal each other, then I would put them in an equation
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+%28-4y%29+%2B+20+=+%28y+%2B+40%29%2F2+, then I would solve for "y"
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+%28%28-4y%29+%2B+20%29%2F1+=+%28y+%2B+40%29%2F2+, then I would show the student the cross multiplication
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We come up with +-8y+%2B+40+=+y+%2B+40+, then just solve, in which we would get +y+=+0+
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Then I would replace "y", and go through the math, and find that +x+=+20+
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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 )
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Solution set = (20,0)
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Please don't answer if you don't know the name of this method
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Thanks ahead of time, Electrified_Levi

Answer by stanbon(75887) 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.