document.write( "Question 1159602: A+B=10
\n" ); document.write( "C/D=4
\n" ); document.write( "A+C=3
\n" ); document.write( "B+D=9
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Algebra.Com's Answer #782655 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
You should say what kind of method you need to use.\r
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\n" ); document.write( "\n" ); document.write( "\"system%28A%2BB=10%2CC=4D%2CA%2BC=3%2CB%2BD=9%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"system%28A%2BB=10%2CA%2B4D=3%2CB%2BD=9%29\", \"A=10-B\";\r
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\n" ); document.write( "\n" ); document.write( "\"system%2810-B%2B4D=3%2CB%2BD=9%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"system%28-B%2B4D=-7%2CB%2BD=9%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"system%28B-4D=7%2CB%2BD=9%29\"-----------this much by just a couple of substitutions. You can solve this for B and D, right? You can then finish on your own for A and C?
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