Question

Six years ago Matt was four times as old as Brown, and in $3$ years the sum of their ages will be $43$ years.

Find their present ages.

Answer 1

Step-1: Form a pair of linear equations:

Given that,

• Six years ago Matt was four times as old as Brown
• In $3$ years the sum of their ages will be $43$ years

Let the present age of Brown and Matt be $\mathrm{x}$ years and $\mathrm{y}$ years, respectively.

Six years ago Matt's age was $\left(\mathrm{y}-6\right)$ years and Brown's age was $\left(\mathrm{x}-6\right)$ years.

Then according to question, $\left(\mathrm{y}-6\right)=4\left(\mathrm{x}-6\right)$

$\Rightarrow$ $\mathrm{y}-6=4\mathrm{x}-24$

$\Rightarrow$ $\mathrm{y}=4\mathrm{x}-18$ $...\left(\mathrm{i}\right)$

After $3$years, Matt's age will be $\left(\mathrm{y}+3\right)$ years and Brown's age was $\left(\mathrm{x}+3\right)$ years.

Then, according to question; $\left(\mathrm{y}+3\right)+\left(\mathrm{x}+3\right)=43$

$\Rightarrow$ $\mathrm{x}+\mathrm{y}+6=43$

$\Rightarrow$ $\mathrm{y}=37-\mathrm{x}$ $...\left(\mathrm{ii}\right)$

Step-2: Find their present ages:

Solve equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, to find the present ages of Matt and Brown

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we can write $4\mathrm{x}-18=37-\mathrm{x}$

$\Rightarrow$ $5\mathrm{x}=55$

$\Rightarrow$ $\mathrm{x}=11$

Substitute $\mathrm{x}=11$ in $\left(\mathrm{ii}\right)$, we get $\mathrm{y}=37-11$

$\Rightarrow$ $\mathrm{y}=26$

Hence, the present age of Brown is $11$ years and present age of Matt is $26$ years.