solving the system

Question

solve the system : $$ y+|x-2|=3 $$, $$ |x+y|= m $$ graphicly when $m$ equals $6$.

I can easily (realtively) skecth the first graph , however, how the bloody hell do you sketch $|x+y|= 6$??

Answer 1

You look for pairs of numbers $x, y$ such that either $x+y = 6$ or $x + y = -6$. Those pairs lie on two diagonal lines, one of which passes through $(0, 6)$ and $(6, 0)$, and the other of which passes through $(0, -6)$ and $(-6, 0)$.

Answer 2

Look at the figure below, blue and green line shows the system $|x + y| = 6$ and the other two lines shows the other system. The black rectangles show the solution of the system. It is $ (2.5, 3.5)$ and $ (-2.5, -3.5)$.

Answer 3

Just think about what absolute value means.

$$\mid x+y\mid = 6$$ means that either $$x+y = 6$$ or $$x+y = -6$$ Just graph both of these functions as a single relation.

Answer 4

We have that \begin{align} |x+y| = 6 \implies x+y = 6\hspace{10px} \text{ or }\hspace{10px} x+y = -6. \end{align} From this we get that by isolating $y$ \begin{align} y = -x+6 \hspace{10px} \text{ or }\hspace{10px} y=-x-6. \end{align} You should be able to sketch those two graphs by hand.