Inequality involving a single absolute value, question solution uses the piecewise nature of the absolute value function.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "You are given the inequality \\[|\\simplify{{a}x+{b}}| \\var{latex(sym)} \\var{c}.\\]
\n", "advice": "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:
\n\\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}
\n\\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}.\\end{alignat}
\n\\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}
\n\n\n
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}\\]
\nThis 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:
\n\nCase 1: $\\simplify{{a}x+{b}}\\ge 0$
\nRearranging $\\simplify{{a}x+{b}}\\ge 0$ for $x$ gives $x \\ge \\simplify[fractionNumbers]{{-b/a}}$ $x \\le \\simplify[fractionNumbers]{{-b/a}}$ and so case 1 is only relevant for $x \\ge \\simplify[fractionNumbers]{{-b/a}}$ $x \\le \\simplify[fractionNumbers]{{-b/a}}$.
\nIn case 1, our inequality is simply $\\simplify{{a}x+{b}} \\var{latex(sym)} \\var{c}$, and so rearranging it for $x$ gives $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,!unitDenominator]{{(c-b)/a}}$ $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.
\nSo 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 $\\simplify[fractionNumbers]{-{b/a}}\\le x\\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}\\var{latex(sym)} x\\le \\simplify[fractionNumbers]{{-b/a}}$ $x\\var{latex(sym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $x\\var{latex(backsym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.
\n\nCase 2: $\\simplify{{a}x+{b}}< 0$
\nRearranging $\\simplify{{a}x+{b}}< 0$ for $x$ gives $x < \\simplify[fractionNumbers]{{-b/a}}$ $x > \\simplify[fractionNumbers]{{-b/a}}$ and so case 2 is only relevant for $x < \\simplify[fractionNumbers]{{-b/a}}$ $x > \\simplify[fractionNumbers]{{-b/a}}$.
\nIn case 2, our inequality is actually $-(\\simplify{{a}x+{b}}) \\var{latex(sym)} \\var{c}$, and so rearranging it for $x$ gives $x \\var{latex(backsym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.
\nSo 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 $\\simplify[fractionNumbers]{{-(b+c)/a}}\\var{latex(sym)} x< \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b}/{a}}$ $\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b/a}}< x \\var{latex(sym)} \\simplify[fractionNumbers,simplifyFractions,unitDenominator]{-{(b+c)/a}}$ $x\\var{latex(backsym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x\\var{latex(sym)}\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.
\n\nTherefore the solution to our original inequality $|\\simplify{{a}x+{b}}| \\var{latex(sym)} \\var{c}$ is
\n\\[\\var[fractionNumbers]{r1}<x<\\var[fractionNumbers]{r2}.\\]
\n\\[\\var[fractionNumbers]{r1}\\le x \\le\\var[fractionNumbers]{r2}.\\]
\n\\[x< \\var[fractionNumbers]{r1} \\text{ or } x> \\var[fractionNumbers]{r2}. \\]
\n\\[x\\le \\var[fractionNumbers]{r1} \\text{ or } x\\ge\\var[fractionNumbers]{r2}. \\]
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\nNote: To type $\\ge$ use >= and for $\\le$ use <=.
Note: If your answer consists of two infinite intervals, such as $x<-1, \\,x>5$, type your inequalities with the word or between them, such as x<-1 or x>5, rather than using a comma between them.