document.write( "Question 1205317: mo bisects < amb , m< amo = 7x-3 and m< bmo = 2x+12 solve for x and find m < amb
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Algebra.Com's Answer #842030 by MathLover1(20849) $\"\"$ $\"About$

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\n" ); document.write( "mo bisects < $\"amb\"$ \r
\n" ); document.write( "\n" ); document.write( "to bisect an angle means to divide it in two $\"equal\"$ parts\r
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\n" ); document.write( "\n" ); document.write( "so, $\"m\"$ < $\"amo\"$ = $\"m\"$ < $\"bmo\"$ and $\"m\"$ < $\"amb\"$ = $\"m\"$ < $\"amo\"$ + $\"m\"$ < $\"bmo+\"$ \r
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\n" ); document.write( "\n" ); document.write( "if \r
\n" ); document.write( "\n" ); document.write( " $\"m\"$ < $\"amo+=+7x-3+\"$ and $\"m\"$ < $\"bmo+=+2x%2B12\"$ , we have\r
\n" ); document.write( "\n" ); document.write( " $\"7x-3=+2x%2B12\"$ ......solve for $\"x+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"7x-2x=+3%2B12\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"5x=+15\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x=3\"$ \r
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\n" ); document.write( "\n" ); document.write( "then\r
\n" ); document.write( "\n" ); document.write( " $\"m\"$ < $\"amo+=+7x-3+=7%2A3-3=21-3=18\"$ \r
\n" ); document.write( "\n" ); document.write( "and $\"m\"$ < $\"bmo+=+2x%2B12=2%2A3%2B12=18\"$ \r
\n" ); document.write( "\n" ); document.write( "then\r
\n" ); document.write( "\n" ); document.write( " $\"m\"$ < $\"amb=18%2B18=36\"$ \r
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