How do you solve #2/(x-1) - 2/3 =4/(x+1)#?
The denominators of an equation containing fractions can be eliminated.
Find the LCM and multiply each term by the LCM of the denominators; the denominators can cancel; this is an alternative to finding a common denominator and converting all the numerators.
#(color(red)(3(x+1)(x-1))xx2)/(x-1) -(2xxcolor(red)(3(x+1)(x-1)))/3 " " = (4xxcolor(red)(3(x+1)(x-1)))/(x+1)#
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To solve the equation 2/(x-1) - 2/3 = 4/(x+1), we can follow these steps:
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Multiply every term in the equation by the least common denominator (LCD) of (x-1), 3, and (x+1), which is 3(x-1)(x+1). This step eliminates the denominators.
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Simplify the equation by distributing and combining like terms.
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Solve for x by isolating the variable on one side of the equation.
Here are the steps in detail:
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Multiply every term by the LCD, 3(x-1)(x+1):
3(x-1)(x+1) * [2/(x-1) - 2/3] = 3(x-1)(x+1) * 4/(x+1)
Simplifying the left side: 3(x-1)(x+1) * 2/(x-1) - 3(x-1)(x+1) * 2/3
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Distribute and combine like terms:
6(x+1) - 2(x-1)(x+1) = 12(x-1)
Simplifying further: 6x + 6 - 2(x^2 - 1) = 12x - 12
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Simplify and solve for x:
6x + 6 - 2x^2 + 2 = 12x - 12
Rearranging the terms: -2x^2 + 6x + 8 = 12x - 12
Combining like terms: -2x^2 - 6x + 12x = -12 - 8
Simplifying further: -2x^2 + 6x = -20
Rearranging the terms: 2x^2 - 6x + 20 = 0
Factoring the quadratic equation is not possible, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values: x = (-(-6) ± √((-6)^2 - 4(2)(20))) / (2(2))
Simplifying: x = (6 ± √(36 - 160)) / 4
x = (6 ± √(-124)) / 4
Since the square root of a negative number is not a real number, this equation has no real solutions.
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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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