I have the following system:
$$ \left\{\begin{matrix} |x+1|+|y-1|=5\\ |x+1|-4y=-4 \end{matrix}\right. $$ I do not know how to handle the absolute value expressions. What can I do?
How can I solve a linear system with absolute values involved?
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systems-of-equations
absolute-value
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0The rule of thumb on this site is that you convince people you have tried to answer your question by showing what you already have. What you already have? Have you tried going over the four cases generated by $x\lessgtr-1$ and $y\lessgtr1$? – 2017-02-03
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0Absolute value? Then it is *not* a linear system. You can though write down several equivalent linear systems... – 2017-02-03
1 Answers
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$$\left\{\begin{matrix} |x+1|+|y-1|=5\\ |x+1|-4y=-4 \end{matrix}\right.$$ from second one $|x+1|=4y-4$ put in first $$4y-4+|y-1|=5\\4y+|y-1|=9 \\4y+|y-1|=9 \to \begin{cases}y\geq 1 \to 4y+y-1=9 \to y=2\checkmark\\y < 1 \to 4y-y+1=9 \to y=\frac83 (not-acceptable) \end{cases}$$
then put $y$ to find $x$ $$|x+1|=4(2)-4=4 \to x=3,-5$$ this mean ,two points $(-5,2),(+3,3)$
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0In your third line from the last one you have a $\;\le\;$ sign where you should probably want a $\;<\;$ one. Nice+1 – 2017-02-03
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1The $\leq$ sign is actually okay, since the absolute value can be defined as $abs(x) = -x$ for $x \leq 0$. – 2017-02-03
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0@khosrotash Why is not acceptable the second $y$ value? – 2017-02-03
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1@JoshuaSalazar: because ,first assume that $y<1 $ then find $y=\dfrac83 \\ \dfrac83 >1$ – 2017-02-03