How do you solve #1/(x-4) + (x-4)/(x-4) = 7 / ( x^2+x-20)#?
Eliminate the denominators by multiplying both sides by
Solve the resulting quadratic.
Use the quadratic formula:
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To solve the equation 1/(x-4) + (x-4)/(x-4) = 7/(x^2+x-20), we can simplify the equation by combining the fractions with a common denominator. The common denominator is (x-4)(x+5). Simplifying the equation, we get (x+5 + (x-4))/(x-4)(x+5) = 7/(x^2+x-20). Combining like terms in the numerator, we have (2x+1)/(x-4)(x+5) = 7/(x^2+x-20). Cross-multiplying, we get (2x+1)(x^2+x-20) = 7(x-4)(x+5). Expanding and simplifying the equation, we have 2x^3 + 3x^2 - 39x - 20 = 7x^2 - 7x - 140. Rearranging the terms, we get 2x^3 - 4x^2 + 32x + 120 = 0. Factoring out 2, we have 2(x^3 - 2x^2 + 16x + 60) = 0. Factoring the cubic polynomial, we find (x+2)(x-6)(x-5) = 0. Therefore, the solutions to the equation are x = -2, x = 6, and x = 5.
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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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