CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 06

Linear Equation in Two Variables

An equation of the form \( Ax + By + C = 0 \) is called a linear equation, where A is the coefficient of x, B is the coefficient of y and C is the constant term; A, B, C are real numbers and A and B cannot both be zero.

If \( A \ne 0, B = 0 \), the equation is of the form Ax + C = 0 (a line parallel to the y-axis). If \( A = 0, B \ne 0 \), it is By + C = 0 (a line parallel to the x-axis). If \( A \ne 0, B \ne 0, C = 0 \), it is Ax + By = 0 (a line through the origin). If \( A \ne 0, B \ne 0, C \ne 0 \), it is Ax + By + C = 0.

It is called a linear equation in two variables because the two unknowns x and y occur only in the first power and the product of the two unknowns does not occur. A single equation in two variables has infinitely many solutions, so we need a pair of equations (simultaneous equations) to find a particular solution.

Standard form: \( a_1x + b_1y + c_1 = 0 \) ... (i) and \( a_2x + b_2y + c_2 = 0 \) ... (ii). There are four methods of solving such a pair: (a) graphical method, (b) elimination by substitution, (c) elimination by equating the coefficients, (d) elimination by cross multiplication.

Graphical Method

Find at least two solutions of each equation by assuming a value of one variable and calculating the other. Plot these points on the same coordinate axes to get two straight lines, one for each equation. Three cases arise:

Case I: The two lines intersect at a point P(a, b). The equations have a unique solution x = a, y = b and are said to be consistent.
Case II: The two lines are parallel. The equations have no solution and are said to be inconsistent.
Case III: The two lines are coincident. The equations have infinitely many solutions and are said to be consistent (dependent).

Elimination by Substitution

Step 1: From the first equation find x in terms of y.
Step 2: Substitute this value of x in the second equation.
Step 3: The second equation becomes an equation in one variable y. Solve for y.
Step 4: Using this value of y in either given equation, find x.

Elimination by Equating the Coefficients

In this method, the coefficients of one variable in both equations are made equal (by multiplying one or both equations by suitable numbers), and then the variable is eliminated by subtraction or addition. We may eliminate either x or y.

Elimination by Cross Multiplication

For \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) with \( \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \):

\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]

so \( x = \dfrac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \dfrac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \).

Solved Examples on the Methods of Solving

Example 3. Solve using the substitution method: x + 4y = 14 and 7x − 3y = 5.
Answer:
\( x + 4y = 14 \) ... (i) and \( 7x - 3y = 5 \) ... (ii). From (i), \( x = 14 - 4y \) ... (iii).
Substitute in (ii): \( 7(14 - 4y) - 3y = 5 \), \( 98 - 28y - 3y = 5 \), \( 93 = 31y \), so y = 3.
Then \( 7x - 3(3) = 5 \), \( 7x = 14 \), x = 2. Solution: x = 2, y = 3.

Example 4. Solve using equating the coefficients: 9x − 4y = 8 and 13x + 7y = 101.
Answer:
Multiply (i) \( 9x - 4y = 8 \) by 7 and (ii) \( 13x + 7y = 101 \) by 4: \( 63x - 28y = 56 \) and \( 52x + 28y = 404 \).
Add: \( 115x = 460 \), so x = 4. Substitute in (i): \( 36 - 4y = 8 \), \( y = 7 \). Solution: x = 4, y = 7.

Example 5. Solve using equating the coefficients: \( \dfrac{bx}{a} - \dfrac{ay}{b} + a + b = 0 \) and \( bx - ay + 2ab = 0 \).
Answer:
The first equation, multiplied by ab, is \( b^2x - a^2y + a^2b + ab^2 = 0 \) ... (i). The second is \( bx - ay + 2ab = 0 \) ... (ii).
Multiply (ii) by b: \( b^2x - aby + 2ab^2 = 0 \). Subtract it from (i): \( -a^2y + aby + a^2b - ab^2 = 0 \), so \( ay(b - a) = ab(b - a) \), hence \( ay = ab \) and y = b (assuming a ≠ b).
Put y = b in (i): \( b^2x - a^2b + a^2b + ab^2 = 0 \), so \( b^2x = -ab^2 \) and x = −a. Solution: x = −a, y = b.

Example 6. Solve using equating the coefficients: \( \dfrac{42}{3x + 4y} + \dfrac{52}{4y - 3x} = 5 \) and \( \dfrac{7}{3x + 4y} + \dfrac{13}{4y - 3x} = 1 \), where \( 3x + 4y \ne 0 \) and \( 4y - 3x \ne 0 \).
Answer:
Let \( \dfrac{1}{3x + 4y} = a \) and \( \dfrac{1}{4y - 3x} = b \). Then \( 42a + 52b = 5 \) ... (i) and \( 7a + 13b = 1 \) ... (ii).
Multiply (ii) by 6: \( 42a + 78b = 6 \) ... (iii). (iii) − (i): \( 26b = 1 \), so \( b = \dfrac{1}{26} \). From (ii): \( 7a = 1 - \dfrac{13}{26} = \dfrac{1}{2} \), so \( a = \dfrac{1}{14} \).
So \( 3x + 4y = 14 \) ... (iv) and \( 4y - 3x = 26 \) ... (v). Adding: \( 8y = 40 \), y = 5. Subtracting (v) from (iv): \( 6x = -12 \), x = −2. Solution: x = −2, y = 5.

Example 7. Solve using the cross multiplication method: 3x + 2y + 25 = 0 and x + y + 15 = 0.
Answer:
\( a_1 = 3, b_1 = 2, c_1 = 25 \); \( a_2 = 1, b_2 = 1, c_2 = 15 \).
\( \dfrac{x}{2 \times 15 - 25 \times 1} = \dfrac{y}{25 \times 1 - 15 \times 3} = \dfrac{1}{3 \times 1 - 2 \times 1} \), that is, \( \dfrac{x}{5} = \dfrac{y}{-20} = \dfrac{1}{1} \).
So x = 5 and y = −20.

Check Your Progress: Methods of Solving

Check Your Progress 1. Solve graphically: 3x + 2y = 12 and 5x − 3y = 1.
Answer:
Table for 3x + 2y = 12: (0, 6), (2, 3), (4, 0). Table for 5x − 3y = 1: (2, 3), (−1, −2), (5, 8). The lines meet at (2, 3). x = 2, y = 3.

Check Your Progress 2. Ram is walking along the line y = 3x and Shyam is walking along the line y = 3x + 5. Represent the situation graphically. Will they meet?
Answer:
Both lines have slope 3 and different intercepts (0 and 5), so they are parallel (a_1/a_2 = b_1/b_2 = 1 ≠ c_1/c_2). No, they will not meet.

Check Your Progress 3. Solve using substitution: (i) 3x + 2y = 13, 5x − 3y = 9 (ii) 2x + 3y = 22, 3x − 2y = 7.
Answer:
(i) From the first, \( x = \dfrac{13 - 2y}{3} \); then \( 5 \cdot \dfrac{13 - 2y}{3} - 3y = 9 \) gives \( 65 - 10y - 9y = 27 \), so y = 2 and x = 3. x = 3, y = 2.
(ii) From the second, \( x = \dfrac{7 + 2y}{3} \); then \( 2 \cdot \dfrac{7 + 2y}{3} + 3y = 22 \) gives \( 14 + 4y + 9y = 66 \), y = 4 and x = 5. x = 5, y = 4.

Check Your Progress 4. Solve using elimination by equating the coefficients: (i) 11x − 6y = 28, 2x + 3y = 1 (ii) 6/x − 5/y = 22, 5/x + 4/y = 2.
Answer:
(i) From 2x + 3y = 1, 6y = 2 − 4x. Then 11x − (2 − 4x) = 28 gives 15x = 30, so x = 2 and y = −1. x = 2, y = −1. (Check: 11(2) − 6(−1) = 28 and 2(2) + 3(−1) = 1.)
(ii) Put \( \dfrac{1}{x} = u \) and \( \dfrac{1}{y} = v \): \( 6u - 5v = 22 \) and \( 5u + 4v = 2 \). Multiply by 4 and 5 and add: \( 24u - 20v + 25u + 20v = 88 + 10 \), \( 49u = 98 \), u = 2; then \( 4v = 2 - 10 = -8 \), v = −2. So x = 1/2, y = −1/2.

Check Your Progress 5. Solve using cross multiplication: (i) 4x + 5y + 9 = 0, 3x + 4y + 8 = 0 (ii) 8x − 7y = 19, 10x − 9y = 23.
Answer:
(i) \( \dfrac{x}{5 \cdot 8 - 4 \cdot 9} = \dfrac{y}{9 \cdot 3 - 8 \cdot 4} = \dfrac{1}{4 \cdot 4 - 3 \cdot 5} \), that is, \( \dfrac{x}{4} = \dfrac{y}{-5} = \dfrac{1}{1} \). x = 4, y = −5.
(ii) Write as 8x − 7y − 19 = 0 and 10x − 9y − 23 = 0: \( \dfrac{x}{(-7)(-23) - (-9)(-19)} = \dfrac{y}{(-19)(10) - (-23)(8)} = \dfrac{1}{8(-9) - 10(-7)} \), that is, \( \dfrac{x}{161 - 171} = \dfrac{y}{-190 + 184} = \dfrac{1}{-2} \), so x = 5 and y = 3. x = 5, y = 3.

Condition for Solvability

For the pair \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):

Solved Examples on Conditions for Solvability

Example 8. Find the value of p for which the system Px − y = 2 and 6x − 2y = 3 has only one solution.
Answer:
\( a_1 = p, b_1 = -1 \); \( a_2 = 6, b_2 = -2 \). For a unique solution \( \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \), so \( \dfrac{p}{6} \ne \dfrac{-1}{-2} \), which gives \( p \ne 3 \). So p can have all real values except 3.

Example 9. Find the value of k for which the system kx + 4y = k − 4 and 16x + ky = k has infinitely many solutions.
Answer:
We need \( \dfrac{k}{16} = \dfrac{4}{k} = \dfrac{k - 4}{k} \).
From \( \dfrac{k}{16} = \dfrac{4}{k} \): \( k^2 = 64 \), so k = ±8. From \( \dfrac{4}{k} = \dfrac{k - 4}{k} \) (k ≠ 0): \( 4 = k - 4 \), so k = 8.
Hence k = 8 (check: 8x + 4y = 4 and 16x + 8y = 8 are the same line).

Example 10. Determine the value of k so that (3k + 1)x + 3y − 2 = 0 and \( (k^2 + 1)x + (k - 2)y - 5 = 0 \) have no solution.
Answer:
For no solution, \( \dfrac{3k + 1}{k^2 + 1} = \dfrac{3}{k - 2} \ne \dfrac{-2}{-5} \).
\( (3k + 1)(k - 2) = 3(k^2 + 1) \Rightarrow 3k^2 - 5k - 2 = 3k^2 + 3 \Rightarrow -5k = 5 \Rightarrow k = -1 \).
For k = −1: \( \dfrac{3}{k - 2} = -1 \) and \( \dfrac{2}{5} \) are different, so the condition \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \) holds. Hence the system has no solution for k = −1.

Check Your Progress: Conditions for Solvability

Check Your Progress 1. Which of the following pairs of equations have exactly one solution, no solution or infinitely many solutions? (a) 5x − y + 6 = 0, 4x + 7y + 1 = 0 (b) 9x + 7y = 3, 7x + 9y = 3 (c) 2x − 5y = 6, 4x − 10y = 3 (d) x + 4y − 4 = 0, 3x + 12y = 12 (e) x + y = 10, x − y = 10.
Answer:
(a) \( \tfrac{5}{4} \ne \tfrac{-1}{7} \): unique solution. (b) \( \tfrac{9}{7} \ne \tfrac{7}{9} \): unique solution.
(c) \( \tfrac{2}{4} = \tfrac{-5}{-10} \ne \tfrac{6}{3} \): no solution.
(d) \( \tfrac{1}{3} = \tfrac{4}{12} = \tfrac{-4}{-12} \): infinitely many solutions. (e) \( \tfrac{1}{1} \ne \tfrac{1}{-1} \): unique solution.

Check Your Progress 2. Find the value of p if 6x + py = 5 and 3x + 4y = 2 have (i) a unique solution (ii) no solution.
Answer:
(i) \( \tfrac{6}{3} \ne \tfrac{p}{4} \), so \( p \ne 8 \). (ii) \( \tfrac{6}{3} = \tfrac{p}{4} \ne \tfrac{5}{2} \): p = 8 and 2 ≠ 5/2, so p = 8.

Check Your Progress 3. Find the value of k if 4x + 2y = k and 2x + 3y = 6 have a unique solution.
Answer:
\( \tfrac{4}{2} \ne \tfrac{2}{3} \) for every k, so the system has a unique solution for every real value of k.

Check Your Progress 4. Find the value of k if kx − y = 5 and x + 3y = 8 have infinitely many solutions.
Answer:
We would need \( \tfrac{k}{1} = \tfrac{-1}{3} = \tfrac{5}{8} \), but \( \tfrac{-1}{3} \ne \tfrac{5}{8} \). So there is no real value of k.

Check Your Progress 5. Find the value of k if 3x + ky = 2 and 5x + y = 7 have no solution.
Answer:
\( \tfrac{3}{5} = \tfrac{k}{1} \ne \tfrac{2}{7} \): This gives k = 3/5, and the common ratio 3/5 differs from 2/7, so the lines are parallel for k = 3/5.