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
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Just set f(x) equal to zero. When you do that the given equation becomes:
.
.
Now get rid of the denominators by multiplying all terms (on both sides) of this equation
by . When you do that multiplication you get:
.
.
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:
.
.
And the equation reduces to:
.
.
Multiply out the terms to get:
.
3x - 6 + 4x - 4 = 0
.
Combine the x terms and the equation becomes:
.
7x – 6 – 4 = 0
.
Combine the -6 with the -4 and the equation further becomes:
.
7x – 10 = 0
.
Eliminate the -10 from the left side by adding +10 to both sides. This makes the equation:
.
7x = 10
.
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:
.
.
Hope this helps you understand the problem.
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