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If you add 3 to my age and divide the sum by 5, you will have my daughter's age. But 3 years ago I was 8 times as old as my daughter was. find their ages now.

I've got (m+3)/5 =d and m-3=8d

My teacher wrote m-3=8(d-3).

But no matter which equations I use. I keep getting decimals, negative numbers, or fractions for the ages.

Please help and explain the steps.

Thank you!

Comments

 

Hi Roxi -- I like positive, whole numbers for ages, too.

Mom's age is altered +/-3, which suggests her age is multiple of 3.

Also, adding 3 makes mom's age a multiple of 5 as well: 3x5 gives mults of 15. That means adding 3 takes mom to age 15 or 30 or 45.

I'd start with mom+3= 30 and girl 6.

With mom currently 27, mom-3= 24, which is 8x girl minus 3, or 3 tears old :)

 

Hi Roxi -- I like positive, whole numbers for ages, too.

Mom's age is altered +/-3, which suggests her age is multiple of 3.

Also, adding 3 makes mom's age a multiple of 5 as well: 3x5 gives mults of 15. That means adding 3 takes mom to age 15 or 30 or 45.

I'd start with mom+3= 30 and girl 6.

With mom currently 27, mom-3= 24, which is 8x girl minus 3, or 3 tears old :)

 

Hi Roxi -- I like positive, whole numbers for ages, too.

Mom's age is altered +/-3, which suggests her age is multiple of 3.

Also, adding 3 makes mom's age a multiple of 5 as well: 3x5 gives mults of 15. That means adding 3 takes mom to age 15 or 30 or 45.

I'd start with mom+3= 30 and girl 6.

With mom currently 27, mom-3= 24, which is 8x girl minus 3, or 3 tears old :)

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1 Answer

Roxi,

     First of all, it seems that you're teacher's equation is correct. Can you see how she got there? "3 years ago" means that you subtract 3 from BOTH your age AND your daughter's, as you were both 3 years younger 3 years ago. The only difference between your version and your teacher's is that you didn't subtract 3 years from your daughter's age.

     Next, it seems that you are making an error somewhere in your algebra. Our overall goal here is to create one equation with one variable and solve for that variable, and then to use the answer to solve for the other variable in our original equations. Let's start by solving for "m" in your teacher's version of the equation, m-3=8(d-3).

m-3=8d-24      distributed the 8

m=8d-21         added 3 to each side

So now we have m. Let's use this definition of "m" in our other equation. If we substitute this "m" with "8d-21" we will have an equation that is all in "d", and we can then find the value for "d".

((8d-21)+3)/5=d       we substituted the "m" for what we solved for "m" above

(8d-21)+3=5d          multiply both sides by 5

8d -18=5d           

-18=-3d              subtract 8d from both sides

d=6                   divide both sides by -3

Now we can simply substitute this "d" for "6" in the above equations. "m" will equal 27 in both. Does this make sense?