ASAP ASAP PLS HELP !!

Question

FOUR YEARS AGO JACK WAS TWICE JAY'S AGE. IN FOUR YEARS, JACK WILL BE ONE AND A QUARTER TIMES JAY'S AGE. HOW OLD ARE JACK AND JAY NOW?

please write a clear explanation because my mind isn't big. thanks :D

Answer 1

4 years ago Jack was twice Jay's age

x-4 = 2(y-4) = 2y-8

x-2y =-8+4 = -4

x-2y =-4

multiply both sides by 4

4x-8y = -16

save this 1st equation to subtract from the next 2nd equation

in 4 years Jack will be 1 1/4 times Jay's age

x+4 = (5/4)(y+4)

4x+16 = 5y +20

4x-5y = 4 is the 2nd equation

4x-8y = -16 is the 1st equation

subtract 1st from 2nd to eliminate one variable

4x-5y - (4x-8y) = 4-(-16)

4x-5y-4x+8y = 4+16

8y-5y =20

3y = 20

y =20/3 = 6 2/3 years old for Jay now

x=2y-4= 13 1/3 - 4= 9 1/3 years old for Jack now

check the answers

4 years ago they were 2 2/3 and 5 1/3

5 1/3 is twice 2 2/3

16/3 = 2(8/3)

in 4 years they will be 10 2/3 and 13 1/3

13 1/3 is 1 1/4 times 10 2/3 = (5/4)(32/3) = 40/3 = 13 1/3