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.
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
Add 2/3 to both sides:
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)
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
Add 2/3 to both sides:
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)
....the question wasn't to solve it, it was to rearrange it to make x the subject