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-rw-r--r--Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex12
1 files changed, 6 insertions, 6 deletions
diff --git a/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex b/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex
index ca72d2d3660..7059840ea97 100644
--- a/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex
+++ b/Master/texmf-dist/doc/support/ketcindy/source/ketmanual/Fig/mxtex02.tex
@@ -5,14 +5,14 @@
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\special{pn 8}%
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-\put(0.0500,5.0000){\hspace*{\Width}\raisebox{\Height}{部分分数への分解 $\frac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\frac{8}{x+2}-\frac{1}{x+1}+x-3$}}%
+\settowidth{\Width}{Decomposition into partial fractions}\setlength{\Width}{0\Width}%
+\settoheight{\Height}{Decomposition into partial fractions}\settodepth{\Depth}{Decomposition into partial fractions}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}%
+\put(0.0500,5.0000){\hspace*{\Width}\raisebox{\Height}{Decomposition into partial fractions}}%
%
%
-\settowidth{\Width}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}%
-\settoheight{\Height}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{部分分数への分解 $\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}%
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+\settowidth{\Width}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Width}{0\Width}%
+\settoheight{\Height}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\settodepth{\Depth}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}\setlength{\Height}{-0.5\Height}\setlength{\Depth}{0.5\Depth}\addtolength{\Height}{\Depth}%
+\put(0.0500,3.5000){\hspace*{\Width}\raisebox{\Height}{$\dfrac{x^3}{\left(x+1\right)\,\left(x+2\right)}=\dfrac{8}{x+2}-\dfrac{1}{x+1}+x-3$}}%
%
%
\end{picture}}% \ No newline at end of file