# Pair of Linear Equations in Two Variables - Class 10 : Notes

Notes for pair of linear equations in two variables chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) An equation in the form ax + by + c = 0, where a, b and c are real numbers, and a≠ 0 and b ≠ 0, is called a linear equation in two variables x and y.
For Example: 2x + 3y + 7 = 0, where a = 2, b = 3, c =5 are real numbers. So, given equation is a linear equation in two variables.

(2) Each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.
For Example2x + 3y = 5 has (1, 1) as its solution. So,(1, 1) will lie on the line 2x + 3y = 5.

(3) The general form for a pair of linear equations in two variables x and y is a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, where a1 , b1 , c1 , a2 , b2 , c2 are all real numbers and a12 + b12 ≠ 0, a22 + b22 ≠ 0.
For Example2x + 3y – 7 = 0 and 9x – 2y + 8 = 0 forms a pair of linear equations.

(4) A pair of linear equations which has no solution is called an inconsistent pair of linear equations. In this case, the lines may be parallel a1/a2 = b1/b2 ≠ c1/c2.
For Examplex + 2y – 4 = 0 and 2x + 4y – 12 = 0 are parallel lines. (5) A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. In this case, the lines may intersect in a single point and a1/a2 ≠ b1/b2.
For Examplex – 2y = 0 and 3x + 4y – 20 intersects each other at unique point (4, 2). (6) A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. In this case, the lines may be coincident and a1/a2 = b1/b2 = c1/c2.
For Example2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 are coincident lines. (7) Algebraic Methods of Solving a Pair of Linear Equations:
(i) Substitution Method
Follow steps given below to understand Substitution Method:
Step 1: Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometime, one can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
Step 3: Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

For ExampleSolve the following pair of equations by substitution method: 7x – 15y = 2 and x + 2y = 3
We can re-write x + 2y = 3 as x = 3 – 2y – (1)
Substituting value of x in 7x – 15y = 2, we get,
7(3 – 2y) – 15y = 2
21 – 14y – 15y = 2
-29y = -19
Thus, y = 19/29.
Now, substituting value of y in (1), we get,
x = 3 – 2(19/29) = 49/29.

(ii) Elimination Method:
Follow steps given below to understand Elimination Method:
Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step 3: Solve the equation in one variable (x or y) so obtained to get its value.
Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.

For Example: Solve the following pair of equations by elimination method: 9x - 4y = 2000 and 7x - 3y = 2000.
Multiplying 9x - 4y = 2000 by 3 and 7x - 3y = 2000 by 4, we get,
27x - 12y = 6000 and 28x - 12y = 8000
Subtracting both these equations, we get,
(28x – 27x) – (12y – 12y) = 8000 – 6000
x = 2000
Substituting value of x in 9x - 4y = 2000, we get,
9(2000) – 4y = 2000
y = 4000.

(iii) Cross Multiplication Method:
Follow steps given below to understand Cross Multiplication Method:
Step 1: Write the given equations in the form a1x + b1 y + c1 = 0 and a2x + b2y + c2 = 0.
Step 2: Take the help of the diagram below And write the equations as shown below Step 3: Find x and y, provided a1b1 – a2b1 ≠ 0.

For ExampleSolve the following pair of equations by cross multiplication method: 2x + 3y - 46 = 0 and 3x + 5y – 74 = 0.
Given equations are in form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Now, with the help of diagram we can rewrite equations as
x/((3)(-74) – (5)(-46) = y/((-46)(3) – (-74)(2) = 1/((2)(5) – (3)(3)
x/(-222 + 230) = y/(-138 + 148) = 1/(10 – 9)
x/8 = y/10 = 1/1
x/8 = 1/1 and y/10 = 1/1
Thus, x = 8 and y = 10.

(8) Equations Reducible to a Pair of Linear Equations in Two Variables:
Let us understand it by an example:
For ExampleSolve 2/x + 3/y = 13 and 5/x – 4/y = -2.
We can rewrite the given equations as,
2(1/x) + 3(1/y) = 13 and 5(1/x) – 4(1/y) = -2
Let us substitute 1/x a = p and 1/y = q, so we get,
2p + 3q = 13 and 5p – 4q = -2
On solving these equations, we get,
p = 2 and q = 3.
We know, p = 1/x = 2 and q = 1/y = 3.
Thus, x = ½ and y = 1/3.