CBSE Class 10 Mathematics Chapter 3: Pair of Linear Equations in Two Variables NCERT Solutions

Solution:

Let the present age of Aftab be $x$ years and the present age of his daughter be $y$ years.

According to the first condition: "Seven years ago, I was seven times as old as you were then."

Seven years ago:

• Aftab's age was $x\; -\; 7$ years.
• Daughter's age was $y\; -\; 7$ years.

So, the equation is:

$$x\; -\; 7\; =\; 7(y\; -\; 7)$$

Expanding this equation:

$$x\; -\; 7\; =\; 7y\; -\; 49$$

Rearranging the terms to form a linear equation:

$$x\; -\; 7y\; =\; -49\; +\; 7$$

$$x\; -\; 7y\; =\; -42\; \backslash quad\; \backslash quad\; (1)$$

According to the second condition: "Also, three years from now, I shall be three times as old as you will be."

Three years hence:

• Aftab's age will be $x\; +\; 3$ years.
• Daughter's age will be $y\; +\; 3$ years.

So, the equation is:

$$x\; +\; 3\; =\; 3(y\; +\; 3)$$

Expanding this equation:

$$x\; +\; 3\; =\; 3y\; +\; 9$$

Rearranging the terms to form a linear equation:

$$x\; -\; 3y\; =\; 9\; -\; 3$$

$$x\; -\; 3y\; =\; 6\; \backslash quad\; \backslash quad\; (2)$$

Algebraic Representation:

The situation can be represented algebraically by the following pair of linear equations:

$$x\; -\; 7y\; =\; -42$$

$$x\; -\; 3y\; =\; 6$$

Graphical Representation:

To represent these equations graphically, we need to find at least two points for each line.

For equation (1): $x\; -\; 7y\; =\; -42$ or $x\; =\; 7y\; -\; 42$

If $y\; =\; 5$, then $x\; =\; 7(5)\; -\; 42\; =\; 35\; -\; 42\; =\; -7$. So, one point is $(-7,\; 5)$.

If $y\; =\; 6$, then $x\; =\; 7(6)\; -\; 42\; =\; 42\; -\; 42\; =\; 0$. So, another point is $(0,\; 6)$.

If $y\; =\; 7$, then $x\; =\; 7(7)\; -\; 42\; =\; 49\; -\; 42\; =\; 7$. So, another point is $(7,\; 7)$.

For equation (2): $x\; -\; 3y\; =\; 6$ or $x\; =\; 3y\; +\; 6$

If $y\; =\; 0$, then $x\; =\; 3(0)\; +\; 6\; =\; 6$. So, one point is $(6,\; 0)$.

If $y\; =\; -1$, then $x\; =\; 3(-1)\; +\; 6\; =\; -3\; +\; 6\; =\; 3$. So, another point is $(3,\; -1)$.

If $y\; =\; -2$, then $x\; =\; 3(-2)\; +\; 6\; =\; -6\; +\; 6\; =\; 0$. So, another point is $(0,\; -2)$.

Now, we plot these points on a graph paper and draw the lines passing through them. The intersection of these two lines will give the solution to the system of equations.

The graph shows two lines. One line passes through points $(-7,\; 5)$, $(0,\; 6)$, and $(7,\; 7)$. The other line passes through points $(6,\; 0)$, $(3,\; -1)$, and $(0,\; -2)$. These lines intersect at a point, which represents the solution. By observing the graph, the intersection point is approximately $(21,\; 9)$. Let's verify this: For $x=21,\; y=9$, $21\; -\; 7(9)\; =\; 21\; -\; 63\; =\; -42$ (satisfies eq 1) and $21\; -\; 3(9)\; =\; 21\; -\; 27\; =\; -6$ (does not satisfy eq 2). There seems to be a discrepancy in the provided graphical interpretation in the source. Let's re-evaluate the points and graph.

Correcting the interpretation based on standard graphical methods:

Plotting the points $(-7,\; 5)$ and $(0,\; 6)$ for the first equation, and $(6,\; 0)$ and $(0,\; -2)$ for the second equation, and drawing the lines. The intersection point is where $x\; -\; 7y\; =\; -42$ and $x\; -\; 3y\; =\; 6$. Subtracting the second equation from the first:

$$(x\; -\; 7y)\; -\; (x\; -\; 3y)\; =\; -42\; -\; 6$$

$$-4y\; =\; -48$$

$$y\; =\; 12$$

Substituting $y\; =\; 12$ into the second equation:

$$x\; -\; 3(12)\; =\; 6$$

$$x\; -\; 36\; =\; 6$$

$$x\; =\; 42$$

So, the unique solution is $(42,\; 12)$. Aftab's age is 42 years and his daughter's age is 12 years. The graphical representation should show the lines intersecting at $(42,\; 12)$.