document.write( "Question 1126805: one number is seven more than twice another. if their sum is decrease by ten the result is twelve find the numbers \n" ); document.write( "

Algebra.Com's Answer #743121 by MathLover1(20850) $\"\"$ $\"About$

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\n" ); document.write( "let one number be $\"x\"$ , and other number be $\"y\"$ \r
\n" ); document.write( "\n" ); document.write( "if $\"x\"$ is $\"seven\"$ $\"+more\"$ than $\"twice\"$ $\"y\"$ , we have\r
\n" ); document.write( "\n" ); document.write( " $\"x=2y%2B7\"$ .......eq.1\r
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\n" ); document.write( "\n" ); document.write( " if their $\"sum\"$ is $\"decrease\"$ by $\"ten\"$ the $\"result\"$ is $\"twelve\"$ ; so, we have \r
\n" ); document.write( "\n" ); document.write( " $\"x%2By-10=12\"$ ....eq.2 \r
\n" ); document.write( "\n" ); document.write( "substitute $\"x\"$ from eq.1\r
\n" ); document.write( "\n" ); document.write( " $\"2y%2B7%2By-10=12\"$ ....eq.2 ...solve or $\"y\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"3y-3=12\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"3y=12%2B3\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"3y=15\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"y=15%2F3\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"y=5\"$ \r
\n" ); document.write( "\n" ); document.write( "now find $\"x\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x=2y%2B7\"$ .......eq.1\r
\n" ); document.write( "\n" ); document.write( " $\"x=2%2A5%2B7\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x=10%2B7\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"x=17\"$ \r
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\n" ); document.write( "\n" ); document.write( "so, one number is $\"17\"$ , and the other number is $\"5\"$ \r
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