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Der svaret ikke er et helt tall skal du svare med brøk. Eksempel: Bruk 1/2 i stedet for 0,5, 3/2 i stedet for 1,5 osv.

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Gitt at $x>\\var{left_ba}$ og $x<\\var{right_ba}$. Det kan skrives som [[0]] $<x<$ [[1]].

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$\\var{left_ba} <x< \\var{right_ba}$ betyr at $\\var{left_ba} <x$ og $x< \\var{right_ba}$. Det kan leses som \"$x$ er mellom $\\var{left_ba}$ og $\\var{right_ba}$\".

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Gitt ulikheten $\\var{left_bb}<\\frac{x}{\\var{c}}<\\var{right_bb}$. Løser vi den for $x$ får vi [[0]]$<x<$ [[1]].

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Vi løser på samme måte som enkle ulikheter, men pass på å gjøre det samme på alle sider av ulikhetstegnene. Målet er å få $x$ alene i midten.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{left_bb}$$<$$\\displaystyle \\frac{x}{\\var{c}}$$<$$\\var{right_bb}$
 
$\\var{left_bb}\\times \\var{c}$$<$$\\displaystyle \\frac{x}{\\var{c}}\\times \\var{c}$$<$$\\var{right_bb}\\times \\var{c}$
 
$\\var{bleft}$$<$$x$$<$$\\var{bright}$
\n

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Gitt $\\var{left_bc}<\\frac{\\simplify{{a}x+{b}}}{\\var{c}}<\\var{right_bc}$. Løsning for $x$ gir [[0]]$<x<$ [[1]].

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Vi løser på samme måte som enkle ulikheter, men pass på å gjøre det samme på alle sider av ulikhetstegnene. Målet er å få $x$ alene i midten.Solving inequalities is similar to solving equations, ensure you do the same thing to all sides. Note that the operations we do are to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{left_bc}$$<$$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}$$<$$\\var{right_bc}$
  
$\\var{left_bc}\\times \\var{c}$$<$$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}\\times \\var{c}$$<$$\\var{right_bc}\\times \\var{c}$
  
$\\var{left_bc*c}$$<$$\\simplify{{a}x+{b}}$$<$$\\var{right_bc*c}$
 
$\\simplify[basic]{{left_bc*c}-{b}}$$<$$\\simplify[basic]{{a}x+{b}-{b}}$ $<$$\\simplify[basic]{{right_bc*c}-{b}}$
 
$\\var{left_bc*c-b}$$<$$\\var{a}x$$<$$\\var{right_bc*c-b}$
 
$\\displaystyle\\frac{\\var{left_bc*c-b}}{\\var{a}}$$<$$\\displaystyle\\frac{\\var{a}x}{\\var{a}}$$<$$\\displaystyle\\frac{\\var{right_bc*c-b}}{\\var{a}}$
 
$\\displaystyle\\simplify{{left_bc*c-b}/{a}}$$<$$x$$<$$\\displaystyle\\simplify{{right_bc*c-b}/{a}}$
\n

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