Problem Solving

Solve : x/(x+4) > 5/3


x < -10

x > -4

-10 < x < -4

x > -4 or x < -10

x > -10

C

LHS has to be greater than 1.6

if x is greater greater than equal to -4 then x/x+4 will be less than 1.6 and if x is less than equal to -10 then too. i checked it with one value each,

hence -10<x<-4

bblast wrote:Solve : x/(x+4) > 5/3


x < -10

x > -4

-10 < x < -4

x > -4 or x < -10

x > -10

C

The answer choices imply critical points at x=-10 and x=-4.
Test values to the left and right of these critical points.

Let x = -13:
-13/(-13 + 4) > 5/3
-13/-9 > 5/3
13/9 > 15/9.
Doesn't work.
Eliminate any answer choice that includes x=-13 in its range.
Eliminate A and D.

Let x=0:
0/(0+4) > 5/3.
0 > 5/3.
Doesn't work.
Eliminate any answer that includes x=0 in its range.
Eliminate B and E.

The correct answer is C.

Hi Mitch,

Can u pls advise my approach is incorrect

x/x+4>5/3
3x>5x+20
-2x>20
x>-10

wasim4gmat wrote:Hi Mitch,

Can u pls advise my approach is incorrect

x/x+4>5/3
3x>5x+20
-2x>20
x>-10

Not quite.
When we divide (or multiply) each side of an inequality by a negative value, we have to change the direction of the inequality.
In the problem above, we don't know whether the denominator (x+4) is positive or negative, so multiplying each side by (x+4) is risky.
One approach is to multiply each side by the SQUARE of the denominator, since the square of a value cannot be negative.
Here is an algebraic solution:

Multiply each side by 3:
x/(x+4) > 5/3
3x/(x+4) > 5.

Multiply each side by the square of the denominator: (x+4)².
Since (x+4)² cannot be negative, we don't have to worry about changing the direction of the inequality.
[3x/(x+4)] * (x+4)² > 5(x+4)²
3x(x+4) > 5(x² + 8x + 16)
3x² + 12x > 5x² + 40x + 80
0 > 2x² + 28x + 80
0 > x² + 14x + 40
0 > (x+4)(x+10).

The critical points are x=-4 and x=-10.
These are the only values that make the right hand side equal to 0.
When x is any other value, the right hand side will be less than or greater than 0.
To determine the range of x, test one value to the left and right of each critical point.

x < -10.
Plug x=-11 into 0 > (x+4)(x+10):
0 > (-11+4)(-11+10).
0 > (-7)(-1).
0 > 7.
Doesn't work.
x < -10 is not part of the range.

-10 < x < -4.
Plug x=-5 into 0 > (x+4)(x+10):
0 > (-5+4)(-5+10).
0 > (-1)(5).
0 > -5.
This works.
-10 < x < -4 is part of the range.

x > -4.
Plug x=0 into 0 > (x+4)(x+10):
0 > (0+4)(0+10).
0 > 40.
Doesn't work.
x > -4 is not part of the range.

Thus, -10 < x < -4.

Plugging in values and eliminating incorrect answer choices seems MUCH easier.

Thanks for solving it Mitch. This is the great way of resolving the equation.

Multiply each side by the square of the denominator: (x+4)².
Since (x+4)² cannot be negative, we don't have to worry about changing the direction of the inequality.
[3x/(x+4)] * (x+4)² > 5(x+4)²
3x(x+4) > 5(x² + 8x + 16)
3x² + 12x > 5x² + 40x + 80
0 > 2x² + 28x + 80
0 > x² + 14x + 40
0 > (x+4)(x+10).



What is source of your learning. Which books do you study.

(x+4)(x+10) <0
Case 1
x+4 < 0 or x+10 > 0

x <-4 or x >10

Case 2
x+4 > 0 or x+10 < 0

x >-4 or x < -10


Next we can try substituting values in expression to see the condition.

x/(x+4) > 5/3

x = -11,9,10