Question

a's age is 4 times that of b's age. 10 years ago, a's age was 19 times that of b's age.find the present ages of a and b.(linear equation in one variable)​

Answer 1

⇝ Given :-

• a's age is 4 times that of b's age.

• 10 years ago, a's age was 19 times that of b's age.

⇝ To Find :-

• Their Present Ages

⇝ Solution :-

Let,

• Present Age of a = 4x years

Accordingly ,

• Present Age of b = x years

❒ 10 Years Ago :

• Age of a = (4x - 10) years

• Age of b = (x - 10) years

✏ According To Question :

$4\text x -10 = 19(\text x - 10)$

$:\longmapsto4\text x - 10 = 19\text x - 190$

$:\longmapsto4\text x - 19\text x = - 190 + 10$

$:\longmapsto \cancel - 15\text x = \cancel - 180$

$:\longmapsto\text x = \frac{180}{15}$

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf x = 12} }}}$

As,

• Present Age of a = 4x years

• Present Age of b = x years

Hence,

$\pink{\begin{cases} \bf Present \: \: Age \: of \: a = 48 \:years \\ \\ \bf Present \: Age \: of \: b = 12 \: years\end{cases}}$

Answer 2

Answer:

age of b = 12

age of a=. 48

Step-by-step explanation:

Let the age of b = x

age of a= 4x

(4x-10)= 19(x-10)

4x-10= 19x-190

15x = 180

x= 180/15= 12