Question:medium

Five years ago, Adil was thrice as old as Bharat. Ten years later Adil shall be twice as old as Bharat. To know the present ages of Adil and Bharat: (i) form the linear equations representing the above information. (ii) show that the system of equations is consistent with unique solution. (iii) find the present ages of Adil and Bharat.

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Always double check your answers by substituting the computed ages back into the original word problem:
- Five years ago, Adil was 45 and Bharat was 15. \( 45 = 3 \times 15 \) (True).
- Ten years later, Adil will be 60 and Bharat will be 30. \( 60 = 2 \times 30 \) (True).
This self-check ensures absolute accuracy.
Updated On: Jul 7, 2026
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Solution and Explanation

Step 1: Set up the two equations by cross-checking each time frame.
Let Adil's present age be $x$ years and let Bharat's present age be $y$ years.
Five years ago, Adil's age was $x - 5$ and Bharat's age was $y - 5$. The question says Adil was thrice as old as Bharat at that time, so:
\[ x - 5 = 3(y - 5) \] \[ x - 5 = 3y - 15 \] \[ x - 3y + 10 = 0 \] This is Equation 1.
Ten years later, Adil's age will be $x + 10$ and Bharat's age will be $y + 10$. The question says Adil will be twice as old as Bharat then, so:
\[ x + 10 = 2(y + 10) \] \[ x + 10 = 2y + 20 \] \[ x - 2y - 10 = 0 \] This is Equation 2.

Step 2: Check consistency using the determinant of the coefficients.
Write down the coefficients of $x$ and $y$ from both equations: for Equation 1, $a_1 = 1$, $b_1 = -3$; for Equation 2, $a_2 = 1$, $b_2 = -2$.
Two lines meet at exactly one point (a unique solution exists) when the value $a_1 b_2 - a_2 b_1$ is not zero.
\[ a_1 b_2 - a_2 b_1 = (1)(-2) - (1)(-3) = -2 + 3 = 1 \] Since this value is $1$, which is not zero, the two lines are not parallel and not the same line, so they cross at one point. The system is consistent and has a unique solution.

Step 3: Solve the pair of equations using cross multiplication.
For two equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the cross multiplication rule gives:
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \] Here $a_1 = 1, b_1 = -3, c_1 = 10$ and $a_2 = 1, b_2 = -2, c_2 = -10$.
Work out each part one at a time.
\[ b_1c_2 - b_2c_1 = (-3)(-10) - (-2)(10) = 30 + 20 = 50 \] \[ c_1a_2 - c_2a_1 = (10)(1) - (-10)(1) = 10 + 10 = 20 \] \[ a_1b_2 - a_2b_1 = (1)(-2) - (1)(-3) = -2 + 3 = 1 \] So:
\[ \frac{x}{50} = \frac{y}{20} = \frac{1}{1} \] This gives $x = 50$ and $y = 20$ directly, without eliminating one variable by subtracting the equations.

Final Answer:
(i) The two equations are $x - 3y + 10 = 0$ and $x - 2y - 10 = 0$.
(ii) Since $a_1b_2 - a_2b_1 = 1 \neq 0$, the system is consistent and has a unique solution.
(iii) Adil's present age is 50 years and Bharat's present age is 20 years.
\[ \boxed{x = 50, \ y = 20} \]
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