# Two years ago pete was three times as old as his cousin claire. two years before that, pete was four

###### Question:

## Answers

6 years if you add the years up and look at the ages

22:11

1. 21:10

2. 20:9

3. 19:8

4. 18:7

5. 17:6

6. 16:5

7. 15:4

8. 14:3

9. 13:2

10. 12:1

Ratios dont add up

4

Step-by-step explanation:

Let P represent Pete's age today. Let C represent Claire's age today.

2 years ago: P - 2 = 3(C - 2) ⇒ P - 2 = 3C - 6 ⇒ P = 3C - 4

4 years ago: P - 4 = 4(C - 4) ⇒ P - 4 = 4C - 16 ⇒ P = 4C - 12

Use substitution to solve the system above:

3C - 4 = 4C - 12

- 4 = C - 12

8 = C

Plug in the C-value above into one of the equations to solve for P:

P = 3C - 4

= 3(8) - 4

= 24 - 4

= 20

So, Pete is 20 yrs old today and Claire is 8 years old today

When will Pete be twice as old as Claire?

20 + x = 2(8 + x)

20 + x = 16 + 2x

20 = 16 + x

4 = x

Their ages will have a 2:1 ratio in 4 years.

P=pete's age now

c=claire's age now

2 year ago pete was 3times as old as his cousin clarie

p-2=3 times (c-2)

also the ratio is 1:3

2 years before that, pete was 4 times as old as claire

p-4=4 times (c-4)

so solve

p-2=3(c-2)

p-4=4(c-4)

distribute

p-2=3c-6

p-4=4c-16

p-2=3c-6

add 2 to both sides

p=3c-4

p-4=4c-16

add 4 to both sides

p=4c-12

so nowe we have

p=3c-4 and p=4c-12

therefor they are equal

3c-4=p=4c-12

3c-4=4c-12

add 12 to both sides

3c+8=4c

subtract 3c from both sides

8=c

subsitute

p-2=3(c-2)

p-2=3(8-2)

p-2=3(6)

p-2=18

addd 2

p=20

so right now, pete is 20 and claire is 8

so we solve by doing

20+what to 8+what=22 to 11

make fraction

and replace what with x

[tex]\frac{20+x}{8+x}= \frac{22}{11}[/tex]

sipmlify 22/11 into 2/1

multiply both sides by 8+x

20+x=[tex]\frac{(2)(8+x)}{1}[/tex]

20+x=2(8+x)

distribute

20+x=16+2x

subtract 16 from both sides

4+x=2x

subtract x from both sides

4=x

the answer is in 4 years

test

20+4=24

8+4=12

24/12=22/11?

2/1=2/1

true

the answer is in 4 years