Linear Equations In Two Variables

1. The equation of the form ax = b or ax + b = 0, where a and b are two real numbers such that x ≠ 0 and x is a variable is called a inear equation in one variable.

2. The general form of a linear equation in two variables is ax + by +c = 0 or ax + by = c where a, b, c are real numbers and a ≠ 0, b ≠ 0 and x, y are variables.

3. The graph of a linear equation in two variables is a straight line.

4. The graph of a linear equation in one variable is a straight line parallel to x-axis for ay = b and parallel to y-axis for ax = b, where a ≠ 0.

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

6. The value of the variable x and y satisfying each one of the equations in a given system of linear equations in x and y simultaneously is called a solution of the system.

Standard form of linear equation:

(Standard form refers to all positive coefficient)

a₁x + b₁y + c₁ = 0 ....(i)

a₂x + b₂y + c₂ = 0 ....(ii)

For solving such equations we have three methods.

(i) Elimination by substitution

(ii) Elimination by equating the coefficients

(iii) Elimination by cross multiplication.

ELIMINATION BY SUBSTITUTION:

(Find the value of any one variable in terms of other and then use it to find other variable from the second equation):

question 1. Solve x + 4y = 14 .....(i)

7x - 3y = 5 ....(ii)

Solution: From equation (i) x = 14 - 4y ....(iii)

Substitute the value of x in equation (ii)

⇒ 7 (14 - 4y) - 3y = 5

⇒ 98 - 28y - 3y = 5

⇒ 98 - 31y = 5

⇒ 93 = 31y

⇒ y = 93/31

⇒ y = 3

Now substitute value of y in equation (iii)

⇒ 7x - 3 (3) = 5

⇒ 7x - 3 (3) = 5

⇒ 7x = 14

⇒ x = 14/7 = 2

So the solution is x = 2 and y = 3

ELIMINATION BY EQUATING THE COEFFICIENTS:

(Eliminating One variable by making the coefficient equal to get the value of one variable and then put it in any equation to find other variable):

question 1. Solve 9x - 4y = 8 ...(i)

13x + 7y = 101 ...(ii)

Solution: Multiply equation (i) by 7 and equation (ii) by 4, we get

adding 63x – 28y = 56

52x + 28y = 404

115x = 460

⇒ .x = 460/115

⇒ x = 4

Substitute x = 4 in equation (i)

9 (4) –0 4y = 8 ⇒ 36 – 8 = 4y ⇒ 28 = 4y

⇒ y = 28/4 = 7

So the solution is x = 4 and y = 7.

COMPARISON METHOD:

(Find the value of one variable from both the equation and equate them to get the value of other variable):

Let any pair of linear equations in two variables is of the form

a₁x + b₁y + c₁ = 0 .........(i)

a₂x + b₂y + c₂ = 0 .....…(ii)

question 1. Solve for x and y : 2x + 3y = 8 , x - 5y = -9.

Solution: 2x + 3y = 8 …(i)

 x - 5y = -9. …(ii)

From equation (i),

2x = 8 - 3y , x  = 8 - 3y/2  …(iii)

From equation (ii),

 x = -9 + 5y…(iv)

From equation (iii) and (iv), we have

Putting value of y in equation (iv), we have

x = 8 - 3 x 2/2

x = 1

Solution x = 1, y = 2.

CROSS MULTIPLICATION METHOD:

Let the equation

a₁x + b₁y + c₁ = 0 …(i)

a₂x + b₂y + c₂ = 0 …(ii)

To obtain the values of x and y, we follow these steps:

1. Multiply Equation (i) by b₂ and (ii) by b1, we get

2. Subtracting Equation (iv) from (iii), we get:

3. Substituting this value of x in (i) or (ii), we get

. …(vi)

We can write the solution given by equations (v) and (vi) in the following form:

  …(vii)

In remembering the above result, the following diagram may be helpful:

For solving a pair of linear equations by this method, we will follow the following steps:

1. Write the given equations in the form (i) and (ii).

2. Taking the help of the diagram above, write equations as given in (viii).

3. Find x and y.

question 1. Solve the following system of equations in x and y

ax + by - a + b = 0, bx - ay - a - b = 0

Solution: a₁ = a, b₁ = b, a₂ = b and b₂ = –a

∴

Hence, the given system of the equation has unique solution.

By cross multiplication, we have

or x = 1 and y = –1

question 2. Solve for x and y.

Solution: Here a1 =1/a, b1 =1/b, a2 = 1/a² and b2 = 1/b²

∴ .

Hence, the given system of equation has unique solution.

By cross multiplication, we have

question 3. Solve the following system of equations by using cross multiplication method:

5x + 3y = 19, 16x - 7y = 11

Solution: Here a₁ = 5, b₁ = 3 and a₂ = 16, b₂ = –7

∴ .

Hence, the given system of equation has a unique solution. By cross multiplication, we have

or x = 2 and y = 3

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