Question 1158969: Hi!
I have been trying to use old, retired ACT exams to prepare to take the actual test, and in checking my work to find what I need to improve on I've run into a problem that I have solved but don't quite understand.
"The solution of the system of equations below is the set of all (x,y)such that 2x-3y=6. What is the value of k?
18x-27y=54
6x+ky=-2k
A.-9
B.-1
C.3
D.6
E.9
I solved by substituting each of the options given for k in the bottom equation, and then using the elimination method for solving systems of equations. However, it was very time consuming and I did not understand how the equation in the paragraph fit in with the other two equations. The answer was A. I did not understand why the answer that made the two equations come out to 0 was correct. I did not even use the equation in the paragraph to solve, and I only got the answer correct when fixing my work later, not when taking the test (I simulated an actual test as much as I could by timing myself). Is there a faster, more accurate method? Perhaps using the calculator? I do own a graphing calculator, but it is not a TI-84.
I hope this makes sense. Thank you for taking the time to help me!
Anna McLeran
Found 3 solutions by josgarithmetic, MathLover1, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source):
Answer by MathLover1(20850)  (Show Source):
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
Hi!
I have been trying to use old, retired ACT exams to prepare to take the actual test, and in checking my work to find what I need to improve on I've run into a problem that I have solved but don't quite understand.
"The solution of the system of equations below is the set of all (x,y)such that 2x-3y=6. What is the value of k?
18x-27y=54
6x+ky=-2k
A.-9
B.-1
C.3
D.6
E.9
I solved by substituting each of the options given for k in the bottom equation, and then using the elimination method for solving systems of equations. However, it was very time consuming and I did not understand how the equation in the paragraph fit in with the other two equations. The answer was A. I did not understand why the answer that made the two equations come out to 0 was correct. I did not even use the equation in the paragraph to solve, and I only got the answer correct when fixing my work later, not when taking the test (I simulated an actual test as much as I could by timing myself). Is there a faster, more accurate method? Perhaps using the calculator? I do own a graphing calculator, but it is not a TI-84.
I hope this makes sense. Thank you for taking the time to help me!
Anna McLeran
18x - 27y = 54 ----- eq (i)
6x + ky = - 2k ----- eq (ii)
18x + 3ky = - 6k --- Multiplying eq (ii) by 3 ----- eq (iii)
18x - 27y = 54 --- eq (i)
Looking at eqs (i) & (iii), and EQUATING terms, you'll see that 3ky = - 27y
Therefore, 
You DON'T have to go any further, but just for the fun of it, when you EQUATE the right-side. you'll also see that - 6k = 54.
Then, we get: 
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