Dr. Garica is twice as old as his son. Twenty years ago he was 4 times as old as his son was then. How old are they now?

Let's let g = Dr. Garcia's age and s = his son's age.

g = 2s, right?

20 years ago, Dr. Garcia was g - 20 years old.  He was also 4(s - 20), 4 times his son's age 7 years ago.

g - 20 = 4(s - 20)

g - 20 = 4s - 80

In the first equation, we have g = 2s.  We are going to substitute this g into the second equation.

2s - 20 = 4s - 80

       60 = 2s

       s = 30

So, his son is 30 and Mr. Garcia is 60. 30 x 2 = 60.

20 years ago, his son would have been 10 and Mr. Garcia would have been 40.  10 x 4 = 40.

Step-by-step explanation:

Brainliest?

Dr. Garcia's age is 60, and his son's age is 30.

Step-by-step explanation:

Let x be Dr. Garcia's age, and y his son's age.

1. Dr. Garcia is twice as old as his son

x = 2 y

2. Twenty years ago, he was 4 times as old as his son was then

x - 20 = 4 (y - 20)

Substitute x in 2 and solve for y:

2 y - 20 = 4 (y - 20)

2 y - 20 = 4 y - 80

- 20 = 2 y - 80

60 = 2 y

y = 30

Put y back into 1:

x = 2 (30)

x = 60

So Dr. Garcia's age is 60, and his son's age is 30.

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

3

step-by-step explanation: