Asked by Heather
solve the following inequality
70x-51<70/x
I get this--but I do not know how to write the final answer in a solution set
70x-51<70/x
70x^2/x - 51x /x - 70/x < 0
70x^2 - 51x - 70/x <0
(10x + 7) (7x - 10)/x <0
x=0
(10x + 7)= 0
(7x - 10)=0
Can you please help me to write the answer in a solution set???
70x-51<70/x
I get this--but I do not know how to write the final answer in a solution set
70x-51<70/x
70x^2/x - 51x /x - 70/x < 0
70x^2 - 51x - 70/x <0
(10x + 7) (7x - 10)/x <0
x=0
(10x + 7)= 0
(7x - 10)=0
Can you please help me to write the answer in a solution set???
Answers
Answered by
Steve
when you get to this point:
70x^2/x - 51x /x - 70/x < 0
multiply by x to get
70x^2 - 51x - 70 < 0 and note that x≠0
since 1/x is not defined for x=0
now you have
(10x+7)(7x-10) < 0
At this point you should have some idea what you are looking for. You know it is a parabola which opens upward, and crosses the x-axis at -7/10 and 10/7.
So, the interval between the roots satisfies the original inequality:
-7/10 < x < 10/7
Algebraically, since
(10x+7)(7x-10) < 0, either
(10x+7) < 0 and (7x-10) > 0
or
(10x+7) > 0 and (7x-10) < 0
solve those and you will find a solution set which agrees to the interval above.
70x^2/x - 51x /x - 70/x < 0
multiply by x to get
70x^2 - 51x - 70 < 0 and note that x≠0
since 1/x is not defined for x=0
now you have
(10x+7)(7x-10) < 0
At this point you should have some idea what you are looking for. You know it is a parabola which opens upward, and crosses the x-axis at -7/10 and 10/7.
So, the interval between the roots satisfies the original inequality:
-7/10 < x < 10/7
Algebraically, since
(10x+7)(7x-10) < 0, either
(10x+7) < 0 and (7x-10) > 0
or
(10x+7) > 0 and (7x-10) < 0
solve those and you will find a solution set which agrees to the interval above.
Answered by
Steve
oops. because x cannot be 0, the final solution is
(-7/10,0) U (0,10/7)
(-7/10,0) U (0,10/7)
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