CBSE Class 10 Mathematics Pair Of Linear Equations In Two Variables Set 02

TOPICS

1. Simultaneous linear equations in two variables.

2. Graphical representation of linear equations.

3. Methods of solving simultaneous linear equations in two variables:

• (a) Graphically
• (b) Algebraically
  • Substitution Method
  • Elimination Method
  • Cross-Multiplication Method

4. Conditions of solvability (OR consistency).

5. Applications or word problems.

INTRODUCTION

In IX class the students have learnt to solve linear equations in one variable. In this chapter we have to discuss linear equations in two variables.

In our daily life we come across many situations in which equations in two variables are required, such as, Shivani went to Sunday market with her friends. They wanted to eat 'gol-gappa' as well as "dahi-bhalla". They took number of plates of "gol-gappa" twice that of "dahi-bhalla". The cost of one plate of gol-gappa was Rs. 5 and cost of one plate of dahi-bhalla was Rs. 10. She spent Rs. 60 which she had spent for market. How would you find out the number of plates of "gol-gappa" and "dahi-bhalla" they took.

We will try to solve the problem with the help of linear equations in two variables. Let the number of plates of gol-gappa that Shivani and her friends took be \( x \) and the number of plates of dahi-bhalla they took be \( y \).

Now the situation can be represented by two equations.
\( x = 2y \) ...(1)
and \( 5x + 10y = 60 \) ...(2)
There are several ways to solve this pair of linear equations.

Basic important concepts to understand

1. Term: A Combination of 4 mathematical elements
(i) Sign \( \oplus \) or \( \ominus \)
(ii) Number (any real number)
(iii) Variable (\( a, b, c, x, y, z, p, \alpha, r \) ...... etc)
(iv) Exponents i.e. powers is known as the term
for example: \( 3x^2 \) is a term because it has all 4 elements required for term i.e. sign \( \oplus \) number 3, variable \( x \), power 2.

Similarly \( 5y^2 \), \( \sqrt{\frac{7}{3}}x \), \( -\frac{1}{40}y \) are some examples of terms.
Any real number say \( \left( \frac{1}{3} \right) \) is also a term because \( \left( \frac{1}{3} \right) \) can be written as \( \frac{1}{3} \times 1 = \frac{1}{3} \times (x)^0 \).
Hence \( \left( \frac{1}{3} \right) \) is also a term because its sign is \( \oplus \), number is \( \left( \frac{1}{3} \right) \), Variable \( x \) and power zero.

Note: All real numbers are known as constant term.

2. Algebraic Expression: A combination of constant and variables, connected by four fundamental arithmetical operations of \( +, -, \times \) and \( \div \) is called algebraic expression.
For example, \( 3 - 7x \), \( 8 + 9y^2 \), \( 6 - 7a + 8b \) are algebraic expressions.

Equation: An algebraic expression is equated to some constant, variable or other algebric expression than that relation is known as equation, for example \( 3x + 7 = 10 \), \( x^2 - 7x + 10 = 0 \), \( 9x - 7y = 10x + 12 \), \( xy + 7 = 20z \).

Linear Equation: If the greatest exponent of the variable(s) in an equation is one, the equation is said to be a linear equation(s).
If the number of variables used in linear equation is one then equation is said to be linear equation in one variable.
For example, \( 3x + 11 = 0 \), \( 9y - 7 = 12 \), \( 6x - 12 = 18 \), \( 3t - 7 = 5 \) are some examples.
Linear equation in one variable is \( ax + b = 0 \) with \( a \neq 0 \), and solution of this equation is the value of \( x \) which makes L.H.S. and R.H.S. equal. In the case of \( ax + b = 0 \), \( x = -\frac{b}{a} \) is its solution.

Linear equation in two variables: Equation of the form: \( ax + by + c = 0 \) with \( a \neq 0 \), \( b \neq 0 \) is known as linear equation in two variables. For example \( 3x - 2y - 1 = 0 \) or \( 3x - 2y = 1 \), \( x + y = 2 \) or \( x + y - 2 = 0 \) are few examples of linear equations in two variables.
As discussed in class IX that graph of linear equation is a straight line so a linear equation in two variable has infinitely many solutions because graph of such equations is a line and a line passes through infinitely many points and each point lies on line is known as its solution so a linear equation in two variables has infinitely many solutions.

Dependent and Independent Variable of a linear equation in two variables: It Can be made very clear by taking an example of linear equation in two variables.
Let \( 3x + 2y = 5 \) is a linear equation in two variables so we can express \( x \) in terms of \( y \) like: \( x = \frac{5 - 2y}{3} \) then in this situation we assume our own the values of \( y \) to get the values of \( x \). Hence here \( y \) is known as independent variable because we are assuming the values of \( y \). So value of \( x \) depends on the value of \( y \) assumed, hence \( x \) is known as dependent variable.
Now for same linear equation in two variables \( 3x + 2y = 5 \).
If \( y \) is expressed in the terms of \( x \) then \( y = \frac{5 - 3x}{2} \).
In this situation we take (assumes) value of \( x \) and for each value of \( x \) then value of \( y \) depend on the value of \( x \) assumed.
Hence in this situation \( x \) is independent variable and \( y \) is dependent variable.
So in a linear equation of type : \( ax + by = c \) (\( a \neq 0, b \neq 0 \)).
Case I If \( x \) is expressed in the term of \( y \) then \( x \) is dependent variable and \( y \) is independent variable.
Case II If \( y \) is expressed in the term of \( x \) then \( y \) is know as dependent variable and \( x \) is known as independent variable.

Solution: Solution is / are the value / values for the variable(s) used in equation which make(s) the two sides L.H.S. and R.H.S. of the equation equal.

System of Simultaneous Equations

Set of two linear equations in two variables written in the form of \( a_1x + b_1y + c_1 = 0 \), with \( a_1 \neq 0, b_1 \neq 0 \) and \( a_2x + b_2y + c_2 = 0 \), with \( a_2 \neq 0, b_2 \neq 0 \) is known as system of simultaneous linear equations in two variables.
To get solution of simultaneous linear equations in two variables two methods are used:
(i) Graphical method (ii) Algebraic method
If two or more pairs of values for \( x \) and \( y \) which satisfy the given equation are joined on graph paper, we find the graph of the given equation.
Every solution \( x = a, y = b \) (where \( a \) and \( b \) are real numbers), of the given equation determines a point \( (a, b) \) which lies on the graph of line.
Every point \( (c, d) \) lies on the line determines a solution \( x = c, y = d \) of the given equation. Thus, line is known as the graph of the given equation.

When \( a \neq 0, b = 0 \) and \( c \neq 0 \) then the equation \( ax + by + c = 0 \) becomes \( ax + c = 0 \) or \( x = \frac{-c}{a} \) then the graph of this equation is a straight line parallel to y-axis and passing through a point \( \left( -\frac{c}{a}, 0 \right) \).
When \( a = 0, b \neq 0 \) and \( c \neq 0 \) then the equation \( ax + by + c = 0 \) becomes \( by + c = 0 \) or \( y = -\frac{c}{b} \) then the graph of the equation is a straight line parallel to x-axis and passing through the point \( \left( 0, -\frac{c}{b} \right) \).
When \( a \neq 0, b = 0 \) and \( c = 0 \) then the equation \( ax = 0 \) or \( x = 0 \). Then the graph is y-axis itself.
When \( a = 0, b \neq 0 \), and \( c = 0 \) then equation becomes \( by = 0 \) or \( y = 0 \). Then the graph of this equation is x-axis itself.
When only \( c = 0 \) then the equation becomes \( ax + by = 0 \). Then the graph of this equation is a line passing through the origin.
The graph of \( x = \text{constant} \) is a line parallel to the y-axis.
(i) The graph of \( y = \text{constant} \) is a line parallel to the x-axis.
(ii) The graph of \( y = \pm x \) is a line passing through the origin.
(iii) The graph of \( y = x \) & \( y = -x \) passes through origin.

By finding common solutions of two (more) equations we mean a pair of value required one for \( x \) and other for \( y \), which satisfy both the equations simultaneously.

There are three possibilities of two given lines in a plane :
(i) When the lines are intersecting: If the lines are intersecting, the equations are consistent and there exist a unique solution, and the point of intersection gives the unique solution (common solution).
(ii) When the lines are parallel: If the lines are parallel, we have no solution and system is said to be inconsistent.
(iii) When the lines are coincident: If both the equations are represented by same line then the two given equations are said to be dependent equations. In this case, there are infinite solutions.

Consistency and Inconsistency of system of simultaneous linear equations (Algebrically)

For system of simultaneous linear equations in two variables:
\( a_1x + b_1y + c_1 = 0, a_1 \neq 0, b_1 \neq 0 \) and \( a_2x + b_2y + c_2 = 0, a_2 \neq 0, b_2 \neq 0 \)

1. If graphs of two intersects then system in said to be consistent with unique solution.
2. If graphs of two lines are parallel then system is said to be in consistent or no common solution.
3. If graphs of two lines are concident then system is known as consistent with infinite solution.

Now if we want to check their conditions of consistency or inconsistency without plotting the graphs then we check the following conditions.
1. If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), then system is said to be consistent with unique solution.
2. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then system is said to be Inconsistent or no solution in common.
3. If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), then system is said to be consistent with infinitely many common solutions.

CONCEPT-2 : SOLUTION OF SYSTEM OF SIMULTANEOUS LINEAR EQUATIONS ALGEBRAICALLY

TYPE - 1: Substitution Method

Algebraic solution of a system of linear equations
Sometimes, 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 accurate result. Algebraic methods given below yield accurate answers.
(i) Method of elimination by substitution, (ii) Method of elimination by equating the coefficients.
(iii) Method of cross multiplication.

Substitution Method
In this method, we first find the value of one variable \( (y) \) in terms of another variable \( (x) \) from one equation. Substitute this value of \( y \) in the second equation. Second equation becomes a linear equation in \( x \) only and it can be solved for \( x \).
Putting the value of \( x \) in the first equation, we can find the value of \( y \).
This method of solving a system of linear equations is known as the method of elimination by substitution.
'Elimination', because we get rid of \( y \) or 'eliminate' \( y \) from the second equation. 'Substitution', because we 'substitute' the value of \( y \) in the second equation.

Working Rule :
Let the two equations be
\( a_1x + b_1y + c_1 = 0 \) ...(1)
\( a_2x + b_2y + c_2 = 0 \) ...(2)
Step 1. Find the value of one variable, say \( y \), in terms of the other i.e. \( x \) from any equation, say (1).
Step 2. Substitute the value of \( y \) obtained in step I in the other equation i.e. equation (2). This equation becomes equation in one variable \( x \) only.
Step 3. Solve the equation obtained in step II to get the value of \( x \).
Step 4. Substitute the value of \( x \) from step III to the equation obtained in step I. From this equation, we get the value of \( y \). In this way, we get the solution i.e. values of \( x \) and \( y \).
Remark : Verification is a must to check the answer.

TYPE - 2: Elimination Method

Method of elimination by equating the coefficients
In this method, the coefficients of any one variable 'y' in both the equations are made equal by multiplying the equations with the suitable non-zero number. By adding/subtracting one of the two variables (y) is eliminated to get an equation in one variable x. The equation obtained in one variable (x) is solved to get the value of the variable (x). The value of this variable (x) is substituted in any of the equation to get the value of the other variable (y). In this method, multiplication is used instead of division. Multiplication is more convenient than division. So this method is preferred than the previous one.

Working Rule
Let the two equations be
\( a_1x + b_1y + c_1 = 0 \) ...(1)
\( a_2x + b_2y + c_2 = 0 \) ...(2)
Step 1. To eliminate \( y \), multiply equation (1) by \( b_2 \) and equation (2) by \( b_1 \).
[To eliminate \( x \), multiply equation (1) by \( a_2 \) and equation (2) by \( a_1 \)].
Step 2. Subtract the equations obtained in step 2. The variable \( y \) is eliminated and we get an equation in one variable \( x \).
Step 3. Solve the equation in one variable \( (x) \) obtained in step 2.
Step 4. Substitute the value of one unknown \( (x) \), found in step 3 in any one of the given equations.
Step 5. Solve the equation obtained in step 4 and get the value of the other unknown \( (y) \). In this way, we get the solution i.e. values of \( x \) and \( y \).

TYPE - 3: Examples based on method of Cross-multiplication

Cross-multiplication Method
By the method of elimination by substitution, only those equations can be solved, which have unique solution. But the method of cross multiplication discussed below is applicable in all the cases; whether the system has a unique solution, no solution or infinitely many solutions.

Working rule for cross multiplication method
Step 1. Convert the given equations in the following form by writing all the terms on the L.H.S. and zero on R.H.S.
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Step 2. There are three columns in the above equations formed by coefficients of \( x, y \) and constant terms.
Step 3. (i) Write down two columns in the denominator of \( x \), ignoring the column of \( x \), i.e., \( \begin{pmatrix} b_1 & c_1 \\ b_2 & c_2 \end{pmatrix} \)
(ii) Again, write two columns in the denominator of \( -y \), leaving the column of \( y \), i.e., \( \begin{pmatrix} a_1 & c_1 \\ a_2 & c_2 \end{pmatrix} \)
(iii) Similarly, write two columns in the denominator of 1, i.e., \( \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \)
Step 4. Now, equate them and mark crossed-arrows from left top to right bottom and left bottom to right top, as shown below:
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{-y}{a_1c_2 - a_2c_1} = \frac{1}{a_1b_2 - a_2b_1} \)
Step 5. Find the value of \( x \) by equating the first and third expressions. The value of \( y \) is obtained by equating the second and third expressions.

CONCEPT-3 : CONDITIONS FOR CONSISTENCY OR SOLVABILITY

Let
\( a_1x + b_1y + c_1 = 0 \) ...(1)
\( a_2x + b_2y + c_2 = 0 \) ...(2)
Multiplying the equation (1) by \( b_2 \) and equation (2) by \( b_1 \), we get
\( a_1b_2x + b_1b_2y + b_2c_1 = 0 \) ...(3)
\( a_2b_1x + b_1b_2y + b_1c_2 = 0 \) ...(4)
Subtracting equation (4) from equation (3), we get
\( x(a_1b_2 - a_2b_1) = b_1c_2 - b_2c_1 \) ...(5)
Similarly,
\( y(a_1b_2 - a_2b_1) = c_1a_2 - c_2a_1 \) ...(6)

Case I. If \( a_1b_2 - a_2b_1 \neq 0 \)

\( \implies \) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Then, from equation (5) and equation (6), we have
\( x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \)
\( y = \frac{a_2c_1 - a_1c_2}{a_1b_2 - a_2b_1} \)
So, we conclude that the system is consistent with a unique solution.

Case II. Suppose that \( a_1b_2 - a_2b_1 = 0 \)
In this case we cannot divide by \( a_1b_2 - a_2b_1 \) to get \( x \) and \( y \). (as \( a_1b_2 - a_2b_1 = 0 \))

\( \implies \) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \)
Let \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = k \implies a_1 = ka_2, b_1 = kb_2 \)
Now, there are two possibilities
Possibility 1. \( c_1 = kc_2 \)
Putting the values of \( a_1, b_1 \) and \( c_1 \) in equation (1), we get
\( k(a_2x + b_2y + c_2) = 0 \)
Thus, the system of equations becomes
\( k(a_2x + b_2y + c_2) = 0 \) ...(7)
\( a_2x + b_2y + c_2 = 0 \) ...(8)
Clearly, every solution of equation (7) is a solution of equation (8) and vice-versa.
Thus, in this case, the given system has infinite solutions.
Therefore, when \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k \), then the given system is consistent with an infinite number of solutions.

Possibility 2. \( c_1 \neq kc_2 \)
On putting the values of \( a_1, b_1 \) and \( c_1 \) in equation (1), we get
\( k(a_2x + b_2y + c_2) \neq 0 \)
or \( a_2x + b_2y + c_2 \neq 0 \)
But, equation (2) is \( a_2x + b_2y + c_2 = 0 \)
These, two equations are contradictory. Hence, the system is inconsistent (no solution).

SUMMARY

Let the system of linear equations in two variables be
\( a_1x + b_1y = c_1 \),
\( a_2x + b_2y = c_2 \)

(a) Consistent system with unique solution. When \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) or \( a_1b_2 - a_2b_1 \neq 0 \)
e.g., \( 2x + 3y = 8 \)
\( 3x + 4y = 10 \)
Here, \( \frac{2}{3} \neq \frac{3}{4} \), so the system is consistent with unique solution.

(b) Consistent system with infinite number of solutions.
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and \( c_1 = kc_2 \)
i.e., \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
e.g., \( 2x + 3y = 8 \)
\( 4x + 6y = 16 \)
Here, \( \frac{2}{4} = \frac{3}{6} = \frac{8}{16} \), so the system is consistent with infinite number of solutions.

(c) Inconsistent system (No solution). \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and \( c_1 \neq kc_2 \)
or \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)