Question 353526
1. Solve the equation in each quadrant.
QI: {{{x>0}}} {{{abs(x)=x}}}
{{{y>0}}} {{{abs(y)=y}}}
{{{x+y=2}}}
{{{y=-x+2}}}
QII: {{{x<0}}} {{{abs(x)=-x}}}
{{{y>0}}} {{{abs(y)=y}}}
{{{-x+y=2}}}
{{{y=x+2}}}
QIII: {{{x<0}}} {{{abs(x)=-x}}}
{{{y<0}}} {{{abs(y)=-y}}}
{{{-x-y=2}}}
{{{y=-x-2}}}
QIII: {{{x>0}}} {{{abs(x)=x}}}
{{{y<0}}} {{{abs(y)=-y}}}
{{{x-y=2}}}
{{{y=x-2}}}
.
.
Graph it but remember you can only graph quadrant by quadrant.
{{{drawing(300,300,-5,5,-5,5,grid(1),circle(0,2,0.2),circle(2,0,0.2),circle(-2,0,0.2),circle(0,-2,0.2),line(2,0,0,2),red(line(0,2,-2,0)),green(line(-2,0,0,-2)),blue(line(0,-2,2,0)),graph(300,300,-5,5,-5,5,0))}}}
.
.
.
2. {{{xy=6}}} and {{{-xy=6}}}
.
.
.
{{{drawing(300,300,-10,10,-10,10,grid(1),graph(300,300,-10,10,-10,10,6/x),graph(300,300,-10,10,-10,10,-6/x))}}}
.
.
.
3. {{{x*abs(y)=4}}}
Since {{{abs(y)}}} is always positive, the function will only exist for {{{x>0}}} since {{{x<0}}} would give a negative value.
{{{drawing(300,300,0,10,-5,5,grid(1),graph(300,300,0,10,-5,5,-4/x),graph(300,300,0,10,-5,5,4/x))}}}
.
.
.
4. In this case, the function only provides positive {{{y}}} values. 
{{{abs(x)*y=10}}}
{{{y=10/abs(x)}}}
{{{drawing(300,300,-10,10,-10,10,grid(1),graph(300,300,-10,10,-10,10,10/abs(x)))}}}