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.
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