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You are given the inequality \\[|\\simplify{{a}x+{b}}| \\var{latex(sym)} \\var{c}.\\]

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approx of the furthest we need to plot in the y direction

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Inequality involving a single absolute value, question solution uses the piecewise nature of the absolute value function.

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one interval.

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two intervals.

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≤

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≥

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The solution to the above inequality

is made up of [[0]]

Given by the condition [[1]] [[2]] $x$ [[3]] [[4]].

Given by the condition, $x$ [[5]] [[6]] or $x$ [[7]] [[8]].

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Short-cut Method

For questions of this type (where the absolute value is 'less than' or 'less than and equal to' something positive) we can set out our working in a shorter way:

\\begin{alignat}{2} &&|\\simplify{{a}x+{b}}| &\\var{latex(sym)} \\var{c}\\\\ \\var{-c} &\\var{latex(sym)}&\\simplify{{a}x+{b}}&\\var{latex(sym)}\\var{c}\\\\ \\var{-c-b} &\\var{latex(sym)} &\\simplify{{a}x}\\quad&\\var{latex(sym)}\\var{c-b}\\\\ \\var[fractionNumbers]{r1}&\\var{latex(sym)}&x\\quad&\\var{latex(sym)}\\var[fractionNumbers]{r2}.\\end{alignat}

\\begin{alignat}{2} &&|\\simplify{{a}x+{b}}| &\\var{latex(sym)} \\var{c}\\\\ \\var{-c} &\\var{latex(sym)}&\\simplify{{a}x+{b}}&\\var{latex(sym)}\\var{c}\\\\ \\var{-c-b} &\\var{latex(sym)} &\\simplify{{a}x}\\quad&\\var{latex(sym)}\\var{c-b}\\\\ \\var[fractionNumbers]{r2}&\\var{latex(backsym)}&x\\quad&\\var{latex(backsym)}\\var[fractionNumbers]{r1}\\\\\\var[fractionNumbers]{r1}&\\var{latex(sym)}&x\\quad&\\var{latex(sym)}\\var[fractionNumbers]{r2}.\\end{alignat}

Cases Method

Recall that the absolute value is defined as the piecewise function \\[|x|=\\begin{cases}x, &\\text{ for } x\\ge 0\\\\ -x, &\\text{ for } x<0. \\end{cases}\\]

This means that $|\\simplify{{a}x+{b}}|$ is actually $\\simplify{{a}x+{b}}$ when $\\simplify{{a}x+{b}}\\ge 0$ but it is $-(\\simplify{{a}x+{b}})$ when $\\simplify{{a}x+{b}}<0$. We have two cases:

Case 1: $\\simplify{{a}x+{b}}\\ge 0$

Rearranging $\\simplify{{a}x+{b}}\\ge 0$ for $x$ gives $x \\ge \\simplify[fractionNumbers]{{-b/a}}$ $x \\le \\simplify[fractionNumbers]{{-b/a}}$ and so case 1 is only relevant for $x \\ge \\simplify[fractionNumbers]{{-b/a}}$ $x \\le \\simplify[fractionNumbers]{{-b/a}}$.

In case 1, our inequality is simply $\\simplify{{a}x+{b}} \\var{latex(sym)} \\var{c}$, and so rearranging it for $x$ gives $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,!unitDenominator]{{(c-b)/a}}$ $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.

So in conclusion for case 1, we require that $x \\ge \\simplify[fractionNumbers]{{-b/a}}$ $x \\le \\simplify[fractionNumbers]{{-b/a}}$ and $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ which is the same as $\\simplify[fractionNumbers]{-{b/a}}\\le x\\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}\\var{latex(sym)} x\\le \\simplify[fractionNumbers]{{-b/a}}$ $x\\var{latex(sym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $x\\var{latex(backsym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.

Case 2: $\\simplify{{a}x+{b}}< 0$

Rearranging $\\simplify{{a}x+{b}}< 0$ for $x$ gives $x < \\simplify[fractionNumbers]{{-b/a}}$ $x > \\simplify[fractionNumbers]{{-b/a}}$ and so case 2 is only relevant for $x < \\simplify[fractionNumbers]{{-b/a}}$ $x > \\simplify[fractionNumbers]{{-b/a}}$.

In case 2, our inequality is actually $-(\\simplify{{a}x+{b}}) \\var{latex(sym)} \\var{c}$, and so rearranging it for $x$ gives $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.

So in conclusion for case 2, we require that $x < \\simplify[fractionNumbers]{-{b/a}}$ $x > \\simplify[fractionNumbers]{-{b/a}}$ and $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ which is the same as $\\simplify[fractionNumbers]{{-(b+c)/a}}\\var{latex(sym)} x< \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b}/{a}}$ $\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b/a}}< x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{-{(b+c)/a}}$ $x\\var{latex(backsym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x\\var{latex(sym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.

Therefore the solution to our original inequality $|\\simplify{{a}x+{b}}| \\var{latex(sym)} \\var{c}$ is

\\[\\var[fractionNumbers]{r1}<x<\\var[fractionNumbers]{r2}.\\]

\\[\\var[fractionNumbers]{r1}\\le x \\le\\var[fractionNumbers]{r2}.\\]

\\[x< \\var[fractionNumbers]{r1}, \\, x> \\var[fractionNumbers]{r2}. \\]

\\[x\\le \\var[fractionNumbers]{r1}, \\, x\\ge\\var[fractionNumbers]{r2}. \\]

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