How do you do x+1/3 = 4/x3?
Submitted : 20180614 07:40:43 Popularity:
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transfer all the x's to one side of the equation, and all the number values to the other. Then solve for X.
x + 1/3 = 4/(x  3)
3x(x  3) + (x  3) =
transfer all the x's to one side of the equation, and all the number values to the other. Then solve for X.
x + 1/3 = 4/(x  3)
3x(x  3) + (x  3) = 12
3x^2  8x  15 = 0
Solutions:
x = 4/3  sqrt(61)/3
x = 4/3 + sqrt(61)/3
You really need to use parentheses to group things. Because division takes precedence, your equation would be interpreted as:
x + (1/3) = (4/x)  3
However, I think you meant to group the terms as:
(x + 1)/3 = 4/(x  3)
Cross multiply to get rid of the denominators:
(x + 1)(x  3) = 3*4
x²  3x + x  3 = 12
x²  2x  15 = 0
Factor that:
(x  5)(x + 3) = 0
That leads to two solutions:
x  5 = 0
x = 5
or
x + 3 = 0
x = 3
Answer:
x = 5 or x = 3
10/3 = 4/x  x
10x = 12  3x^2
3x^2+10x12 = 0
x = [10±√244]/6
x = (5 ± √61)/3
You have to isolate x on one side of the equals sign. Consider this:
X + 2 = 10
By subtracting 2 from both sides of the equation, you isolate X.
X + 2 = 10
X + 2 ( 2) = 10 ( 2)
X = 8
So you have to get x's on one side and nonx's on the other. You could start by subtracting 1/3 from both sides:
x + 1/3 = 4/x  3
x + 1/3 ( 1/3) = 4/x  3 ( 1/3)
Combine like terms
x + (1/3  1/3) = 4/x + (3 1/3)
x = 4/x  2 2/3
Now the left side of the equation has no nonx values, however the right side still contains both x values and nonx values, so you must do the same thing you just did, but with the "4/x" on the right side.
I'll leave off here and let you solve the rest of the way.
x+1/3 = 4/x3
ambiguous. best guess
x + (1/3) = (4/x)  3
x – (4/x) = –3 – (1/3) = –10/3
x² + (10/3)x – 4 = 0
x = 0.937, −4.27
another guess
(x+1) / 3 = 4 / (x3)
12 = (x+1)(x–3)
x² – 2x –3 = 12
x² – 2x – 15 = 0
(x – 5)(x + 3) = 0
x = 5, –3


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