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I have a problem graphing the following functions. How can I express my equations as "$y$ is a function of $x$" and take $x$ value to get the $y$ value?

$1.\,|x|+|y| =1$

$2.\,|x+y| = 1$

Thank you

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First, they are not functions. The graphs you make will fail the vertical line test. For (most) values of $x$ (in the domain) there will be two values of $y$ that make the equation true.

The way to graph them is to unpack the absolute value signs. The second is easier because it has only one absolute value sign. It says $x+y=1$ or $x+y=-1$. Each of those is a line in the plane and your graph is the union of those two lines.

For the first you have two absolute value signs, so $2^2=4$ possibilities to unpack them. One is $x+y=1, x\ge 0, y\ge 0$. This is a line segment in the first quadrant. The other three pieces will be one in each quadrant. In each quadrant you know the sign of $x,y$ so can decide whether the equation should involve them with a $+$ sign or $-$ sign. Can you carry on?

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For $1$: $$\vert x\vert +\vert y\vert=1 \Leftrightarrow \vert y\vert=1-\vert x\vert$$ Now check for what values $y$ is positive and negative, $$y=\begin{cases} 1-\vert x\vert, & \text{if $y\geq0$} \\ \vert x\vert-1, & \text{if $y<0$} \end{cases}$$ I'll let you work out the rest (checking the values for $x$).

For $2$ same logic applies: $$y=\begin{cases} -x-1, & \text{if $y<-x$} \\ 1-x, & \text{if $y>-x$} \end{cases}$$

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