CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Notes Set 03

Introduction

In Class IX, we have read about linear equations in two variables. A linear equation is a rational and integral equation of the first degree.
For example, the equations \( 3x + 2y = 7 \), \( 2x - \sqrt{3}\,y = \sqrt{5} \), \( y - 4x = \sqrt{3} \) are linear equations in two variables, since in each case
(i) Neither x nor y is under a radical sign, i.e., the variables x and y are rational.
(ii) Neither x nor y is in the denominator.
(iii) The exponent of x and y in each term is one.

In general, \( ax + by + c = 0 \), where \( a, b, c \in R \) and a and b are non-zero, is a linear equation in two variables. A linear equation in two variables has an infinite number of solutions. The graph of a linear equation in two variables is always a straight line. In this chapter, we study systems of linear equations in two variables, the solution of such systems, and the graphical and algebraic methods of solving them. At the end of the chapter we discuss some applications of linear equations in two variables to simple problems.

Historical Facts

Diophantus, the last genius of Alexandria and the best algebraic mathematician of the Greco-Roman era, made a unique contribution to the development of algebra and the history of mathematics. He was born in the 3rd century and lived for 84 years. His age is described in a Greek collection of epigrams:
“He spent one-sixth of his life in childhood, his beard grew after one-twelfth more, after another one-seventh he married, five years later his son was born, the son lived to half the father’s age, and Diophantus died four years after his son.”
Let his age be x years. Then \( \frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x \Rightarrow 9x = 756 \Rightarrow x = 84 \) years.
He is known as the father of algebra. Arithmetica is his famous book.

Recall

(i) Equation: A statement of equality of two algebraic expressions which involve one or more unknown quantities is known as an equation.
(ii) Linear Equation: An equation in which the maximum power of the variable is one is called a linear equation.
(iii) Linear Equation in One Variable: An equation of the form \( ax + b = 0 \), where x is a variable, a and b are real numbers and \( a \neq 0 \), is called a linear equation in one variable.
(iv) Linear Equation in Two Variables: An equation of the form \( ax + by + c = 0 \), where a, b, c are real numbers, \( a \neq 0,\ b \neq 0 \) and x, y are variables, is called a linear equation in two variables. Any pair of values of x and y which satisfies the equation \( ax + by + c = 0 \) is called a root or solution of it.
Example: (x = 1, y = 1) is a solution of \( 4x - y - 3 = 0 \).
Remark: A linear equation in two variables has an infinite number of solutions.

(v) Graph of a Linear Equation in Two Variables: Assume \( y - x = 2 \) is a linear equation in two variables. The following table gives the abscissa and ordinate of points on the line represented by the equation \( y - x = 2 \).

Plotting the points (1, 3), (2, 4) and (3, 5) on the graph paper and drawing the line joining them, we obtain the graph of the line represented by the given equation, as shown in the figure.

Simultaneous Linear Equations in Two Variables

A pair of linear equations in two variables is said to form a system of simultaneous linear equations.
General form:

where \( a_1, a_2, b_1, b_2, c_1, c_2 \) are real numbers, \( a_1^2 + b_1^2 \neq 0 \) and \( a_2^2 + b_2^2 \neq 0 \), and x, y are variables.
Each of the following pairs of linear equations forms a system of two simultaneous linear equations in two variables:
(i) \( x - 2y = 3,\ 2x + 5y = 5 \)    (ii) \( 3x + 5y + 7 = 0,\ 5x + 2y + 9 = 0 \)

Solution of the System of Equations

Consider the system of simultaneous linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \). A pair of values of the variables x and y satisfying each one of the equations in a given system of two simultaneous linear equations in x and y is called a solution of the system.

Example: x = 2, y = 3 is a solution of the system of simultaneous linear equations \( 2x + y = 7,\ 3x + 2y = 12 \).
The given equations are \( 2x + y = 7 \) ...(i) and \( 3x + 2y = 12 \) ...(ii).
Put x = 2, y = 3 in the LHS of equation (i): LHS = 2 × 2 + 3 = 7 = RHS.
Put x = 2, y = 3 in the LHS of equation (ii): LHS = 3 × 2 + 2 × 3 = 12 = RHS.
The values x = 2, y = 3 satisfy both equations (i) and (ii). Hence x = 2, y = 3 is a solution of the given system.

A system of simultaneous linear equations can have:

Remark: An equation involving two variables cannot give the values of both the variables. For the values of both the variables we require two equations. Similarly, for three variables we require three equations, and so on, i.e., to find n variables we need n equations.

Homogeneous System of Equations

A system of simultaneous equations is said to be homogeneous if all of the constant terms are zero.
General form: \( a_1x + b_1y = 0 \) and \( a_2x + b_2y = 0 \).
A homogeneous equation of the form \( ax + by = 0 \) is a line passing through the origin. Therefore, the system is always consistent.
(i) When \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the system of equations has only one solution (x = 0, y = 0).
(ii) When \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system of equations has infinitely many solutions.

Example 1. On comparing the ratios \( \frac{a_1}{a_2},\ \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find out whether the following pairs of linear equations are consistent or inconsistent.
(i) \( 3x + 2y = 5,\ 2x - 3y = 7 \)    (ii) \( 2x - 3y = 8,\ 4x - 6y = 9 \)
Answer:
(i) We have \( 3x + 2y = 5 \Rightarrow 3x + 2y - 5 = 0 \) and \( 2x - 3y = 7 \Rightarrow 2x - 3y - 7 = 0 \).
\[ \frac{a_1}{a_2} = \frac{3}{2},\quad \frac{b_1}{b_2} = \frac{2}{-3},\quad \frac{c_1}{c_2} = \frac{5}{7} \]
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Therefore, the given pair of linear equations is consistent.
(ii) We have \( 2x - 3y = 8 \Rightarrow 2x - 3y - 8 = 0 \) and \( 4x - 6y = 9 \Rightarrow 4x - 6y - 9 = 0 \).
\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2},\quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2},\quad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9} \]
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Therefore, the given pair of linear equations is inconsistent.

Example 2. For what value of k does the system of equations \( x + 2y = 5,\ 3x + ky + 15 = 0 \) have (i) a unique solution (ii) no solution?
Answer:
We have \( x + 2y - 5 = 0 \) and \( 3x + ky + 15 = 0 \).
(i) The condition for a unique solution is \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
\[ \frac{1}{3} \neq \frac{2}{k} \Rightarrow k \neq 6 \]
Hence, for all real values of k except 6, the given system of equations has a unique solution.
(ii) The condition for no solution is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
\[ \frac{1}{3} = \frac{2}{k} \neq \frac{-5}{15} \]
From \( \frac{1}{3} = \frac{2}{k} \) we get k = 6, and from \( \frac{2}{k} \neq \frac{-1}{3} \) we get \( k \neq -6 \).
Hence the given system of equations has no solution when k = 6.

Example 3. Find the value of k for which the system of equations \( 4x + 5y = 0,\ kx + 10y = 0 \) has infinitely many solutions.
Answer:
The given system is of the form \( a_1x + b_1y = 0,\ a_2x + b_2y = 0 \) with \( a_1 = 4,\ a_2 = k,\ b_1 = 5,\ b_2 = 10 \).
If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system has infinitely many solutions.
\[ \frac{4}{k} = \frac{5}{10} \Rightarrow k = 8 \]

Example 4. Find the values of a and b for which the given system of equations has an infinite number of solutions:
\( 2x + 3y = 7;\ (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1 \)
Answer:
We have \( 2x + 3y - 7 = 0 \) and \( (a + b + 1)x + (a + 2b + 2)y - \{4(a + b) + 1\} = 0 \).
The condition for an infinite number of solutions is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
\[ \frac{2}{a + b + 1} = \frac{3}{a + 2b + 2} = \frac{-7}{-\{4(a + b) + 1\}} \]
\[ \Rightarrow \frac{2}{a + b + 1} = \frac{3}{a + 2b + 2} \quad \text{and} \quad \frac{3}{a + 2b + 2} = \frac{7}{4(a + b) + 1} \]
\( \Rightarrow 2a + 4b + 4 = 3a + 3b + 3 \) and \( 12a + 12b + 3 = 7a + 14b + 14 \)
\( \Rightarrow a - b = 1 \) ...(i) and \( 5a - 2b = 11 \) ...(ii)
Multiplying (i) by 2, we get \( 2a - 2b = 2 \) ...(iii)
Subtracting (iii) from (ii), we get \( 3a = 9 \Rightarrow a = 3 \).
Put a = 3 in (i): \( 3 - b = 1 \Rightarrow b = 2 \).
Hence, the given system of equations has infinitely many solutions when a = 3 and b = 2.

Graphical Method of Solving a System of Simultaneous Linear Equations

To solve a system of two linear equations graphically:
(i) Draw the graph of the first equation.
(ii) On the same pair of axes, draw the graph of the second equation.
(iii)(a) If the two lines intersect at a point, read the coordinates of the point of intersection to obtain the solution, and verify your answer.
(b) If the two lines are parallel, there is no point of intersection. Write the system as inconsistent; hence there is no solution.
(c) If the two lines have the same graph, then write the system as consistent with an infinite number of solutions.

Algebraic Method of Solving a Pair of Linear Equations in Two Variables

Sometimes the graphical method does not give an accurate answer. While reading the coordinates of a point on a graph paper we are likely to make an error. So we require some precise method to obtain an accurate result. The algebraic methods are given below:
(i) Method of elimination by substitution.
(ii) Method of elimination by equating the coefficients.
(iii) Method of cross-multiplication.

Algebraic Solution by Substitution Method

To solve a pair of linear equations in two variables x and y by the substitution method, we follow these steps:
Step I: Write the given equations \( a_1x + b_1y + c_1 = 0 \) ...(i) and \( a_2x + b_2y + c_2 = 0 \) ...(ii).
Step II: Choose one of the two equations and express y in terms of x (or x in terms of y), i.e., express one variable in terms of the other.
Step III: Substitute this value of y obtained in Step II in the other equation to get a linear equation in x.
Step IV: Solve the linear equation obtained in Step III and get the value of x.
Step V: Substitute this value of x in the relation obtained in Step II and find the value of y.

Example 6. Solve for x and y: \( 4x + 3y = 24,\ 3y - 2x = 6 \).
Answer:
\( 4x + 3y = 24 \) ...(i)    \( 3y - 2x = 6 \) ...(ii)
From equation (i), we get \( y = \frac{24 - 4x}{3} \) ...(iii)
Substituting in equation (ii), we get \( 3\left(\frac{24 - 4x}{3}\right) - 2x = 6 \)
\[ \Rightarrow 24 - 4x - 2x = 6 \Rightarrow -6x = -18 \Rightarrow x = 3 \]
Substituting x = 3 in (iii), we get \( y = \frac{24 - 12}{3} = \frac{12}{3} = 4 \).
Hence, x = 3, y = 4.

Example 7. Solve the following pair of linear equations by the substitution method: \( \sqrt{2}x + \sqrt{3}y = 0 \) and \( \sqrt{3}x - \sqrt{8}y = 0 \).
Answer:
\( \sqrt{2}x + \sqrt{3}y = 0 \) ...(i)    \( \sqrt{3}x - \sqrt{8}y = 0 \) ...(ii)
From (i), we get \( y = \frac{-\sqrt{2}x}{\sqrt{3}} \) ...(iii)
Substituting this value of y in (ii), we get \( \sqrt{3}x - \sqrt{8}\left(\frac{-\sqrt{2}x}{\sqrt{3}}\right) = 0 \)
\[ \Rightarrow \sqrt{3}x + \frac{4x}{\sqrt{3}} = 0 \Rightarrow 3x + 4x = 0 \Rightarrow 7x = 0 \Rightarrow x = 0 \]
Substituting x = 0 in (iii), we get \( y = \frac{-\sqrt{2} \times 0}{\sqrt{3}} = 0 \).
Hence, the solution is x = 0, y = 0.

Algebraic Solution by Elimination Method

To solve a pair of linear equations in x and y by the elimination method, we follow these steps:
Step I: Write the given equations \( a_1x + b_1y + c_1 = 0 \) ...(i) and \( a_2x + b_2y + c_2 = 0 \) ...(ii).
Step II: Multiply the given equations by suitable numbers so that the coefficients of one of the variables become numerically equal.
Step III: If the numerically equal coefficients are opposite in sign, add the new equations; otherwise subtract them.
Step IV: Solve the linear equation in one variable obtained in Step III and get the value of one variable.
Step V: Substitute the value of the variable obtained in Step IV in either of the two equations and find the value of the other variable.

Example 8. Solve the following pair of linear equations by the elimination method: \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \).
Answer:
\( 3x + 4y = 10 \) ...(i)    \( 2x - 2y = 2 \) ...(ii)
Multiplying (ii) by 2, we get \( 4x - 4y = 4 \) ...(iii)
Adding (i) and (iii), we get \( 7x = 14 \Rightarrow x = 2 \).
Putting x = 2 in (ii), we get \( 2 \times 2 - 2y = 2 \Rightarrow y = 1 \).
Hence, the solution is x = 2, y = 1.

Example 9. Solve: \( ax + by = c,\ bx + ay = 1 + c \).
Answer:
\( ax + by = c \) ...(i)    \( bx + ay = 1 + c \) ...(ii)
Adding (i) and (ii), we get \( (a + b)x + (a + b)y = 2c + 1 \Rightarrow x + y = \frac{2c + 1}{a + b} \) ...(iii)
Subtracting (ii) from (i), we get \( (a - b)x - (a - b)y = -1 \Rightarrow x - y = \frac{-1}{a - b} \) ...(iv)
Adding (iii) and (iv), we get
\[ 2x = \frac{2c + 1}{a + b} - \frac{1}{a - b} = \frac{(2c + 1)(a - b) - (a + b)}{a^2 - b^2} = \frac{2ac - 2bc - 2b}{a^2 - b^2} \]
\[ \Rightarrow x = \frac{ac - bc - b}{a^2 - b^2} \]
Subtracting (iv) from (iii), we get
\[ 2y = \frac{2c + 1}{a + b} + \frac{1}{a - b} = \frac{(2c + 1)(a - b) + (a + b)}{a^2 - b^2} = \frac{2ac - 2bc + 2a}{a^2 - b^2} \]
\[ \Rightarrow y = \frac{ac - bc + a}{a^2 - b^2} \]
Hence, \( x = \frac{ac - bc - b}{a^2 - b^2} \) and \( y = \frac{ac - bc + a}{a^2 - b^2} \), where \( a^2 \neq b^2 \).

Algebraic Solutions by Cross-Multiplication Method

Consider the system of linear equations \( a_1x + b_1y + c_1 = 0 \) ...(i) and \( a_2x + b_2y + c_2 = 0 \) ...(ii). To solve it by the cross-multiplication method, we follow these steps.
Step I: Write the coefficients as follows:

The arrows between the two numbers indicate that they are to be multiplied. The products with upward arrows are to be subtracted from the products with downward arrows. To apply the above formula, all the terms must be on the left of the equal sign in the system of equations. Equations (i) and (ii) then give

Case I: If \( a_1b_2 - a_2b_1 \neq 0 \), then x and y have finite values and the system of equations has a unique solution.
Case II: If \( a_1b_2 - a_2b_1 = 0 \), i.e., \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), two cases arise.
(a) If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda\ (\lambda \neq 0) \), then \( a_1 = a_2\lambda,\ b_1 = b_2\lambda,\ c_1 = c_2\lambda \). Putting these values in equation (i):
\( a_2\lambda x + b_2\lambda y + c_2\lambda = 0 \Rightarrow \lambda(a_2x + b_2y + c_2) = 0 \), but \( \lambda \neq 0 \), so \( a_2x + b_2y + c_2 = 0 \) ...(ii).
So (i) and (ii) are dependent, and there is an infinite number of solutions.
(b) If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then \( a_1b_2 - a_2b_1 = 0 \), but \( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \) and \( y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \) have a non-zero numerator, so \( x = \frac{\text{finite value}}{0} \) and \( y = \frac{\text{finite value}}{0} \) do not exist. So the system of equations is inconsistent.

Example 10. Solve by the cross-multiplication method: \( x + 2y + 1 = 0 \) and \( 2x - 3y - 12 = 0 \).
Answer:
We have \( x + 2y + 1 = 0 \) and \( 2x - 3y - 12 = 0 \). By the cross-multiplication method, we have

\[ \frac{x}{2 \times (-12) - (-3) \times 1} = \frac{y}{1 \times 2 - (-12) \times 1} = \frac{1}{1 \times (-3) - 2 \times 2} \]
\[ \Rightarrow \frac{x}{-24 + 3} = \frac{y}{2 + 12} = \frac{1}{-3 - 4} \Rightarrow \frac{x}{-21} = \frac{y}{14} = \frac{1}{-7} \]
\[ \Rightarrow x = \frac{-21}{-7} = 3 \quad \text{and} \quad y = \frac{14}{-7} = -2 \]
Hence, the solution is x = 3, y = −2.

Example 11. Solve by the cross-multiplication method: \( (a - b)x + (a + b)y = 2(a^2 - b^2),\ (a + b)x - (a - b)y = 4ab \).
Answer:
Writing the equations in the standard form, we get
\[ (a - b)x + (a + b)y - 2(a^2 - b^2) = 0 \]
\[ (a + b)x - (a - b)y - 4ab = 0 \]
Applying the cross-multiplication method, we get
\[ \frac{x}{(a + b)(-4ab) - \{-(a - b)\}\{-2(a^2 - b^2)\}} = \frac{y}{\{-2(a^2 - b^2)\}(a + b) - (-4ab)(a - b)} = \frac{1}{(a - b)\{-(a - b)\} - (a + b)(a + b)} \]
Simplification of the expression under x:
\[ -4ab(a + b) - 2(a - b)(a^2 - b^2) = -2(a + b)\left[2ab + (a - b)^2\right] = -2(a + b)(a^2 + b^2) \]
Simplification of the expression under y:
\[ -2(a^2 - b^2)(a + b) + 4ab(a - b) = -2(a - b)\left[(a + b)^2 - 2ab\right] = -2(a - b)(a^2 + b^2) \]
Simplification of the expression under 1:
\[ -(a - b)^2 - (a + b)^2 = -2(a^2 + b^2) \]
Hence,
\[ \frac{x}{-2(a + b)(a^2 + b^2)} = \frac{y}{-2(a - b)(a^2 + b^2)} = \frac{1}{-2(a^2 + b^2)} \]
\[ \Rightarrow \frac{x}{a + b} = \frac{y}{a - b} = \frac{1}{1} \]
Hence, x = a + b, y = a − b.

Equations of the Form ax + by = c and bx + ay = d, where a ≠ b

To solve the equations \( ax + by = c \) ...(i) and \( bx + ay = d \) ...(ii), where \( a \neq b \), we follow these steps:
Step I: Add (i) and (ii) and obtain \( (a + b)x + (b + a)y = c + d \), i.e., \( x + y = \frac{c + d}{a + b} \) ...(iii)
Step II: Subtract (ii) from (i) and obtain \( (a - b)x - (a - b)y = c - d \), i.e., \( x - y = \frac{c - d}{a - b} \) ...(iv)
Step III: Solve (iii) and (iv) to get x and y.

Example 12. Solve for x and y: \( 47x + 31y = 63,\ 31x + 47y = 15 \).
Answer:
\( 47x + 31y = 63 \) ...(i) and \( 31x + 47y = 15 \) ...(ii)
Adding (i) and (ii), we get \( 78x + 78y = 78 \Rightarrow x + y = 1 \) ...(iii)
Subtracting (ii) from (i), we get \( 16x - 16y = 48 \Rightarrow x - y = 3 \) ...(iv)
Adding (iii) and (iv), we get \( 2x = 4 \Rightarrow x = 2 \).
Putting x = 2 in (iii), we get \( 2 + y = 1 \Rightarrow y = -1 \).
Hence, the solution is x = 2, y = −1.

Equations Reducible to Linear Equations in Two Variables

Equations which contain the variables only in the denominators are called reciprocal equations. These equations can be of the following types and can be solved by the methods mentioned below.
Type I: \( \frac{a}{u} + \frac{b}{v} = c \) and \( \frac{a'}{u} + \frac{b'}{v} = c' \), where a, b, c, a', b', c' are real numbers. Put \( \frac{1}{u} = x \) and \( \frac{1}{v} = y \), and find the values of x and y by any method described earlier. Then \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \).
Type II: \( au + bv = cuv \) and \( a'u + b'v = c'uv \), where a, b, c, a', b', c' are real numbers. Divide both equations by uv, and the equations can be converted to the form explained in Type I.
Type III: \( \frac{a}{lx + my} + \frac{b}{cx + dy} = k \) and \( \frac{a'}{lx + my} + \frac{b'}{cx + dy} = k' \). Put \( \frac{1}{lx + my} = u \) and \( \frac{1}{cx + dy} = v \). Then the equations become \( au + bv = k \) and \( a'u + b'v = k' \). Find the values of u and v, and put \( lx + my = \frac{1}{u} \) and \( cx + dy = \frac{1}{v} \). Again solve for x and y by any method explained earlier.

Example 13. Solve for x and y: \( \frac{3a}{x} - \frac{2b}{y} + 5 = 0 \) and \( \frac{a}{x} + \frac{3b}{y} - 2 = 0 \) (\( x \neq 0,\ y \neq 0 \)).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations can be written as
\( 3au - 2bv = -5 \) ...(i)    \( au + 3bv = 2 \) ...(ii)
Multiplying (i) by 3 and (ii) by 2, we get \( 9au - 6bv = -15 \) ...(iii) and \( 2au + 6bv = 4 \) ...(iv)
Adding (iii) and (iv), we get \( 11au = -11 \Rightarrow u = \frac{-1}{a} \).
Put \( u = \frac{-1}{a} \) in (ii): \( a\left(\frac{-1}{a}\right) + 3bv = 2 \Rightarrow 3bv = 3 \Rightarrow v = \frac{1}{b} \).
But \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Therefore \( \frac{1}{x} = \frac{-1}{a} \Rightarrow x = -a \) and \( \frac{1}{y} = \frac{1}{b} \Rightarrow y = b \).
Hence, the solution is x = −a, y = b.

Example 14. Solve \( \frac{57}{x + y} + \frac{6}{x - y} = 5 \) and \( \frac{38}{x + y} + \frac{21}{x - y} = 9 \).
Answer:
Let \( \frac{1}{x + y} = p \) and \( \frac{1}{x - y} = q \). Then the given equations can be written as
\( 57p + 6q - 5 = 0 \) and \( 38p + 21q - 9 = 0 \).
By the cross-multiplication method, we have

\[ \frac{p}{6 \times (-9) - 21 \times (-5)} = \frac{q}{(-5) \times 38 - (-9) \times 57} = \frac{1}{57 \times 21 - 38 \times 6} \]
\[ \Rightarrow \frac{p}{-54 + 105} = \frac{q}{-190 + 513} = \frac{1}{1197 - 228} \Rightarrow \frac{p}{51} = \frac{q}{323} = \frac{1}{969} \]
\[ \Rightarrow p = \frac{51}{969} = \frac{1}{19} \quad \text{and} \quad q = \frac{323}{969} = \frac{1}{3} \]
But \( \frac{1}{x + y} = p \) and \( \frac{1}{x - y} = q \). Therefore \( \frac{1}{x + y} = \frac{1}{19} \Rightarrow x + y = 19 \) ...(i) and \( \frac{1}{x - y} = \frac{1}{3} \Rightarrow x - y = 3 \) ...(ii).
Adding (i) and (ii), we get \( 2x = 22 \Rightarrow x = 11 \). Put x = 11 in (i): \( 11 + y = 19 \Rightarrow y = 8 \).
Hence, the solution is x = 11, y = 8.

Example 15. Solve for x and y: \( \frac{7x - 2y}{xy} = 5,\ \frac{8x + 7y}{xy} = 15 \).
Answer:
\( \frac{7x - 2y}{xy} = 5 \Rightarrow \frac{7}{y} - \frac{2}{x} = 5 \) ...(i)
\( \frac{8x + 7y}{xy} = 15 \Rightarrow \frac{8}{y} + \frac{7}{x} = 15 \) ...(ii)
Putting \( \frac{1}{y} = u \) and \( \frac{1}{x} = v \), we get \( 7u - 2v = 5 \) ...(iii) and \( 8u + 7v = 15 \) ...(iv)
Multiplying (iii) by 7 and (iv) by 2 and adding, we get \( 49u - 14v = 35 \) and \( 16u + 14v = 30 \), so \( 65u = 65 \Rightarrow u = 1 \), i.e., \( \frac{1}{y} = 1 \) or y = 1.
Substituting u = 1 in (iii), we get \( 7 - 2v = 5 \Rightarrow v = 1 \), i.e., \( \frac{1}{x} = 1 \) or x = 1.
Hence, x = 1, y = 1.

Applications of Linear Equations in Two Variables

In this section, we study some applications of simultaneous linear equations in solving a variety of word problems related to day-to-day life situations. The following examples are self-explanatory and give some insight into the solution of such problems.

Type I: Based on Articles and Their Costs / Quantities

Example 16. 7 audio cassettes and 3 video cassettes cost Rs. 1110, while 5 audio cassettes and 4 video cassettes cost Rs. 1350. Find the cost of an audio cassette and a video cassette.
Answer:
Let the cost of an audio cassette and a video cassette be Rs. x and Rs. y respectively.
The cost of 7 audio cassettes and 3 video cassettes is Rs. 1110, so \( 7x + 3y = 1110 \) ...(i)
The cost of 5 audio cassettes and 4 video cassettes is Rs. 1350, so \( 5x + 4y = 1350 \) ...(ii)
Multiplying (i) by 4 and (ii) by 3, we get \( 28x + 12y = 4440 \) ...(iii) and \( 15x + 12y = 4050 \) ...(iv)
Subtracting (iv) from (iii), we get \( 13x = 390 \Rightarrow x = 30 \).
Putting x = 30 in (i), we get \( 210 + 3y = 1110 \Rightarrow 3y = 900 \Rightarrow y = 300 \).
Hence, the cost of an audio cassette is Rs. 30 and that of a video cassette is Rs. 300.

Type II: Based on Numbers

Example 17. The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.
Answer:
Let the digit at the ten’s place be x and that at the unit’s place be y. Then \( x + y = 12 \) ...(i)
The two-digit number is \( 10x + y \). According to the question,
\[ (10y + x) = (10x + y) + 18 \Rightarrow 9y - 9x = 18 \Rightarrow y - x = 2 \]
...(ii)
Adding (i) and (ii), we get \( 2y = 14 \Rightarrow y = 7 \). Put y = 7 in (i): \( x + 7 = 12 \Rightarrow x = 5 \).
Hence, the required number is \( 10 \times 5 + 7 \), i.e., 57.

Type III: Based on Fractions

Example 18. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
Answer:
Let the required fraction be \( \frac{x}{y} \). Then
\[ \frac{x + 1}{y - 1} = 1 \Rightarrow x + 1 = y - 1 \Rightarrow x - y = -2 \] ...(i)
\[ \frac{x}{y + 1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x - y = 1 \] ...(ii)
Subtracting (i) from (ii), we get x = 3. Put x = 3 in (i): \( 3 - y = -2 \Rightarrow y = 5 \).
Hence, the fraction is \( \frac{3}{5} \).

Type IV: Based on Ages

Example 19. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of the father and the son.
Answer:
Let the present ages of the father and the son be x years and y years respectively.
Two years ago, the father’s age = (x − 2) years and the son’s age = (y − 2) years.
Therefore \( (x - 2) = 5(y - 2) \Rightarrow x - 5y = -8 \) ...(i)
Two years later, the father’s age = (x + 2) years and the son’s age = (y + 2) years.
Therefore \( (x + 2) = 3(y + 2) + 8 \Rightarrow x + 2 = 3y + 6 + 8 \Rightarrow x - 3y = 12 \) ...(ii)
Subtracting (i) from (ii), we get \( 2y = 20 \Rightarrow y = 10 \). Putting y = 10 in (ii), we get \( x - 30 = 12 \Rightarrow x = 42 \).
Hence, the present ages of the father and the son are 42 years and 10 years respectively.

Type V: Based on Geometrical Applications

Example 20. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer:
Let the larger angle be \( x^\circ \) and the smaller angle be \( y^\circ \). Then
\( x + y = 180 \) ...(i) and \( x = y + 18 \Rightarrow x - y = 18 \) ...(ii)
Adding (i) and (ii), we get \( 2x = 198 \Rightarrow x = 99 \). Putting x = 99 in (i), we get \( 99 + y = 180 \Rightarrow y = 81 \).
Hence, the required angles are 99° and 81°.

Type VI: Based on Time, Distance and Speed

Formulae to be used:
1. (a) \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)   (b) \( \text{Distance} = \text{Speed} \times \text{Time} \)   (c) \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
2. Let the speed of a boat in still water = u km/h and the speed of the current = v km/h. Then (a) the speed of the boat downstream = (u + v) km/h, and (b) the speed of the boat upstream = (u − v) km/h.

Example 21. A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer:
Let the speeds of the train and the car be x km/h and y km/h respectively. Time = Distance / Speed.
In the first case he covers 250 km by train and 120 km by car, so \( \frac{250}{x} + \frac{120}{y} = 4 \) ...(i)
In the second case he covers 130 km by train and 240 km by car, and takes 4 hours 18 minutes, where 18 minutes = \( \frac{18}{60} \) hour. So
\[ \frac{130}{x} + \frac{240}{y} = 4 + \frac{18}{60} = \frac{43}{10} \]...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then \( 250u + 120v = 4 \) ...(iii) and \( 130u + 240v = \frac{43}{10} \) ...(iv)
Multiplying (iii) by 2, we get \( 500u + 240v = 8 \) ...(v)
Subtracting (iv) from (v), we get \( 370u = 8 - \frac{43}{10} = \frac{37}{10} \Rightarrow u = \frac{1}{100} \).
Putting \( u = \frac{1}{100} \) in (iii), we get \( \frac{5}{2} + 120v = 4 \Rightarrow 120v = \frac{3}{2} \Rightarrow v = \frac{1}{80} \).
Therefore \( \frac{1}{x} = \frac{1}{100} \Rightarrow x = 100 \) and \( \frac{1}{y} = \frac{1}{80} \Rightarrow y = 80 \).
Hence, the speed of the train is 100 km/h and that of the car is 80 km/h.

Type VII: Miscellaneous

Example 22. 8 men and 12 boys can finish a piece of work in 10 days, while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
Answer:
Let one man alone finish the work in x days and one boy alone finish it in y days. Then the work done by one man in one day = \( \frac{1}{x} \) and the work done by one boy in one day = \( \frac{1}{y} \).
According to the question, \( 10\left(\frac{8}{x} + \frac{12}{y}\right) = 1 \Rightarrow \frac{8}{x} + \frac{12}{y} = \frac{1}{10} \) ...(i)
Also, \( 14\left(\frac{6}{x} + \frac{8}{y}\right) = 1 \Rightarrow \frac{6}{x} + \frac{8}{y} = \frac{1}{14} \) ...(ii)
Multiplying (i) by 2 and (ii) by 3, we get \( \frac{16}{x} + \frac{24}{y} = \frac{1}{5} \) ...(iii) and \( \frac{18}{x} + \frac{24}{y} = \frac{3}{14} \) ...(iv)
Subtracting (iii) from (iv), we get \( \frac{2}{x} = \frac{3}{14} - \frac{1}{5} = \frac{1}{70} \Rightarrow x = 140 \).
Putting x = 140 in (ii), we get \( \frac{6}{140} + \frac{8}{y} = \frac{1}{14} \Rightarrow \frac{8}{y} = \frac{1}{14} - \frac{3}{70} = \frac{2}{70} = \frac{1}{35} \Rightarrow y = 280 \).
Hence, one man alone can finish the work in 140 days and one boy alone can finish it in 280 days.

Synopsis

(i) Two linear equations in the same two variables are called a pair of linear equations in two variables, or briefly, a linear pair. The most general form of a linear pair is \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), where \( a_1, a_2, b_1, b_2, c_1, c_2 \) are real numbers such that \( a_1^2 + b_1^2 \neq 0 \) and \( a_2^2 + b_2^2 \neq 0 \).
(ii) A pair of linear equations in two variables can be represented and solved by (a) the graphical method or (b) the algebraic method.
(iii) Graphical method: The graph of the pair of linear equations in two variables is represented by a pair of lines.
(a) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair is consistent.
(b) If the lines coincide, then there are infinitely many solutions, each point on the line being a solution. In this case, the pair is consistent (dependent).
(c) If the lines are parallel, then the pair has no solution and is called inconsistent.
(iv) Algebraic method: We have discussed the following methods for finding the solutions of a pair of linear equations: (a) substitution method, (b) elimination method, (c) cross-multiplication method.
(v) If \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), then the following situations can arise:
(a) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \): in this case the pair of linear equations is consistent (unique solution).
(b) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \): in this case the pair of linear equations is inconsistent.
(c) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \): in this case the pair of linear equations is dependent and consistent.
(vi) There are several situations which can be mathematically represented by two equations that are not linear to start with. But we alter them so that they are reduced to a linear pair.