Question 1049300
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I have been attempting this problem and I have no clue as to where to go. I only have on problem left and I was not able to
solve it for this section and was wondering if anyone could please provide assistance on how I should go about solving this.

1.) Determine the rate of change for each equation.

3/2x - 5/4y = 15

Here is what I have attempted to do with the problem: I have subtracted the 3/2x to the other side and ended up with -5/4y =
-3/2x +15 which is what I am currently stuck on. I feel like I went about this the correct way, but I could be wrong.
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{{{(3/2)x - (5/4)y = 15}}}

-5/4y = -3/2x +15 "....what I am currently stuck on.
You could continue by DIVIDING each side of the equation by the coefficient of y, {{{- 5/4}}}, or MULTIPLYING by its RECIPROCAL, {{{- 4/5}}}. 
      {{{(- 5/4)y(- 4/5) = (- 3/2)x * (- 4/5) + 15 * (- 4/5)}}} ------ Multiplying each side by {{{- 4/5}}}
{{{(cross(- 5)/cross(4))y(cross(- 4)/cross(5)) = (3/cross(- 2))x * (2cross(- 4)/5) + 3cross(15) * (- 4/cross(5))}}}
                         {{{y = (6/5)x - 12}}}, wherein, m, or {{{6/5}}} is your required RATE of CHANGE!

However, it's less complex, less time-consuming and less of a headache, in this author's opinion, when you CLEAR/GET rid of the
fractions, first and foremost, as follows: {{{(3/2)x - (5/4)y = 15}}}
                                                                                 6x - 5y = 60 ---- Multiplying by LCD, 4
                                                                                      - 5y = - 6x + 60
                                                                            {{{((- 5)/(- 5))y = ((- 6)/(- 5))x + 60/(- 5)}}} ---- Dividing each side by - 5
                                                                                         {{{y = (6/5)x - 12}}}, wherein, m, or {{{6/5}}} is your required RATE of CHANGE!</font></font></font></b></pre>