Question 1185746
<br>
{{{abs(x-5) = 5-x}}} for x<5
{{{(x-5) = x-5}}} for x>5<br>
{{{abs(3-2x) = 3-2x}}} for x<3/2
{{{abs(3-2x) = 2x-3}}} for x>3/2<br>
So we need to look for solutions for x in three intervals: (-infinity,3/2), (3/2,5), and (5, infinity).<br>
(1) (-infinity,3/2)<br>
{{{abs(x-5)=abs(3-2x)+2}}}
{{{5-x=3-2x+2}}}
{{{5-x=5-2x}}}
{{{x=0}}}<br>
x=0 is in the interval we are working in, so x=0 is a solution -- or, rather, (x,y)=(0,5) is a solution.<br>
(2) (3/2,5)<br>
{{{abs(x-5)=abs(3-2x)+2}}}
{{{5-x=2x-3+2}}}
{{{5-x=2x-1}}}
{{{6=3x}}}
{{{x=2}}}<br>
x=2 is also in the interval we are working in, so (x,y)=(2,3) is a solution.<br>
(3) (5,infinity)<br>
{{{abs(x-5)=abs(3-2x)+2}}}
{{{x-5=2x-3+2}}}
{{{x-5=2x-1}}}
{{{x=-4}}}<br>
x=-4 is not in the interval we are working in, so this is not another solution.<br>
ANSWER: (x,y)=(0,5) and (x,y)=(2,3)<br>
A graph, showing the graphs of the two absolute value functions intersecting at (0,5) and (2,3)....<br>
{{{graph(400,400,-2,8,-2,8,abs(x-5),abs(3-2x)+2)}}}<br>