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Øving i å løse lineære likninger. Løsninger angis som heltall eller brøk.

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Solve linear equations with one unknown. Including brackets and fractions.

", "licence": "None specified"}, "statement": "

Løs likningene. Oppgi svarene som heltall eller brøk.

", "advice": "

Oppgave a)

\n

\\[\\simplify{{a}x+{b} = {c}} \\]

\n

Først samler vi alle  $x$-leddene på den ene siden av likningen.

\n

For å få det til må vi {add[0]} $\\var{abs(b)}$ {add[1]} begge sider:

\n

\\[\\simplify[all,!collectNumbers]{{a}x+{b} - {b} = {c} - {b}} \\]

\n

\\[\\var{a}x  = \\simplify{{c-b}} \\]

\n

For å få $x$ alene deler vi på koeffisienten til $x$ (tallet foran $x$) på begge sider.

\n

Vi deler på $\\var{a}$:

\n

\\[  x = {\\simplify{({c}-{b})/{a}}} \\]

\n

\n

Oppgave b)

\n

 \\[ \\frac{\\simplify{{d}x + {f}}}{\\var{g}} = \\var{h} \\]

Først vil vi kvitte oss med brøken på venstre side.  Det gjør vi ved å multiplisere begge sidene med $\\var{g}$:

\\[ \\begin{split} \\frac{\\simplify{{d}x + {f}}}{\\var{g}} \\cdot \\var{g}  &= \\var{h} \\cdot \\var{g} \\\\\\\\ \\simplify{{d}x + {f}} &= \\var{h*g} \\end{split} \\]

Det neste vi gjør er å samle $x$-leddene på den ene siden av likningen.

Vi må da {add2[0]} $\\var{abs(f)}$ {add2[1]} begge sider:

\\[ \\begin{split} \\var{d}x &= \\simplify[]{{h*g}-{f}} \\\\ &= \\simplify[]{{h*g-f}} \\end{split}\\]

Til slutt må vi dele på $\\var{d}$ for å få $x$ alene på venstre siden.

\n


\\[x = \\simplify[fractionNumbers]{{(h*g-f)/d}}  \\]

\n

\n

Oppgave c)

\n

\\[ \\simplify{{b}({c}x+{g})} = \\var{d} \\]

\n

I oppgave b) var $x$-uttrykket delt på $\\var{g}$, og vi måtte da multiplisere for å få bort nevneren.  Her er $x$-uttrykket multiplisert med et tall, og da må vi dividere for å bli kvitt det-
Vi deler begge sider på $\\var{b}$:

\\[ \\begin{split} \\frac{\\simplify{{b}({c}x+{g})}}{ \\var{b}} &= \\frac{\\var{d}}{\\var{b}} \\\\ \\\\ \\simplify{{c}x+{g}} &= \\simplify[fractionNumbers]{{d/b}} \\end{split} \\]


Det neste vi gjør er å samle $x$-leddene på den ene siden av likningen.

Vi må da {add3[0]} $\\var{abs(g)}$ {add3[1]} begge sider:

\n

\\[ \\begin {split} \\simplify[all,!collectNumbers]{{c}x + {g} - {g}} &= \\simplify[fractionNumbers,!cancelTerms]{{d/b} - {g}}\\\\ \\var{c}x &= \\simplify[fractionNumbers]{{d/b-g}} \\end{split} \\]

Til slutt må vi dele på $\\var{c}$for å få $x$ alene på venstre siden.
\\[ x =  \\simplify[fractionNumbers]{{(d/b-g)/c}} \\]

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$\\simplify{{a}x+{b} = {c}}$

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\n

$x=$ [[0]]

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$\\dfrac{\\simplify{{d}x + {f}}}{\\var{g}} = \\var{h}$

\n

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$x=$ [[0]]

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$\\simplify{{b}({c}x+{g})} = \\var{d}$

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\n

$x=$ [[0]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Løs likningene. Oppgi svaret som et helt tall eller en brøk.

", "advice": "

Oppgave a)

\n

I $\\simplify{{l}({m}w-{n}) = {p}w+{q}}$, må vi først løse opp parentesene, deretter samler vi alle $w$'ene på venstre side og alle tallene på høyre side. Til slutt deler vi på tallet foran $w$ (koeffisienten til $w$) for å få $w$ alene.

\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
$\\simplify{{l}({m}w-{n})}$$=$$\\simplify{{p}w+{q}}$ 
 
$\\simplify{{l*m}w-{n*l}}$$=$$\\simplify{{p}w+{q}}$
 
$\\simplify[!cancelTerms,unitFactor]{{l*m}w-{n*l}-{p}w}$$=$$\\simplify[!cancelTerms,unitFactor]{{p}w+{q}-{p}w}$(trekker fra $\\var{p}w$ på begge sider)
 
$\\simplify{{l*m-p}w-{n*l}}$$=$$\\var{q}$
 
$\\var{l*m-p}w-\\var{n*l}+\\var{n*l}$$=$$\\var{q}+\\var{n*l}$(legger til $\\var{n*l}$ på begge sider)
 
$\\var{l*m-p}w$$=$$\\var{q+n*l}$
 
$\\displaystyle{\\frac{\\var{l*m-p}w}{\\var{l*m-p}}}$$=$$\\displaystyle{\\frac{\\var{q+n*l}}{\\var{l*m-p}}}$(deler på $\\var{l*m-p}$ på begge sider)
 
$w$$=$$\\displaystyle{\\simplify{{q+n*l}/{l*m-p}}} $
\n

\n

Oppgave b)

\n

I uttrykket $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}=\\var{g}$ må vi multiplisere med $(y-\\var{f})$ for å bli kvitt brøken, deretter får vi alle $y$'ene på ene siden og tallene på andre siden, og til slutt deler vi på koeffisienten til $y$ for å få $y$ alene..

\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
$\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}$$=$$\\var{g}$ 
 
$\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}\\cdot(y-\\var{f})$$=$$\\var{g}\\cdot (y-\\var{f})$(multipliserer med (y-\\var{f}) )
 
$\\var{d}y$$=$$\\simplify[unitFactor]{{g}y+{-g*f}}$
 
$\\simplify[!cancelTerms,unitFactor]{{d}y+{-g}y}$ $=$$\\simplify[!cancelTerms,unitFactor]{{-g*f}}$(legger til $\\simplify{{-g}y}$ på begge sider
 
$\\var{d-g}y$$=$$\\var{-g*f}$
 
$\\displaystyle{\\frac{\\var{d-g}y}{\\var{d-g}}}$$=$$\\displaystyle{\\frac{\\var{-g*f}}{\\var{d-g}}}$(deler på $\\var{d-g}$ på begge sider)
 
 $y$$=$$\\displaystyle{\\simplify{{-g*f}/{d-g}}}$
\n

Oppgave c)

\n

I uttrykket $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}=\\var{right}}$ må vi multiplisere begge sider med $\\var{denom1}$ og by $\\var{denom2}$ for å bli kvitt alle brøkene, deretter får vi alle $x$'ene på ene siden og tallene på andre siden,og til slutt deler vi begge sidene på koeffisienten til $x$ for å få $x$ alene.

\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\n\n\n\n\n\n\n\n\n\n\n\n
$\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}}$$=$$\\var{right}$ 
 
$\\displaystyle{\\left(\\frac{x+\\var{add}}{\\var{denom1}}\\right)\\cdot\\var{denom1}+\\left(\\frac{x}{\\var{denom2}}\\right)\\cdot\\var{denom1}}$$=$$\\var{right}\\cdot \\var{denom1}$(multipliserer alle ledd med $\\var{denom1}$)
 
$\\displaystyle{x+\\var{add}+\\frac{\\var{denom1}x}{\\var{denom2}}}$$=$$\\var{r1}$
 
$\\displaystyle{(x+\\var{add})\\cdot\\var{denom2}+\\left(\\frac{\\var{denom1}x}{\\var{denom2}}\\right)\\cdot\\var{denom2}}$ $=$$\\var{r1}\\cdot\\var{denom2}$(multipliserer alle ledd med $\\var{denom2}$)
 
$\\displaystyle{\\var{denom2}x+\\var{a2}+\\var{denom1}x}$$=$$\\var{r12}$
 
$\\var{sumdeno}x+\\var{a2}$$=$$\\var{r12}$(samler ledd av samme type)
 
 $\\var{sumdeno}x$$=$$\\var{r12}-\\var{a2}$(samler ledd av samme type)
 
$\\var{sumdeno}x$$=$$\\var{top}$
 
$x$$=$$\\displaystyle{\\simplify{{top}/({sumdeno})}} $(deler med koeffisienten til $x$)
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Løs  $\\simplify{{l}({m}w-{n}) = {p}w+{q}}$

\n

$w=$ [[0]]

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Gitt $\\displaystyle{\\frac{\\var{d}y}{y-\\var{f}}}=\\var{g}$,  

\n

$y=$ [[0]].

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Løs $\\displaystyle{\\frac{x+\\var{add}}{\\var{denom1}}+\\frac{x}{\\var{denom2}}=\\var{right}}$.

\n

$x=$ [[0]]

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