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∙ 9y ago^^^ I have to write a Algebraic Statement
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∙ 9y agoWrite and solve two simultaneous equations. a = Mary's age b = Mike's age Age now: b = 3a Age in 4 years: b + 4 = 2(a + 4)
John is 12, and Tim is eight. Four years ago they would've been eight and four.
she will be 20 years old
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Mary would be two years. This is a division problem 4 (John's age)/2 (two times older)=2(mary's age.)
Mary is 16 years old and her mother is 30 years old. Since Mary's age is half of Johan's age, Johan is twice Mary's age, making him 32 years old.
Mary = 26 Susan = 18
Write and solve two simultaneous equations. a = Mary's age b = Mike's age Age now: b = 3a Age in 4 years: b + 4 = 2(a + 4)
John is 12, and Tim is eight. Four years ago they would've been eight and four.
ken is 9 and john is 18. 4 years ago ken was 5 and John was 14. they had an age difference of 9 years.
This would depend on how old she was when he was born. For example if she was one year old when he was born, then at two years old she would be twice his age, If she was four at his birth, then when she is eight she will be twice his age. (See the related question which was the actual exercise.)
John will be 17 years old. Their ages will always be 3 years apart. And 20-3 = 17.
well wom would be 30 mary is 24 celia is 15 and jack is 16 so 1 year
42/4 = 10.510.5*3=31.5
she will be 20 years old
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