Right now Seth's age is 4/5 the age of his brother Eric. Twenty-one years ago, Eric was twice as old as Seth. What are their ages now?

          Present Time                                                                

Let X= Eric's age                                                            (4/5)X= Seth's age

Question:  What are their ages now?

                                             Past (21 years ago)

X-21 =Eric's age                                                                (4/5)X-21=Seth's age

2*[4/5(X-21]=Eric's age

Therefore, X-21= 2*[4/5(X)-21]=Eric's age           Substitution

X-21= 8/5 X  - 42         Solve for "X" by adding 42 to both sides.

X-21+42=(8/5) X

X+21 = (8/5)X              Subtract "X" from both sides.

21=(3/5)X                    Multiply both sides of equation by reciprocal of (3/5), which is 5/3

21*(5/3)= X                  Finish the problem to find value of "X," which is Eric's age.  

                                   Then find  4/5 (X)= Seth's age  

Seth age = 4/5 × Eric's age

seth age = x

eric age = y

x = 4y/5 eqa 1

2(x-21) = y-21 eqa 2

2(4y/5 - 21) = y-21

8y/5 - 42 = y -21

8y/5 - y = 21

1.6 y - y = 21

0.6 y = 21

y = 35 years Eric's age

x = 28 years Seth's age

sorry i dont know my friend

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: