Question 1148536
<br>
Find equivalent expressions for the given function<br>
{{{abs(2x-abs(60-2x))}}}<br>
for different ranges of values for x.<br>
Case 1: x>30<br>
When x>30, 2x>60, and (60-2x) is negative; that means<br>
{{{abs(60-2x) = 2x-60}}}<br>
and then<br>
{{{abs(2x-abs(60-2x)) = abs(2x-(2x-60)) = abs(60) = 60}}}<br>
So on the interval x>30, the only solution to the given equation is x=60.<br>
Case 2: x<30<br>
When x<30, 2x<60, and (60-2x) is positive; that means<br>
{{{abs(60-2x) = 60-2x}}}<br>
and then<br>
{{{abs(2x-abs(60-2x)) = abs(2x-(60-2x)) = abs(60-4x)}}}<br>
{{{abs(60-4x)}}} is positive for x<15 and negative for x>15, so we need to break case 2 into two sub-cases.<br>
Case 2a: x<15<br>
{{{abs(2x-abs(60-2x)) = abs(2x-(60-2x)) = abs(60-4x) = 60-4x}}}<br>
So on the interval x<15, the solution to the given equation will be when<br>
{{{x = 60-4x}}}
{{{5x = 60}}}
{{{x = 12}}}<br>
Case 2b: x>15 and x<30<br>
{{{abs(2x-abs(60-2x)) = abs(2x-(60-2x)) = abs(60-4x) = 4x-60}}}<br>
So on the interval (15,30), the solution to the given equation will be when<br>
{{{x = 4x-60}}}
{{{60 = 3x}}}
{{{x = 20}}}<br>
ANSWER: There are three solutions to the equation: x=12, x=20, and x=60.<br>
A graph....<br>
{{{graph(400,200,-10,90,-10,90,abs(2x-abs(60-2x)),x)}}}<br>