3 x Jane's age = 8 x Beth's age. Difference between ages is 15 years. How old are Jane and Beth?

3j=8b

j-b=15

j=15+b substitute

3(15+b)=8b

45+3b=8b

5b=45

b=9

j=15+b=15+9=24

3*24=72

8*9=72

beth is 9

jane is 24

Jane: 3x and Beth: 8x and difference means subtraction:

First: 8x - 3x = 15

Second: combine "like" terms:

5x = 15

Third: divide both sides by 5:

5x/5 = 15/5

x = 3

Take (3) and replace it with "x" for Jane's age and Beth's age separately:

Jane = 3(3) = 9

Beth: 3(8) = 24

j = jane's age

b = beth's age

3j = 8b

j-b = 15

solve j-b = 15 for j

j-b=15

j = b+15

substitute j into the other equation

3j = 8b

j = b+15

3(b+15) = 8b

solve for b

3(b+15) = 8b

3b +45 = 8b

-3b -3b

45 = 5b

b = 9

substitute 9 for b in one of the equations

j - b = 15

j - 9 = 15

+9 +9

j = 24

Jane is 24 years old

Beth is 9 years old

Let x be Jane's age and y Beth's age. Then 3x = 8y, and therefore x > y. So x - y = 15. One can now combine these 2 equations:

3x - 8y =0

x - y = 15.

Solving the previous system yields (by substitution for instance)

x = 24 and y= 9.

let Jane,s age be x ,and Beth,s age y

3x=8y then x=8/3y

x-y=15 8/3y-y=15 5/3y=15 (*3/5)

y=9

x=8/3*9=24

Jan,sage 24 Beth,s age 9

let janes age be x

beth's age=3x/8

difference=x-3x/8

=5x/8

given 5x/8=15

multiplying by 8/5

x=15*8/5

=24

jane's age=24 years and beth's age=9 years

jane is 9 and beth is 24

3x = 8x -15

+15 -3x = -3x+15

----------------------

15 = 5x

/5 = /5

x = 3

3*3 = 9, 8*3 = 24, 24 - 9 = 15

3J = 8B

J-B=15

Isolate: J= 15+B

Substitute: 3(15+B)=8B

Solve:

45+3B=8B

5B=45

B=9

and

J=24

Then test:

3J = 8B

3(24) = 8(9)

72 = 72

J-B=15

24-9=15

15=15

Correct

3J=8B

J-B=15

SoBeth is 9 Jane 24

(3J = 8B

(J - B = 15

3J = 8B

J = 8B/3

8B/3 - B = 15

5B = 45

B = 45 : 5

B = 9

J = 8B/3

J = 8.9/3

J = 24

Solution: Jane is 24 y.o and Beth, 9 y.o.

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