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Given that $x>\\var{left_ba}$ and $x<\\var{right_ba}$, we can write [[0]] $<x<$ [[1]].

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$\\var{left_ba} <x< \\var{right_ba}$ means $\\var{left_ba} <x$ and $x< \\var{right_ba}$. You could read this as \"$x$ is between $\\var{left_ba}$ and $\\var{right_ba}$\".

Given $\\var{left_bb}<\\frac{x}{\\var{c}}<\\var{right_bb}$, solving for $x$ gives [[0]]$<x<$ [[1]].

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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

$\\var{left_bb}$	$<$	$\\displaystyle \\frac{x}{\\var{c}}$	$<$	$\\var{right_bb}$

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

$\\var{bleft}$	$<$	$x$	$<$	$\\var{bright}$

Given $\\var{left_bc}<\\frac{\\simplify{{a}x+{b}}}{\\var{c}}<\\var{right_bc}$, solving for $x$ gives [[0]]$<x<$ [[1]].

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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

$\\var{left_bc}$	$<$	$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}$	$<$	$\\var{right_bc}$

$\\var{left_bc}\\times \\var{c}$	$<$	$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}\\times \\var{c}$	$<$	$\\var{right_bc}\\times \\var{c}$

$\\var{left_bc*c}$	$<$	$\\simplify{{a}x+{b}}$	$<$	$\\var{right_bc*c}$

$\\simplify[basic]{{left_bc*c}-{b}}$	$<$	$\\simplify[basic]{{a}x+{b}-{b}}$	$<$	$\\simplify[basic]{{right_bc*c}-{b}}$

$\\var{left_bc*c-b}$	$<$	$\\var{a}x$	$<$	$\\var{right_bc*c-b}$

$\\displaystyle\\frac{\\var{left_bc*c-b}}{\\var{a}}$	$<$	$\\displaystyle\\frac{\\var{a}x}{\\var{a}}$	$<$	$\\displaystyle\\frac{\\var{right_bc*c-b}}{\\var{a}}$

$\\displaystyle\\simplify{{left_bc*c-b}/{a}}$	$<$	$x$	$<$	$\\displaystyle\\simplify{{right_bc*c-b}/{a}}$

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