NCERT Solutions
Class 10 - Mathematics - Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Exercise 3.1

(i) Forming the equations

Let \(x\) be the number of girls and \(y\) be the number of boys.

Total students are 10, so

\[ x + y = 10. \]

The number of girls is 4 more than the number of boys, so

\[ x = y + 4 \Rightarrow x - y = 4. \]

Thus the required pair of linear equations is

\[ x + y = 10, \qquad x - y = 4. \]

Solving graphically (conceptual steps)

1. For \(x + y = 10\), choose convenient values:

\(x = 0 \Rightarrow y = 10\); \(x = 10 \Rightarrow y = 0\). Plot \((0,10)\) and \((10,0)\) and draw the line.

2. For \(x - y = 4\), choose convenient values:

\(x = 4 \Rightarrow y = 0\); \(x = 6 \Rightarrow y = 2\). Plot \((4,0)\) and \((6,2)\) and draw the line.

3. The two lines intersect at the point \((7,3)\), obtained either from the graph or by solving the equations algebraically.

So \(x = 7, y = 3\).

Conclusion

Girls = 7 and Boys = 3.

(ii) Forming the equations

Let \(x\) be the cost (in rupees) of one pencil and \(y\) be the cost (in rupees) of one pen.

Cost of 5 pencils and 7 pens is ₹ 50:

\[ 5x + 7y = 50. \]

Cost of 7 pencils and 5 pens is ₹ 46:

\[ 7x + 5y = 46. \]

Thus the required pair of equations is

\[ 5x + 7y = 50, \qquad 7x + 5y = 46. \]

Solving graphically (conceptual steps)

1. For \(5x + 7y = 50\), take convenient pairs: if \(x = 5\), then \(25 + 7y = 50 \Rightarrow y = \frac{25}{7}\); if \(y = 5\), then \(5x + 35 = 50 \Rightarrow x = 3\). Plot two such points (for easy graph use integer point \((3,5)\)) and draw the line.

2. For \(7x + 5y = 46\), take convenient pairs: if \(x = 4\), then \(28 + 5y = 46 \Rightarrow y = \frac{18}{5}\); if \(y = 4\), then \(7x + 20 = 46 \Rightarrow x = \frac{26}{7}\). Plot two points and draw the line.

3. The two lines intersect at \((3,5)\), which can also be verified algebraically by solving the pair:

Multiply \(5x + 7y = 50\) by 7 and \(7x + 5y = 46\) by 5:

\[ 35x + 49y = 350, \quad 35x + 25y = 230. \]

Subtracting, \((35x + 49y) - (35x + 25y) = 350 - 230 \Rightarrow 24y = 120 \Rightarrow y = 5. \]

Substitute in \(5x + 7y = 50\):

\[ 5x + 7 \times 5 = 50 \Rightarrow 5x + 35 = 50 \Rightarrow 5x = 15 \Rightarrow x = 3. \]

Conclusion

Cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.

For a pair of equations in two variables

\[ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0, \]

the nature of the pair of lines is decided as follows:

• If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines intersect at a single point.
• If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel and distinct.
• If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the lines are coincident.

(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)

Here

\(a_1 = 5, b_1 = -4, c_1 = 8;\) \(a_2 = 7, b_2 = 6, c_2 = -9.\)

Compute the ratios:

\[ \frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}, \quad \frac{c_1}{c_2} = \frac{8}{-9} = -\frac{8}{9}. \]

Clearly \(\frac{5}{7} \neq -\frac{2}{3}\). Hence

The two lines intersect at a point.

(ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)

Here

\(a_1 = 9, b_1 = 3, c_1 = 12;\) \(a_2 = 18, b_2 = 6, c_2 = 24.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}. \]

All three ratios are equal, so

The two lines are coincident.

(iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)

Here

\(a_1 = 6, b_1 = -3, c_1 = 10;\) \(a_2 = 2, b_2 = -1, c_2 = 9.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}. \]

We have \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) but this common value is not equal to \(\frac{c_1}{c_2}\). Therefore

The two lines are parallel.

For a pair of linear equations in two variables

\[ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0, \]

we have:

• If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \): lines intersect; the pair is consistent with a unique solution.
• If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \): lines are parallel; the pair is inconsistent (no solution).
• If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \): lines are coincident; the pair is consistent with infinitely many solutions.

(i) \(3x + 2y = 5\) and \(2x - 3y = 7\)

Write them in standard form \(a_1x + b_1y + c_1 = 0\):

\(3x + 2y - 5 = 0\) and \(2x - 3y - 7 = 0\).

Here \(a_1 = 3, b_1 = 2, c_1 = -5;\) \(a_2 = 2, b_2 = -3, c_2 = -7\).

Ratios:

\[ \frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3}. \]

Since \(\frac{3}{2} \neq -\frac{2}{3}\), the lines intersect at a point.

The system is consistent (unique solution).

(ii) \(2x - 3y = 8\) and \(4x - 6y = 9\)

Standard form:

\(2x - 3y - 8 = 0\) and \(4x - 6y - 9 = 0\).

\(a_1 = 2, b_1 = -3, c_1 = -8;\) \(a_2 = 4, b_2 = -6, c_2 = -9.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}. \]

We have \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) but this common value is not equal to \(\frac{c_1}{c_2}\). Therefore the lines are parallel, so

The system is inconsistent (no solution).

(iii) \(\frac{3}{2}x + \frac{5}{3}y = 7\) and \(9x - 10y = 14\)

First equation: multiply by 6 to clear denominators:

\[ 6 \left( \frac{3}{2}x + \frac{5}{3}y \right) = 6 \cdot 7 \Rightarrow 9x + 10y = 42. \]

So the pair becomes \(9x + 10y - 42 = 0\) and \(9x - 10y - 14 = 0\).

Here \(a_1 = 9, b_1 = 10, c_1 = -42;\) \(a_2 = 9, b_2 = -10, c_2 = -14.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{9}{9} = 1, \quad \frac{b_1}{b_2} = \frac{10}{-10} = -1. \]

Since \(1 \neq -1\), the lines intersect at one point.

The system is consistent (unique solution).

(iv) \(5x - 3y = 11\) and \(-10x + 6y = -22\)

Standard form:

\(5x - 3y - 11 = 0\); \(-10x + 6y + 22 = 0\).

Here \(a_1 = 5, b_1 = -3, c_1 = -11;\) \(a_2 = -10, b_2 = 6, c_2 = 22.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}. \]

All three ratios are equal; hence the two equations represent the same line.

The system is consistent with infinitely many solutions.

(v) \(\frac{4}{3}x + 2y = 8\) and \(2x + 3y = 12\)

First equation: multiply by 3:

\[ 4x + 6y = 24 \Rightarrow 4x + 6y - 24 = 0. \]

Second equation: \(2x + 3y - 12 = 0\).

Thus \(a_1 = 4, b_1 = 6, c_1 = -24;\) \(a_2 = 2, b_2 = 3, c_2 = -12.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{4}{2} = 2, \quad \frac{b_1}{b_2} = \frac{6}{3} = 2, \quad \frac{c_1}{c_2} = \frac{-24}{-12} = 2. \]

All three are equal, so the lines are coincident.

The system is consistent with infinitely many solutions.

Use the criteria based on \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2} \).

(i) \(x + y = 5\) and \(2x + 2y = 10\)

Second equation simplifies by dividing through by 2:

\[ 2x + 2y = 10 \Rightarrow x + y = 5. \]

Thus both equations represent the same line. Here

\[ \frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}. \]

All ratios are equal, so the pair is consistent with infinitely many solutions.

Graphical solution: Both equations give the same line \(x + y = 5\). Every point on this line, such as \((0,5), (5,0), (2,3)\), is a solution.

(ii) \(x - y = 8\) and \(3x - 3y = 16\)

Write in standard form: \(x - y - 8 = 0\); \(3x - 3y - 16 = 0\).

Ratios:

\[ \frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}. \]

We have \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\); therefore the lines are parallel and distinct.

The pair is inconsistent (no solution).

(iii) \(2x + y - 6 = 0\) and \(4x - 2y - 4 = 0\)

Rewrite the second equation:

\[ 4x - 2y - 4 = 0. \]

Here \(a_1 = 2, b_1 = 1, c_1 = -6;\) \(a_2 = 4, b_2 = -2, c_2 = -4.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}. \]

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at a unique point, so the pair is consistent.

Solving algebraically for the point of intersection (which will match the graphical solution):

From \(2x + y - 6 = 0\), \(y = 6 - 2x\).

Substitute into \(4x - 2y - 4 = 0\):

\[ 4x - 2(6 - 2x) - 4 = 0 \Rightarrow 4x - 12 + 4x - 4 = 0 \Rightarrow 8x - 16 = 0 \Rightarrow x = 2. \]

Then \(y = 6 - 2x = 6 - 4 = 2\).

Graphically, the lines intersect at \((2,2)\).

(iv) \(2x - 2y - 2 = 0\) and \(4x - 4y - 5 = 0\)

Here \(a_1 = 2, b_1 = -2, c_1 = -2;\) \(a_2 = 4, b_2 = -4, c_2 = -5.\)

Ratios:

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}. \]

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and distinct.

The pair is inconsistent (no solution).

Let the length of the rectangular garden be \(l\) metres and the width be \(b\) metres.

Given that the length is 4 m more than the width:

\[ l = b + 4. \]

Perimeter of a rectangle is \(2(l + b)\). Half the perimeter is therefore \(l + b\).

It is given that half the perimeter is 36 m:

\[ l + b = 36. \]

So we have the system

\[ l = b + 4, \qquad l + b = 36. \]

Substitute \(l = b + 4\) in \(l + b = 36\):

\[ (b + 4) + b = 36 \Rightarrow 2b + 4 = 36 \Rightarrow 2b = 32 \Rightarrow b = 16. \]

Then

\[ l = b + 4 = 16 + 4 = 20. \]

Therefore, the dimensions of the garden are: length \(20\) m and breadth \(16\) m.

The given equation is

\[ 2x + 3y - 8 = 0. \]

Its coefficients are \(a_1 = 2, b_1 = 3, c_1 = -8\).

(i) Intersecting lines

For intersecting lines, we need

\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2}. \]

Take another equation, for example

\[ 3x + 2y - 7 = 0. \]

Here \(a_2 = 3, b_2 = 2\). Then

\[ \frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{3}{2}. \]

Since these ratios are unequal, the two lines intersect at a unique point.

(ii) Parallel lines

For parallel but distinct lines, we require

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. \]

Choose an equation with proportional \(a\) and \(b\) but different constant term. Multiply the whole given equation by 2 and change the constant term:

\[ 4x + 6y - 16 = 0 \quad \text{(same ratios as original)}. \]

Now \(a_2 = 4, b_2 = 6, c_2 = -16\). We have

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}. \]

To make them distinct, keep \(a_2, b_2\) proportional but alter \(c_2\). One convenient choice shown in the textbook is

\[ 2x + 3y - 12 = 0. \]

Here

\[ \frac{a_1}{a_2} = \frac{2}{2} = 1, \quad \frac{b_1}{b_2} = \frac{3}{3} = 1, \quad \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}. \]

Thus \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), so the two lines are parallel and distinct.

(iii) Coincident lines

For coincident lines, all three ratios must be equal:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. \]

Take an equation which is a non-zero multiple of the given equation, for example multiply by 2:

\[ 4x + 6y - 16 = 0. \]

Now

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}. \]

All three are equal, so the two equations represent the same line and are coincident.

A set of acceptable answers is therefore:

(i) \(3x + 2y - 7 = 0\) (intersecting with the given line),

(ii) \(2x + 3y - 12 = 0\) (parallel to the given line),

(iii) \(4x + 6y - 16 = 0\) (coincident with the given line).

Consider the two lines:

\(L_1: x - y + 1 = 0\) and \(L_2: 3x + 2y - 12 = 0\), together with the x-axis \(y = 0\).

1. Intersection of \(L_1\) with the x-axis

For the x-axis, \(y = 0\). Substitute into \(x - y + 1 = 0\):

\[ x - 0 + 1 = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1. \]

So \(L_1\) meets the x-axis at the point \((-1, 0)\).

2. Intersection of \(L_2\) with the x-axis

Put \(y = 0\) in \(3x + 2y - 12 = 0\):

\[ 3x + 0 - 12 = 0 \Rightarrow 3x = 12 \Rightarrow x = 4. \]

So \(L_2\) meets the x-axis at \((4, 0)\).

3. Intersection of \(L_1\) and \(L_2\)

Solve the system

\[ x - y + 1 = 0 \quad \text{and} \quad 3x + 2y - 12 = 0. \]

From the first equation, express \(y\) in terms of \(x\):

\[ x - y + 1 = 0 \Rightarrow y = x + 1. \]

Substitute in the second equation:

\[ 3x + 2(x + 1) - 12 = 0 \Rightarrow 3x + 2x + 2 - 12 = 0 \Rightarrow 5x - 10 = 0 \Rightarrow 5x = 10 \Rightarrow x = 2. \]

Then

\[ y = x + 1 = 2 + 1 = 3. \]

Hence \(L_1\) and \(L_2\) intersect at the point \((2, 3)\).

4. Vertices of the triangle

The triangle is bounded by the two lines and the x-axis. Its three vertices are therefore:

• Intersection of \(L_1\) with x-axis: \((-1, 0)\).
• Intersection of \(L_2\) with x-axis: \((4, 0)\).
• Intersection of \(L_1\) and \(L_2\): \((2, 3)\).

Thus, the vertices of the triangle are \((-1, 0)\), \((4, 0)\) and \((2, 3)\). When these points are plotted and joined, the enclosed region is the required shaded triangular region.