document.write( "Question 1204153: EF is bisected by AC at D. Find AC if EF = 2x + 22, AC = 3x-5, and DE = 2x-4. \n" ); document.write( "
Algebra.Com's Answer #840210 by mananth(16946)\"\" \"About 
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DE + DF = EF\r
\n" ); document.write( "\n" ); document.write( "DE=DF\r
\n" ); document.write( "\n" ); document.write( "2x - 4 + 2x - 4 = 2x + 22\r
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\n" ); document.write( "\n" ); document.write( "4x - 8 = 2x + 22\r
\n" ); document.write( "\n" ); document.write( "2x - 8 = 22\r
\n" ); document.write( "\n" ); document.write( "2x = 30\r
\n" ); document.write( "\n" ); document.write( "(2x)/2 = 30/2\r
\n" ); document.write( "\n" ); document.write( "x = 15\r
\n" ); document.write( "\n" ); document.write( "we can find the value of AC:\r
\n" ); document.write( "\n" ); document.write( "AC = 3x - 5
\n" ); document.write( "AC = 3(15) - 5
\n" ); document.write( "AC = 45 - 5
\n" ); document.write( "AC = 40\r
\n" ); document.write( "\n" ); document.write( "So, AC = 40.\r
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