Five year ago a mother was seven times as old as her daughter .five years hence she will be three times as old as her daughter .Find their present ages .the question of linear equation in one variable
Answers
As,5 years ago mother's age was 7 times her daughter's age, so,mother's will be y-5=7(x-5)
Similarly,5 years from now the mother's age will be y+5=3(x+5)
By solving both the equations we get x=10 & y=40.
So the present ages of mother and her daughter are 40 & 10 years respectively.
Answer:
the present age of daughter=10 years
present age of mother=40 years
Step-by-step explanation:
Let the present age of daughter=x
five years ago age of daughter=x-5
five years ago mother was seven times as old as her daughter so,
five year ago mother age=7×(x-5)
mother present age=7×(x-5)+5
=7x-35+5
=7x-30
Five years hence,
daughter age=x+5
mother age=7x-30+5
=7x-25
five years hence mother will be three times as old as her daughter.so,
7x-25=3×(x+5)
7x-25=3x+15
7x-3x=15+25
4x=40
x=40÷4
x=10
present age of daughter=10 years
present age of mother=7x-30
=7×10-30
=70-30
=40 years