# Solving an equation with absolute values

The equation I am trying to solve is this : $\newcommand\abs[1]{|#1|}\abs{3y+7}=\abs{2y-1}$.

My conventional approach is to split this into three intervals with $1/2$ and $-7/3$ being the two "split" points.

My question is: is there a more "elegant" or quick way to solve this?

Thank you.

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$\pm(3y+7)=\pm(2y-1)$; then solve the various equations you get out of this. (perhaps not the most elegant way...) –  David Mitra Dec 31 '11 at 18:18
Sometimes it is useful to plot graphs for the functions. (For example if you tried to solve equation like $|x-1|+|x+1|=4$, the picture help you a lot). wolframalpha.com/input/… wolframalpha.com/input/?i=plot+%7Cx+%2B+1%7C%2B+%7Cx-1%7C –  Martin Sleziak Dec 31 '11 at 18:23

The equality holds iff $(3y+7)^2=(2y-1)^2$. Now we could expand, and use the Quadratic Formula. But note that $$(3y+7)^2-(2y-1)^2=[(3y+7)+(2y-1)][(3y+7)-(2y-1)].$$ So we need to solve the two linear equations $5y+6=0$ and $y+8=0$. Of course this is just a complicated rephrasing of David Mitra's simpler solution.

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Hint:

Apart from David Mitra's approach, another way is to square both sides and solve the resulting quadratic equation.

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MathMathCokie, you should be fine now since the details to my hint has been provided by Andre Nicolas. –  smanoos Dec 31 '11 at 18:30

It follows from the facts that $|a|$ is either $a$ or $-a$, and $|a|=|-a|$, that your equation is equivalent to $$\tag{1} \pm(3y+7)=\pm(2y-1).$$ Now $y$ is a solution of equation $(1)$ if and only if it is a solution of $$\tag{2} 3y+7=2y-1$$ or a solution of $$\tag{3} -(3y+7)=2y-1$$ (the other choices for signs in $(1)$ produce an equation that is either equivalent to equation $(2)$ or to equation $(3)$ ).

So, you need to solve equations $(2)$ and $(3)$. The union of their solution sets will be the solution set of the original equation.

Incidently, I think the other answers are more elegant...

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In fact your approach is generally the most efficient way to proceed, i.e. split the x-axis into intervals where the expressions have constant sign. The same idea generalizes to higher dimensions, yielding an algorithm to decide the truth of any first order multivariate statements about the reals - as Tarski showed by his celebrated method of quantifier elimination for real-closed fields. This has an algorithmic form that is essentially a higher-dimensional cylindrical version of the finite interval decomposition for the above one dimensional case. You can find some other simple examples here. For more google "cylindrical algebraic decomposition" or CAD algorithm.

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I am a new user so cant post images here.

Wolfram Mathematica plot |3y+7|=|2y-1|

Important features:

1. the horizontal axis of the graph shown is variable y

2. the vertical axis of the graph shown is variable x

Observe the curves:

x = +(3y+7)       eq (1)

x = -(3y+7)       eq (2)

x = +(2y-1)       eq (3)

x = -(2y-1)       eq (4)

x is always greater than 0 because of the mod function |x| application

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