document.write( "Question 481338: Four family members Tom, Melanie, Bob and Irene have a collective age of 150 years. Melanie is one third Tom’s age and Bob’s age is twice Melanie’s. Irene’s age is as much as Tom’s and Melanie’s put together. What is each person age? Solve by any convenient algebraic method. How would you check your solution?\r
\n" ); document.write( "\n" ); document.write( "I've got the answer: tom=45, melanie=15, bob=30, irene=60 but i could not show the steps or solution. will you do it for me?
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Algebra.Com's Answer #329630 by lwsshak3(11628)\"\" \"About 
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Four family members Tom, Melanie, Bob and Irene have a collective age of 150 years. Melanie is one third Tom’s age and Bob’s age is twice Melanie’s. Irene’s age is as much as Tom’s and Melanie’s put together. What is each person age? Solve by any convenient algebraic method. How would you check your solution?
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\n" ); document.write( "let x=Tom's age
\n" ); document.write( "x/3 =Melanie's age (1/3 Tom's age)
\n" ); document.write( "2x/3=Bob's age (twice Melanie's age)
\n" ); document.write( "(x+x/3) =Irene's age (Tom and Melanie ages)
\n" ); document.write( "..
\n" ); document.write( "x+x/3+2x/3+x+x/3=150
\n" ); document.write( "3x+x/3=150
\n" ); document.write( "9x+x=450
\n" ); document.write( "10x=450
\n" ); document.write( "x=45 (Tom's age)
\n" ); document.write( "45/3=15 (Melanie's age)
\n" ); document.write( "2*15=30 (Bob's age)
\n" ); document.write( "45+15=60 (Irene's age)
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