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.

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**(1) An equation in the form ax + by + c = 0, where a, b and c are real numbers, and aâ‰ 0 and b â‰ 0, is called a linear equation in two variables x and y.
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**(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.
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**(3) The general form for a pair of linear equations in two variables x and y is a _{1}x + b_{1} y + c_{1 }= 0 and a_{2}x + b_{2} y + c_{2} = 0, where a_{1} , b_{1} , c_{1} , a_{2} , b_{2} , c_{2} are all real numbers and a_{1}^{2} + b_{1}^{2} â‰ 0, a_{2}^{2} + b_{2}^{2} â‰ 0.**

**(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 a _{1}/a_{2} = b_{1}/b_{2} â‰ c_{1}/c_{2}.**

Step 1: Find the value of one variable, say

Step 2: Substitute this value of

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 Example:Â **Solve the following pair of equations by substitution method: 7x â€“ 15y = 2 and x + 2y = 3

We can re-write x + 2y = 3 as x = 3 â€“ 2y â€“ (1)

Substituting value of x in 7x â€“ 15y = 2, we get,

7(3 â€“ 2y) â€“ 15y = 2

21 â€“ 14y â€“ 15y = 2

-29y = -19

Thus, y = 19/29.

Now, substituting value of y in (1), we get,

x = 3 â€“ 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:Â **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 â€“ 27x) â€“ (12y â€“ 12y) = 8000 â€“ 6000

x = 2000

Substituting value of x in 9x - 4y = 2000, we get,

9(2000) â€“ 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 a

Step 2: Take the help of the diagram belowAnd write the equations as shown below

Step 3: Find x and y, provided a

** For Example:Â **Solve the following pair of equations by cross multiplication method: 2x + 3y - 46 = 0 and 3x + 5y â€“ 74 = 0.

Given equations are in form a

x/((3)(-74) â€“ (5)(-46) = y/((-46)(3) â€“ (-74)(2) = 1/((2)(5) â€“ (3)(3)

x/(-222 + 230) = y/(-138 + 148) = 1/(10 â€“ 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:

We can rewrite the given equations as,

2(1/x) + 3(1/y) = 13 and 5(1/x) â€“ 4(1/y) = -2

Let us substitute 1/x a = p and 1/y = q, so we get,

2p + 3q = 13 and 5p â€“ 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 = Â½ and y = 1/3.

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