I am very lost with how to figure this problem out, not sure where to start, fro

Question

I am very lost with how to figure this problem out, not sure where to start, from the below pictures I have found the first 3 charges,

From the below data I have found the first three charges which are

1-(4,9)
2-(3,-6)
3--(12,4)

My random distances are

1-(5,5)
2-(5-5)
3-(-12,-5)

TASK 2. Determine the values of the three charges [Warning: the calculation required here could be time consuming. Be sure to give yourself enough time to do it carefully.] (a.) Measure the potential at three points of your choosing. Call those field points A, B, and C. Think about this for a moment before you start, because judicious choices of these three field points will simplify the calculation of the charge values. (b.) Solve the following set of equations (see Eq. (1) above) for the charges (Q1, Q2, and Q3) VA = k(Q1/rA1 + Q2/rA2 + Q3/rA3) VB = k(Q1/rB1 + Q2/rB2 + Q3/rB3) VC = k(Q1/rC1 + Q2/rC2 + Q3/rC3) where k ? 8.99?109 V?m/C), VA, VB, and VC are the potentials you measured at your three chosen field points, and rA1 is the straight-line distance from charge #1 to the field point A, etc. (c.) Enter the values you calculated in Data Table II below and click the Check button. Perfect score is three points. You may change your answers and check them twice more, if necessary (i.e., three tries allowed)

Q1= Q2= Q3=

Explanation / Answer

In general, to solve an equation for a given variable, you need to "undo" whatever has been done to the variable. You do this in order to get the variable by itself; in technical terms, you are "isolating" the variable. This results in "(variable) equals (some number)", where (some number) is the answer they're looking for. For instance: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved Solve x + 6 = –3 I want to get the x by itself; that is, I want to get "x" on one side of the "equals" sign, and some number on the other side. Since I want just x on the one side, this means that I don't like the "plus six" that's currently on the same side as the x. Since the 6 is added to the x, I need to subtract to get rid of it. That is, I will need to subtract a 6 from the x in order to "undo" having added a 6 to it. This brings up the most important consideration with equations: No matter what kind of equation you're dealing with -- linear or otherwise -- whatever you do to the one side, you must do the exact same thing to the other side! Equations are like toddlers in this respect

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