In 1985 Maria was 3 times as old as Sue. Five years earlier the sum of their ages was 18. When was Maria born?

Maria was born in 1964

Step-by-step explanation:

Let Sue's age be x

Maria's age be 3x

Five years ago, their ages were

x-5, 3x-5

So

(x-5) + (3x-5) = 18

x-5 + 3x - 5 = 18

4x -10 = 18

4x = 18+10

x = 28/4 = 7

Sue's age x-5 = 7-5 = 2

Maria's age = 3x-5 = 3(7)-5 = 21-5=16

Maria was born in 1980 - 16 = 1964

1964

Step-by-step explanation:

to solve this question we should assign an equation for each year :

let x be Maria's age and y Sue's age.

Now let's solve each equation by trying to minimise the number of variables we are dealing with

how ?

let's multiply x+y= 18  by -1

we get -x-y=-18

Now let's add this last equation to x-3y-10 = 0 to get rid of x

so Sue's age is 2 .

let's calculate x the age of Maria by replacing y by 2 in x+y= 18

Let's check : in 1985 Maria's age is 21 and sue's age is 7

so it's true.

Now let's calculate Maria's year of birth :

In 1985 Maria was 21 so :

Maria was born in 1964

the barter

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)