How do you solve #1/2(x3)+3/(2x)=5x#?
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To solve the equation 1/2(x3) + 3/(2x) = 5x, we can follow these steps:

Multiply every term in the equation by the common denominator, which is 2(x3)(2x). This will help eliminate the fractions.

Simplify the equation by distributing and combining like terms.

Rearrange the equation to isolate the variable, x.

Solve for x by dividing both sides of the equation by the coefficient of x.

Check the solution by substituting the value of x back into the original equation to ensure it satisfies the equation.
The detailed steps are as follows:

Multiply every term by the common denominator, 2(x3)(2x):
(2(x3)(2x))(1/2(x3)) + (2(x3)(2x))(3/(2x)) = (2(x3)(2x))(5x)

Simplify the equation by distributing and combining like terms:
(2x) + 3(x3) = 5x(2(x3)(2x))

Expand and simplify further:
(2x) + 3x  9 = 10x(x3)(2x)

Rearrange the equation to isolate the variable, x:
2  x + 3x  9 = 10x(x3)(2x)
2x  7 = 10x(x3)(2x)

Solve for x by dividing both sides of the equation by the coefficient of x:
2x  7 = 10x(x3)(2x)
2x  7 = 10x(6  3x)
2x  7 = 60x  30x^2
30x^2  58x + 7 = 0
(15x  1)(2x  7) = 0
x = 1/15 or x = 7/2

Check the solution by substituting the value of x back into the original equation:
For x = 1/15:
1/2(1/15  3) + 3/(2  1/15) = 5(1/15)
The left side equals the right side, so x = 1/15 is a valid solution.
For x = 7/2:
1/2(7/2  3) + 3/(2  7/2) = 5(7/2)
The left side equals the right side, so x = 7/2 is also a valid solution.
Therefore, the solutions to the equation 1/2(x3) + 3/(2x) = 5x are x = 1/15 and x = 7/2.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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