How do you solve #1/2(x-3)+3/(2-x)=5x#?
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To solve the equation 1/2(x-3) + 3/(2-x) = 5x, we can follow these steps:
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Multiply every term in the equation by the common denominator, which is 2(x-3)(2-x). This will help eliminate the fractions.
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Simplify the equation by distributing and combining like terms.
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Rearrange the equation to isolate the variable, x.
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Solve for x by dividing both sides of the equation by the coefficient of x.
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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:
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Multiply every term by the common denominator, 2(x-3)(2-x):
(2(x-3)(2-x))(1/2(x-3)) + (2(x-3)(2-x))(3/(2-x)) = (2(x-3)(2-x))(5x)
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Simplify the equation by distributing and combining like terms:
(2-x) + 3(x-3) = 5x(2(x-3)(2-x))
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Expand and simplify further:
(2-x) + 3x - 9 = 10x(x-3)(2-x)
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Rearrange the equation to isolate the variable, x:
2 - x + 3x - 9 = 10x(x-3)(2-x)
2x - 7 = 10x(x-3)(2-x)
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Solve for x by dividing both sides of the equation by the coefficient of x:
2x - 7 = 10x(x-3)(2-x)
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
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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(x-3) + 3/(2-x) = 5x are x = 1/15 and x = 7/2.
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When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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