MHB Is the inequality correctly solved by multiplying each fraction by 100?

  • Thread starter Thread starter mathdad
  • Start date Start date
AI Thread Summary
The inequality (9/10) < (3x - 1)/-2 < 91/100 can be approached by multiplying each fraction by 100, which maintains the direction of the inequalities since 100 is positive. However, the negative sign in the denominator complicates the process, leading to the transformation 90 < -50(3x - 1) < 91. This simplifies to 90 < -150x + 50 < 91, which further reduces to -90/150 > x > -41/150. The final solution indicates that x must be less than -3/5 and greater than -41/150, confirming the inequality is correctly solved.
mathdad
Messages
1,280
Reaction score
0
Solve the inequality.

(9/10) < (3x - 1)/-2 < 91/100

Do I start by multiplying each fraction by 100?
 
Mathematics news on Phys.org
That would be one reasonable way of starting. As is so often the case, you can, you don't have to. And, since 100 is positive, multiplying by 100 does not change the direction of the inequalities: 90< -50(3x- 1)< 91.

But that "-" in the "-2" is going to cause problems!
 
90< -50(3x- 1)< 91

90 < -150x + 50 < 91

90 < -150x < 91 - 50

90 < -150x < 41

-90/150 > x > -41/150

-3/5 > x > -41/150

Correct?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top