SOLUTION: Find the real values of x such that f(x)=0 (directions).
f(x)=3/x-1 + 4/x-2 - - I don't even know where to begin on this problem. the answer they gave is 10/7 and I'm really con
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-> SOLUTION: Find the real values of x such that f(x)=0 (directions).
f(x)=3/x-1 + 4/x-2 - - I don't even know where to begin on this problem. the answer they gave is 10/7 and I'm really con
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Question 84530This question is from textbook College Agebra
: Find the real values of x such that f(x)=0 (directions).
f(x)=3/x-1 + 4/x-2 - - I don't even know where to begin on this problem. the answer they gave is 10/7 and I'm really confused on that.... This question is from textbook College Agebra
You can put this solution on YOUR website! Just set f(x) equal to zero. When you do that the given equation becomes:
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Now get rid of the denominators by multiplying all terms (on both sides) of this equation
by . When you do that multiplication you get:
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Note that the right side of the equation remains zero because 0 times anything is still
zero. Note also that you can now cancel the terms in the denominator with the same term
in the numerator. The result is:
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And the equation reduces to:
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Multiply out the terms to get:
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3x - 6 + 4x - 4 = 0
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Combine the x terms and the equation becomes:
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7x – 6 – 4 = 0
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Combine the -6 with the -4 and the equation further becomes:
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7x – 10 = 0
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Eliminate the -10 from the left side by adding +10 to both sides. This makes the equation:
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7x = 10
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Finally, solve for x by dividing both sides of the equation by 7 which is the multiplier
of the x term. The division results in the answer:
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Hope this helps you understand the problem.
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