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Given $x+\\var{a}=\\var{b}$, we subtract $\\var{a}$ from both sides to get $x$ by itself.

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$x+\\var{a}$$=$$\\var{b}$
$x+\\var{a}-\\var{a}$$=$$\\var{b}-\\var{a}$
$x$$=$$\\var{ans1}$
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Given $x+\\var{a}=\\var{b}$, we can rearrange the equation to that find $x=$ [[0]].

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Given $y-\\var{c}=\\var{d}$, we add $\\var{c}$ to both sides to get $y$ by itself.

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$y-\\var{c}$$=$$\\var{d}$
$y-\\var{c}+\\var{c}$$=$$\\var{d}+\\var{c}$
$y$$=$$\\var{ans2}$
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Given $y-\\var{c}=\\var{d}$,  $y=$ [[0]].

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Given $\\var{f}+z=\\var{g}$, we subtract $\\var{f}$ from both sides to get $z$ by itself.

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$\\var{f}+z$$=$$\\var{g}$
$\\simplify[basic]{{f}+z-{f}}$$=$$\\simplify[basic]{{g}-{f}}$
$z$$=$$\\var{ans3}$
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Rearrange $\\var{f}+z=\\var{g}$ to determine the value of $z$.

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$z=$ [[0]]

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Given $\\var{h}=\\var{j}+a$, we subtract $\\var{j}$ from both sides to get $a$ by itself.

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$\\var{h}$$=$$\\var{j}+a$
$\\simplify[basic]{{h}-{j}}$$=$$\\simplify[basic]{{j}+a-{j}}$
$\\var{ans4}$$=$$a$
$a$$=$$\\var{ans4}$
\n

\n

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Solve $\\var{h}=\\var{j}+a$ for $a$.

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$a=$ [[0]]

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