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# An Exam Question

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Rearrange the formula y = 3x-2/x+1 to make x the subject.

I found this in a test paper, and am struggling to finish it. I have gotten to yx+y=3x-2 but don't know where to go from there.

Guest Jan 16, 2016

#1
+10

Solve for x:
y = 3 x+1-2/x

y = 3 x+1-2/x is equivalent to 3 x+1-2/x = y:
3 x+1-2/x = y

Bring 3 x+1-2/x together using the common denominator x:
(3 x^2+x-2)/x = y

Multiply both sides by x:
3 x^2+x-2 = x y

Subtract x y from both sides:
-2+x+3 x^2-x y = 0

Collect in terms of x:
-2+3 x^2+x (1-y) = 0

Divide both sides by 3:
-2/3+x^2+1/3 x (1-y) = 0

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

Add 1/36 (1-y)^2 to both sides:
x^2+1/3 x (1-y)+1/36 (1-y)^2 = 2/3+1/36 (1-y)^2

Write the left hand side as a square:
(x+(1-y)/6)^2 = 2/3+1/36 (1-y)^2

Take the square root of both sides:
x+(1-y)/6 = sqrt(2/3+1/36 (1-y)^2) or x+(1-y)/6 = -sqrt(2/3+1/36 (1-y)^2)

Subtract (1-y)/6 from both sides:
x = sqrt(2/3+1/36 (1-y)^2)+(y-1)/6 or x+(1-y)/6 = -sqrt(2/3+1/36 (1-y)^2)

Subtract (1-y)/6 from both sides:
Answer: | x = sqrt(2/3+1/36 (1-y)^2)+(y-1)/6       or x = (y-1)/6-sqrt(2/3+1/36 (1-y)^2)

Guest Jan 16, 2016
Sort:

#1
+10

Solve for x:
y = 3 x+1-2/x

y = 3 x+1-2/x is equivalent to 3 x+1-2/x = y:
3 x+1-2/x = y

Bring 3 x+1-2/x together using the common denominator x:
(3 x^2+x-2)/x = y

Multiply both sides by x:
3 x^2+x-2 = x y

Subtract x y from both sides:
-2+x+3 x^2-x y = 0

Collect in terms of x:
-2+3 x^2+x (1-y) = 0

Divide both sides by 3:
-2/3+x^2+1/3 x (1-y) = 0

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

Add 1/36 (1-y)^2 to both sides:
x^2+1/3 x (1-y)+1/36 (1-y)^2 = 2/3+1/36 (1-y)^2

Write the left hand side as a square:
(x+(1-y)/6)^2 = 2/3+1/36 (1-y)^2

Take the square root of both sides:
x+(1-y)/6 = sqrt(2/3+1/36 (1-y)^2) or x+(1-y)/6 = -sqrt(2/3+1/36 (1-y)^2)

Subtract (1-y)/6 from both sides:
x = sqrt(2/3+1/36 (1-y)^2)+(y-1)/6 or x+(1-y)/6 = -sqrt(2/3+1/36 (1-y)^2)

Subtract (1-y)/6 from both sides:
Answer: | x = sqrt(2/3+1/36 (1-y)^2)+(y-1)/6       or x = (y-1)/6-sqrt(2/3+1/36 (1-y)^2)

Guest Jan 16, 2016
#2
+5

....the question wasn't to solve it, it was to rearrange it to make x the subject

Guest Jan 16, 2016
#3
+5

IT IS NOT SOLVING IT!. IT IS FINDING "x" IN TERMS OF "y".  YOU SEE "y" IN THE ANSWER?

Guest Jan 16, 2016
#4
+78719
+5

Good work,  guest #1   ......!!!!

The key to this problem is to see that (1/3)(1-y)   is actually treated as a coefficient on "x" and then completing the square based on that coefficient

CPhill  Jan 16, 2016

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