document.write( "Question 386758: Mona's age is 4 times Bailey's age. The sum of their ages is 15. Find the age of each. \n" ); document.write( "
Algebra.Com's Answer #273359 by gwendolyn(128)![]() ![]() You can put this solution on YOUR website! First, we'll assign variables to both ages. \n" ); document.write( "let M=Mona's age \n" ); document.write( "let B=Bailey's age\r \n" ); document.write( "\n" ); document.write( "Then, we'll come up with two equations to solve our two variables. \n" ); document.write( "Mona's age is 4 times Bailey's age. This can be shown algebraically as below: \n" ); document.write( "M=4B\r \n" ); document.write( "\n" ); document.write( "We also know that Mona's age plus Bailey's age is 15. This, in an algebraic statement, can be written as below: \n" ); document.write( "M+B=15 \n" ); document.write( "We can substitute the value of M in terms of B from the first equation into the second equation: \n" ); document.write( "4B+B=15 \n" ); document.write( "5B=15 \n" ); document.write( "Then, we can divide both sides of the w equation by 5: \n" ); document.write( " \n" ); document.write( "B=3\r \n" ); document.write( "\n" ); document.write( "Therefore, Bailey is 3 years old. We also want to know Mona's age, so we can simply plug in Bailey's age to either equation and find it. \n" ); document.write( "M+B=15 \n" ); document.write( "M+3=15 \n" ); document.write( "M=12 \n" ); document.write( "Therefore, Mona is 12 and Bailey is 3. \n" ); document.write( " |