two years ago,my age was four times the age of my son.eight years ago ,my age was ten times the age of my son .find the age of my son now
Answers
Answer:
11
Step-by-step explanation:
x = father's current age
y = son's current age
.
Two years ago...
x-2
y-2
.
Father's age was 4 times son's age:
(x-2) = 4(y-2)
.
Eight years ago...
x-8
y-8
Father's age was 10 times son's age:
(x-8) = 10(y-8)
.
So, we have two equations and two unknowns.
That suggests a system of linear equations can be used to solve them.
.
x-2 = 4y -8
x -4y = -6
.
x-8 = 10y -80
x -10y = -72
.
x -4y = -6
x -10y = -72
------------- subtract
6y = 66
divide both sides by 6
y = 11
.
So, the son is now 11 years old.
.
To check this answer...
substitute y=11 to find x
.
(x-2) = 4(11-2)
x -2 = 44 - 8
x = 38
.
check these values using the third equation
.
x-8 = 10(y-8)
x-8 = 30
y-8 = 3
10*3 = 30
Correct.
.
Answer: The son is now 11 years old.
Answer:
Answer is given below
Step-by-step explanation:
two years ago my age was four times that of my son eight years ago my age was ten times that of my son.find the age of my son
Let S be the current age of my son.
Let M be the current age of me.
(M - 2) = 4(S - 2) (two years ago my age was four times that of my son)
(M - 8) = 10(S - 8) (eight years ago my age was ten times that of my son)
Simplifying the above two equations gives us,
M - 4S = -6
M - 10S = -72
Subtracting 2nd eqn from 1st,
6S = 66
S = 11
M = 4S - 6 = 44 - 6 = 38
Our current ages are: Me 38 yrs, Son 11 yrs