Question #ccc07

2 Answers
Dec 16, 2016

#x-2y+6=0#

Explanation:

The standard form of an equation of this type is #ax+bx+c=0#, where #x and y# are variables(not having fixed value) and #a, b and c# are integers(-3,-2,-1,0,1,2,3...).
But, in this question, we need to take #y# to the other side in order to leave only 0 on that side.
To do this, subtract #y# from both sides. This does not affect the equation.

#y-y=1/2x+3-y#
#1/2x-y+3=0#

Now, we have to convert #1/2# to an integer, and to do this we multiply it by 2. For the equation to remain unaffected, we have to multiply all terms with 2. So,

#cancel(2)^1xx1/(cancel(2))x-2xxy+2xx3=2xx0#
#=>x-2y+6=0#

This is the standard form of the equation.

Dec 16, 2016

#-1x + 2y = 6#

Explanation:

The standard form for a linear equation is:

#Ax + By = C# where #A#, #B# and #C# are integers.

Therefore, we must multiply each side of the given equation by #-2# to eliminate the fraction and keep the equation balanced:

#-2y = -2(1/2x + 3)#

#-2y = -1x - 6#

Now we can isolate the #x# and #y# terms on one side of the equation:

#1x - 2y = 1x - 1x - 6#

#1x - 2y = 0 - 6#

#1x - 2y = -6#