The linear equation of two variables is of the form ax + by + c = 0. Since the graph of the linear equation is a straight line, the points which lie on this line form the solution to the given equation. The solutions also satisfy the equation.

If we have more than one equation, we can discuss about common solutions as follows.

ax + b - b = 0 - b [ since b - b = 0 ]

=> ax = -b

=> $ \frac{ax}{a}$ = $\frac{-b}{a}$

=> x = -$\frac{b}{a}$

Hence the final solution is x = - b/a

we get a = 3, b = 12

Therefore the solution , x = - $\frac{b}{a}$ = - $\frac{12}{3}$ = - 4

a

a

The above two equations can be solved for the variables x and y.

Since the graph of the above equations are straight line, the above pair of equations will satisfy any one of the following conditions.

(1) Two lines intersect at a point.

(2) The two lines do not intersect at all ( two lines are parallel )

(3) the two lines coincide.

The following are the methods used to solve the linear equations of two variables.

1. substitution Method.

2. Elimination Method.

3. Cross multiplication Method.

4. Graphical Method.

5. Matrix Method.

Let us first number the equations as (1) and (2)

2x + y = 7 -------------------------------(1)

3 x - y = 3 -------------------------------(2)

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On adding (1) and (2), we get 5x = 10 [ since + y - y = 0 ]

x = 10/5

x = 2

Substituting x=2, in Equation (1), we get,

2(2) + y = 7

4 + y = 7

y = 7 - 4 = 3

=> y = 3

Hence the

Let us first number the equations. 3x + y = 10 ---------------------(1)

6x + 2y = -5 ---------------------(2)

Multiplying Equation (1) by 2, we get 2 ( 3x + y ) = 2 ( 10)

=> 6x + 2y = 20 ----------------------(3)

6x + 2y = -5 -----------------------(2)

(-) (-) (+)

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Subtracting (2) from (3), we get 0 = 25 , which is a false statement.

Hence the above

Hence there will be infinite number of common solutions.

Let us first number the equations . 3x + 2y = 10 ----------------------------(1)

6x + 4 y = 20 ---------------------------(2)

Multiplying Equation (1) by 2, we get, 2 ( 3x + 2y ) = 2 ( 10)

6x + 4y = 20 ----------------------------(3)

6x + 4y = 20 ---------------------------(2)

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Subtracting (2) from (3), we get 0 = 0, which is true.

This is true for any values of x.

Since the values of y depends on the values of x, there will be infinite number of values for x and the corresponding values of y.

Hence

a. 4x + 15 = -1

b. 5 + 3x = 11

2. Solve the following pairs of equations, write the common solution/solutions if any.

a. 3x + 5y = -2 ; 4x - y = 5

b. 4x + 3y = 10, 12 x + 9 y = 30.

c. 8x + 3y = 11, 16 x + 6y = 22.

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