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m signifies 'minus'

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gcd(gtl,gtj)

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old version if(factorise(abs(ktf))[len(factorise(abs(ktf)))-1]=0,1,primes[len(factorise(abs(ktf)))-1])

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t signifies 'times'

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Find the $x$ and $y$ values that satisfy both of the following equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify{{a}x+{b}}$               $(1)$
$y$$=$$\\simplify{{c}x+{d}}$               $(2)$
\n

$x=$ [[0]],   $y=$ [[1]]

\n

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There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify{{a}x+{b}}$               $(1)$
$y$$=$$\\simplify{{c}x+{d}}$               $(2)$
\n

Substitute the expression for $y$ given in $(1)$ into $(2)$:
\\[\\simplify{{a}x+{b} ={c}x+{d}}\\]

\n

Collect like terms:
\\[\\simplify{{a-c}x={d-b}}\\]

\n

Solve for $x$:
\\[x=\\var{xans}\\]

Now we know the $x$ value we can determine the corresponding $y$ value by substituting $x=\\var{xans}$ into either equation $(1)$ or $(2)$, below we substitute into $(1)$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify[!collectnumbers]{{a}({xans})+{b}}$
$=$$\\var{yans}$
\n

Therefore the values that satisfy equations $(1)$ and $(2)$ are $x=\\var{xans}$ and $y=\\var{yans}$.

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Solve the following simultaneous equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\simplify{{f}x+{g}y}$$=$$\\var{-h}$               $(3)$
$\\simplify{{j}x+{k}y}$$=$$\\var{-l}$               $(4)$
\n

$x=$ [[0]],   $y=$ [[1]]

\n

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There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\simplify{{f}x+{g}y}$$=$$\\var{-h}$               $(3)$
$\\simplify{{j}x+{k}y}$$=$$\\var{-l}$               $(4)$
\n

Solve one of the equations for one of the variables. Here we solve equation $(3)$ for $y$:

\n

\\begin{align}\\var{g}y&=\\simplify{{-h}-{f}x}\\\\y&=\\simplify{({-h}-{f}x)/({g})}\\quad(5)\\end{align}

\n

\\begin{align}y&=\\simplify{({-h}-{f}x)/({g})}\\quad(5)\\end{align}

\n

Substitute this expression for $y$ given in $(5)$ into $(4)$:

\n

\\[\\simplify[all,!collectnumbers]{{j}x+{k}(({-h}-{f}x)/({g})) = {-l}}\\]

\n

Collect like terms:
\\[\\simplify[fractionnumbers]{{j-f*k/g}x={-l+h*k/g}}\\]

\n

Solve for $x$:
\\[x=\\var{xbans}\\]

Now we know the $x$ value we can determine the corresponding $y$ value by substituting $x=\\var{xbans}$ into equation $(5)$:

\n

\\begin{align}y&=\\simplify[unitdenominator,!collectnumbers]{({-h}-{f}({xbans}))/({g})}\\\\&=\\var{ybans}\\end{align}

\n

Therefore the values that satisfy equations $(3)$ and $(4)$ are $x=\\var{xbans}$ and $y=\\var{ybans}$.

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Solve the following system of equations.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\simplify{{m}a+{n}b+{p}}$$=$$0$               $(5)$
$\\simplify{{q}a+{r}b+{s}}$$=$$0$               $(6)$
\n

$a=$ [[0]],   $b=$ [[1]]

\n

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There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\simplify{{m}a+{n}b+{p}}$$=$$0$               $(5)$
$\\simplify{{q}a+{r}b+{s}}$$=$$0$               $(6)$
\n

Solve one of the equations for one of the variables. Here we solve equation $(5)$ for $b$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{n}b$$=$$\\simplify{{-m}a+{-p}}$
 
$b$$=$$\\displaystyle{\\simplify{({-m}a+{-p})/({n})}}$               $(7)$
\n

Substitute this expression for $b$ given in $(7)$ into $(6)$:

\n

\\[\\simplify[all,!collectnumbers]{{q}a+{r}*(({-m}a+{-p})/{n}) + {s}=0}\\]

\n

Collect like terms:
\\[\\simplify[fractionnumbers]{{q-r*m/n}a={-s+r*p/n}}\\]

\n

Solve for $a$:
\\[a=\\simplify[fractionnumbers]{{xcans}}\\]

Now we know the $a$ value we can determine the corresponding $b$ value by substituting $a=\\simplify[fractionnumbers]{{xcans}}$ into equation $(7)$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$b$$=$$\\displaystyle{\\simplify[unitdenominator,!collectnumbers,fractionnumbers]{({-m}*({xcans})+{-p})/({n})}}$
 
$=$$\\simplify[fractionnumbers]{{ycans}}$
\n

Therefore the values that satisfy equations $(5)$ and $(6)$ are $a=\\simplify[fractionnumbers]{{xcans}}$ and $b=\\simplify[fractionnumbers]{{ycans}}$.

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