Question

a) 2 years ago, my age was 4 1/2 times the age of my son then. 6 years ago, my age
was twice the square of the age of my son then. Find the present age of my son.​

Answer 1

Basic Concept Used :-

Writing Systems of Linear Equation from Word Problem.

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

Solution :-

• Let my present age be 'x' years

and

• Let present age of son be 'y' years.

According to statement,

Two years ago,

• My age = (x - 2) years

• Son age = (y - 2) years

So,

$\rm :\longmapsto\:(x - 2) = 4\dfrac{1}{2}(y - 2)$

$\rm :\longmapsto\:(x - 2) = \dfrac{9}{2}(y - 2)$

$\rm :\longmapsto\:2x - 4 = 9y - 18$

$\rm :\longmapsto\:2x = 9y - 14 - - (1)$

According to statement,

Six years ago,

• My age = (x - 6) years

• Son age = (y - 6) years

So,

$\rm :\longmapsto\:(x - 6) = 2 {(y - 6)}^{2}$

On multiply both sides by 2, we get

$\rm :\longmapsto\:(2x - 12) = 4 {(y - 6)}^{2}$

On substituting the value of 2x from equation (1), we get

$\rm :\longmapsto\:9y - 14 - 12 = 4 {(y }^{2} + 36 - 12y)$

$\rm :\longmapsto\:9y - 26= 4{y}^{2} + 144 - 48y$

$\rm :\longmapsto\:4{y}^{2} - 57y + 170 = 0$

$\rm :\longmapsto\:4{y}^{2} - 40y - 17y + 170 = 0$

$\rm :\longmapsto\:4y(y - 10) - 17(y - 10) = 0$

$\rm :\longmapsto\:(y - 10)(4y - 17) = 0$

$\rm :\implies\:y = 10 \: \: \: \: or \: \: \: \: y = \dfrac{17}{4} \{rejected \}$

Hence

• Present age of my son = 10 years