Mrs Bruce lacks 1 year from being 5 times as old as her son. Five years from now she will lack 1 from being 3 times as old as her son will be then. Find each of their ages. I know the first equation is x+1= 5y and I know to substitue x = 5y - 1 in the second equation. But I can't figure out the second equation.

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Let's say x is Mrs. Bruce's age and y is her son's. Then right now x=5y-1. In five years (x+5), she will be 3(y+5)-1 or:

x+5=3y+15-1=3y+14.

Substituting the equation for x into here that's:

5y-1+5=3y+14

2y=10

y=5

and x=24

Checking our work, in five years, her son will be ten and she will be 29. Check!

Hi Michael;

B=Mrs. Bruce's Age

N=Son's age

B+5=[3(N+5)]-1

B+5=3N+15-1

B+5=3N+14

B=3N+9

Let's subtract the second equation from the first...

0=2N-10

10=2N

5=N

Let's plug-in N=5 into the first equation...

B=5N-1

B=5(5)-1

B=25-1

B=24

Let's plug-in N=5 into the second equation to verify...

B=3N+9

B=3(5)+9

B=15+9

B=24

Mrs. Bruce is originally 24 years old.

Her son is originally 5 years old.

Mrs. Bruce is one year less than being five times as old as her son.

In five years...

Mrs. Bruce will be 29 years old.

Her son will be 10 years old.

Mrs. Bruce will be one year less than three times as old as her son.

thank you

I love this stuff! Any more?

She is 24, (as 25 is 5 x 5) in 5 years she will be 29, (one year shy of 30 and he will be 10).

This example is easier to do by 'common sense' as the language LACKS tells you she is 'younger than' and ages mother and son can be are limited. If it were a more wide ranging problem an equation would need to be used. For this though, picking an age to see what the answer would be (and adjusting that age as needed until the sum works) is faster.

Hey Michael -- key feature: "5 years from now" the 'Mom offset' is still one (-1) year ...

suggests multiples of 5 ... kid 5, Mom 24 => kid 10, Mom 29 ... Best wishes, sir :)

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