document.write( "Question 1107223: Twice Betty’s age increased by three times Lorena’s age equals 65. Five times Betty’s age decreased by twice Lorena’s age equals 20.\r
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document.write( "What is Lorena’s age? \n" );
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Algebra.Com's Answer #722240 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Twice Betty’s age increased by three times Lorena’s age equals 65. \n" ); document.write( "2b + 3L = 65 \n" ); document.write( " Five times Betty’s age decreased by twice Lorena’s age equals 20. \n" ); document.write( "5b - 2L = 20 \n" ); document.write( "5b = 2L + 20 \n" ); document.write( "divide both sides by 5 \n" ); document.write( "b = .4L + 4 \n" ); document.write( " What is Lorena’s age? \n" ); document.write( "In the 1st equation, replace b with (.4L+4) \n" ); document.write( "2(.4L+4) + 3L = 65 \n" ); document.write( ".8L + 8 + 3L = 65 \n" ); document.write( ".8L + 3L = 65 - 8 \n" ); document.write( "3.8L = 57 \n" ); document.write( "L = 57/3.8 \n" ); document.write( "L = 15 yrs is Lorena's age\r \n" ); document.write( "\n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |