How do you solve for y in #x-2y=2#?

1 Answer
May 28, 2016

Answer:

#y=x/2-1#

Explanation:

#color(brown)("When you see a question of this type ")#your objective is to end up with something like #y="'something'"#. You do this in stages. First isolating all the terms that contain your target variable on one side of the equals and everything else on the other side. You then manipulate as appropriate.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Using the rule: What you do to one side of an equation you do to the other")#

#color(blue)("Step 1: Isolate the terms with "y" in them ")#

Subtract #color(blue)(x)# from both sides

#color(brown)((x-2y)color(blue)(-x)=(2)color(blue)(-x)#

The purpose of the brackets is to make it easier to see what is happening. They serve no other purpose.

#0-2y=-x+2#

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#color(blue)("Step 2: Make the terms with "y" in it positive ")#

Multiply both sides by -1. Note that minus times minus makes positive

#+2y=+x-2#

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#color(blue)("Step : Divide both sides by 2 - removes the 2 from "2y)#

Divide by 2 is the same as multiply by #color(blue)(1/2)#

#color(brown)(color(blue)(1/2xx)2y" "=" "(color(blue)(1/2xx)x)-(color(blue)(1/2xx)2))#

#2/2xxy" "=" "(x/2)-(2/2)#

But #2/2=1" and " 1xxy=y#

#y=x/2-1#