Young Writers Society

Because I know you all want to help the poor little Algebra 1 student who missed class on Friday and has a quiz tomorrow. *smiles sweetly*

Okey-dokey. Problem one:

A man is now 6 times as old as his son. In 6 years, the father will be 3 times as old as the son will be then. Find their present ages.

^ I know for a fact that no one can be -12/7 years old. So. Any suggestions?

Number 2:

Marie is one-ninth as old as her mother. In 3 years, she will be one-fifth as old as her mother will be then. Find their present ages.

^ Last I checked, people can't have a negative age. But that's what I came out with. O.o

Thanks for taking a look, guys. *dies*

Cass, hope I can help you with these...

Problem one:

x = father's age, y = son's age

equation one
x = 6y the father's age is six times the sons age

equation two
x + 6 = 3(y + 6) the father's age plus six years is three times the son's age in six years

x + 6 = 3y + 18 the 3 is factored in

x = 3y + 12 subtract six from both sides

combined equations

6y = 3y + 12 now that we know what both x's equal, we can combine to two y equations through substitution

3y = 12 subtract 3y from both sides

y = 4 divide both sides by three... now that we have the age of the son, substitute that back into equation one to find x, or the father's age


first equation

x = 6y or x = 6(4); x = 24

Your ages are 4 and 24.



Problem two:

x = mother's age
y = Marie's age

first equation

(1/9)x = y Marie's age is one-ninth of her mother's age

second equation

(1/5) * (x + 3) = y + 3 ok, the mother's age in three years (y + 3) equal to one fifth (1/5) of Marie's age in three years (x + 3)

It may be easier to get rid of the fractions at this point:

first equation

x = 9y multiply by nine on both sides

second equation

x + 3 = 5(y + 3) multiply by five on both sides

x + 3 = 5y + 15 factor in the 5

x = 5y + 12 subtract three from both sides

Again now that we have the two x's, we can substitute

combined equation
9y = 5y + 12

4y = 12 subtract 5y from both sides

y = 3 divide by four on both sides... now that we have y, or Marie's age, we can substitute it back into the first equation to find x

first equation

x = 9y or x = 9(3); x = 27

So Marie is 3 (y = 3) and her mother is 27 (x = 27).


If you have a question on any of this, feel free to PM me, or just post your questions here. Hope that helps!

Wow! *is blown away* Thanks so much for the help! I'll let you know if I think of any questions, but for now, I'm all set.