Question #6c2dd

1 Answer
Oct 12, 2017

#-5x+15y=-6#

Explanation:

Standard form can be complex, so let's take it one step at a time.

Let's start out with the equation that's in #y=mx+b# form already:

#y=1/3x-2/5#

Since we don't want to have to deal with fractions later one, let's take care of them right now. We must find the least common denominator of both these fractions. In this case, it's #15#. See below:

#3*5=15#
#5*3=15#

Because of this, we can multiply both the fractions by #15# to get rid of the fractions. This setup will look something like:

#y=15(1/3x-2/5)#

Now simply distribute the #15# to "cancel" out the fractions:

#15*1/3x=15/3x=5x#

and

#15*-2/5=-30/5=-6#

Don't forget, that whatever we do to an equation, has to be applied to EVERYTHING in it. Therefore we must multiply #y# by #15# as well:

#15*y=15y#

When we create real numbers involving fractions, we multiply the denominator usually by the numerator then divide by the same denominator number. However, the common denominator in this case is #15#, so we multiply #15# by the numerator, then divide by the already given denominators.

Now we have an equation without fractions in it, such as:

#15y=5x-6#

Standard form states that #ax+by=c#, therefore we must rearrange our equation now. Simply subtract #5x# from both sides:

#15y-5x=-6#

Now our equation is in standard form, and is complete! Let's check though. Consolidate #y# again and divide by #15# to get #y# alone again:

#15y=5x-6#
#15y/15#
#5x/15#
#-6/15#

Now we've got:

#y=5x/15-6/15#

Since we can simplify both these fractions let's go ahead and do that as our check. The first fraction can be divided by #5# in both the numerator and denominator, while the second fraction can be divided by #3#:

#5/5x=x# and #15/5=3#. Final Fraction: #1/3x#

#-6/3=-2# and #15/3=5#. Final Fraction: #-2/5#

Now we can write it, once again, in slope-intercept form:

#y=1/3x-2/5#

Does this check out? Yes! Therefore, our answer is correct.

Answer: Standard Form: #15y-5x=-6#