Equations Reducible to a Pair of Linear Equations in Two Variables

We have learned that if an equation can be written in the form of ax+by+c=0, where a and b are not both zero, and a, b, c are real numbers, then the equation is called a linear equation in two variables x and y. Each solution (x, y) of the linear equation ax+by+c =0, corresponds to a point on a line representing the equation or vice versa. In this article, we are going to learn the general form of a pair of linear equations, and how the equations reducible to a pair of linear equations in two variables in detail.

General Form of Pair of Linear Equations in Two variables

The general form for the pair of linear equations in two variables x and y is given as:

a1x+b1y+c1 =0 and a2x+b2y+c2 =0

Where a1, a2, b1, b2, c1, c2 are real numbers and a12+b12 ≠ 0, and a22+b22 ≠ 0.

An example for the pair of linear equations in two variables is 5x =y and -7x+2y+3 =0.

The graphical representation of linear equations in two variables is a straight line. Thus, for the pair of linear equations in two variables, there will be two straight lines, and they have the following three possibilities. They are:

Case (1): Two lines intersect

Case (2) Two lines are parallel to each other

Case (3) Two lines are coincident

Equations Reducible to a Pair of Linear Equations in Two Variables Examples

In some cases, the pair of equations in two variables are not linear, but they can be reduced to the pair of linear equations in two variables using some suitable substitution process. Now, let us discuss how the equations are reducible to a pair of linear equations in two variables with the help of examples.

Example 1:

Solve the following pair of linear equations:

(2/a) + (3/b) = 13

(5/a) – (4/b) = -2

Solution:

Given pair of equations:

(2/a) + (3/b) = 13

(5/a) – (4/b) = -2

The above equations can be written as follows:

2(1/a) + 3(1/b) = 13 …(1)

5(1/a) – 4(1/b) = -2 …(2)

The equations (1) and (2) are not in the linear form ax+by+c =0.

So, substitute (1/a) = x and (1/b) = y in the equations (1) and (2)

Hence, the equation (1) and (2) becomes,

2x+3y = 13 …(3)

5x-4y= -2 …(4)

Thus, the equations (3) and (4) are the pair of linear equations in two variables x and y.

Now, solve the equations (3) and (4), we get x = 2 and y = 3.

We know that x = 1/a and y = 1/b

Therefore,

⇒2 = 1/a

⇒a = ½

Similarly,

⇒ 3 = 1/b

⇒ b= ⅓.

Therefore, the values of a and b are ½ and ⅓, respectively.

Verification:

To check whether the values of a and b are correct or wrong, substitute the values in the given equations.

(2/a) + (3/b) = 13

Now, substitute a = ½ and b= ⅓ in the above equation.

⇒ (2/(½)) + (3/(⅓)) = 13

⇒ 4+9 = 13

⇒ 13 =13

L.H.S = R.H.S

Similarly, substitute the values in the equation (5/a) – (4/b) = -2

⇒ (5/(½)) – (4/(⅓)) = -2

⇒ 10- 12 = -2

⇒ -2 = -2

L.H.S = R.H.S

As the values of a and b satisfy both the equations, we can conclude that the value of a is ½ and b= ⅓ are correct.

Example 2:

Solve the given pair of equations by reducing them to a pair of linear equations:

[5/(a-1)]+ [1/(b-2)] = 2

[6/(a-1)]- [3/(b-2)] = 1

Solution:

Given pair of equations:

[5/(a-1)]+ [1/(b-2)] = 2 … (1)

[6/(a-1)]- [3/(b-2)] = 1… (2)

The above given equations are not in linear form. To convert them into a linear form, substitute 1/(a-1) = x and 1/(b-2) = y in (1) and (2). Hence, we get

5x+y = 2 … (3)

6x-3y = 1…(4)

Hence, the equations (3) and (4) are in the form of a pair of linear equations in two variables. Now, solve the equations (3) and (4), and we get

x = ⅓ and y=⅓

So, we know that x = 1/(a-1)

⇒ (⅓) = 1/(a-1)

⇒ a-1 = 3

⇒ a = 4

Similarly, 1/(b-2) = y

⇒ (⅓) = 1/(b-2)

⇒ b-2 = 3

⇒ b = 3+2

⇒ b = 5

Hence, the values of a and b are 4 and 5, respectively.

Verification:

Substitute a = 4 and b = 5 in equation (1) and (2).

Equation (1):

[5/(a-1)]+ [1/(b-2)] = 2

⇒[5/(4-1)]+ [1/(5-2)] = 2

⇒(5/3) + (⅓) = 2

⇒(6/3) = 2

⇒ 2 = 2

Therefore, L.H.S = R.H.S.

Equation (2):

[6/(a-1)]- [3/(b-2)] = 1

⇒ [6/(4-1)]- [3/(5-2)] = 1

⇒ (6/3) – (3/3) = 1

⇒ 2 – 1 = 1

⇒ 1 = 1

Hence, L.H.S = R.H.S

Therefore, the values a = 4 and b = 5 satisfy both equations.

Practice Problems

Solve the pair of equations by reducing them into a pair of linear equations in two variables:

  1. (1/2x) + (1/3y) = 2 and (1/3x) + (1/2y) = (13/6).
  2. [5/(x-1)] + [1/(y-2)] = 2 and [6/(x-1)]-[3/(y-2)] = 1.
  3. (7x-2y)/xy = 5 and (8x+7y)xy = 15.

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