Facts that Matter: Pair of Linear Equations in Two Variables

Table of Contents
1. Linear Equation in Two Variable
2. General Form of a Pair of Linear Equations in Two Variables
3. Solution of a Pair of Linear Equations in Two Variables
4. Algebraic Methods
5. Graphical Method
6. Consistency: Conditions and Tests
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Linear Equation in Two Variable

A relation in two variables that can be written in the form

\(ax + by + c = 0\)

where a, b and c are real numbers and at least one of a or b is non-zero, is called a linear equation in two variables. The variables are usually denoted by x and y.

General Form of a Pair of Linear Equations in Two Variables

A pair of linear equations in two variables is written as

\(a_{1}x + b_{1}y + c_{1} = 0\)

\(a_{2}x + b_{2}y + c_{2} = 0\)

where \(a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}\) are real numbers and for each equation the coefficients of \(x\) and \(y\) are not both zero.

Solution of a Pair of Linear Equations in Two Variables

A solution of the pair of linear equations in two variables is an ordered pair \((x,y)\) which satisfies both equations simultaneously. Geometrically, each linear equation represents a straight line in the plane; a solution corresponds to a point common to both lines.

There are three algebraic possibilities for solutions of a pair of linear equations:

• Exactly one solution (the two lines intersect at a single point).
• No solution (the two lines are parallel and distinct).
• Infinitely many solutions (the two lines coincide).

There are three standard algebraic methods to solve a pair of linear equations and a graphical method. 

Algebraic Methods

Each method is useful in different situations; choose the one which simplifies calculations for the given coefficients.

Types of Algebraic Methods

Substitution Method

Step 1: Solve one of the equations for one variable (for example, express \(y\) in terms of \(x\)).

Step 2: Substitute this expression into the other equation to obtain an equation in one variable and solve it.

Final Step: Substitute the found value back into the earlier expression to obtain the remaining variable.

Example: Solve the pair of equations using substitution method

\(x + y = 5\)

\(x - y = 1\)

Sol.

Express \(y\) in terms of \(x\) from the first equation.

\(y = 5 - x\)

Substitute \(y = 5 - x\) into the second equation.

\(x - (5 - x) = 1\)

Simplify the equation.

\(x - 5 + x = 1\)

\(2x - 5 = 1\)

\(2x = 6\)

\(x = 3\)

Substitute \(x = 3\) into \(y = 5 - x\).

\(y = 5 - 3\)

\(y = 2\)

Answer: \((x,y) = (3,2)\).

Elimination Method

Step 1: Multiply the equations by suitable constants so the coefficients of one variable become equal in magnitude.

Step 2: Add or subtract the equations to eliminate that variable and obtain an equation in one variable.

Step 3: Solve the single variable equation.

Final Step: Substitute this value into one of the original equations to find the other variable.

Example: Solve the pair of equations using Elimination method

\(2x + 3y = 13\)

\(3x - 2y = 4\)

Sol.

Multiply the first equation by 3 and the second by 2 so that coefficients of \(x\) match:

\(6x + 9y = 39\)

\(6x - 4y = 8\)

Subtract the second equation from the first to eliminate \(x\).

\((6x + 9y) - (6x - 4y) = 39 - 8\)

\(13y = 31\)

\(y = \dfrac{31}{13} = \dfrac{31}{13}\)

Substitute \(y\) into the original first equation.

\(2x + 3\left(\dfrac{31}{13}\right) = 13\)

\(2x + \dfrac{93}{13} = 13\)

\(2x = 13 - \dfrac{93}{13}\)

\(2x = \dfrac{169 - 93}{13}\)

\(2x = \dfrac{76}{13}\)

\(x = \dfrac{38}{13}\)

Answer: \(\left(\dfrac{38}{13}, \dfrac{31}{13}\right)\).

Determinant Method (Cramer's Rule)

Write the pair in the form

\(a_{1}x + b_{1}y = -\,c_{1}\)

\(a_{2}x + b_{2}y = -\,c_{2}\)

Define the determinant

\(\Delta = \begin{vmatrix} a_{1} & b_{1} \\[4pt] a_{2} & b_{2} \end{vmatrix} = a_{1}b_{2} - a_{2}b_{1}\)

Also define

\(\Delta_{x} = \begin{vmatrix} -c_{1} & b_{1} \\[4pt] -c_{2} & b_{2} \end{vmatrix} = -c_{1}b_{2} + c_{2}b_{1}\)

\(\Delta_{y} = \begin{vmatrix} a_{1} & -c_{1} \\[4pt] a_{2} & -c_{2} \end{vmatrix} = -a_{1}c_{2} + a_{2}c_{1}\)

If $\Delta \neq 0$, the system has a unique solution.

\(x = \dfrac{\Delta_{x}}{\Delta}\)

\(y = \dfrac{\Delta_{y}}{\Delta}\)

If \(\Delta = 0\) and at least one of \(\Delta_x, \Delta_y\) is non-zero, the system has no solution.
If \(\Delta = 0\) and \(\Delta_x = \Delta_y = 0\), the system has infinitely many solutions.

Example:  Solve the pair of equations using Determinant Method

\(2x + y = 5\)

\(3x - 2y = 4\)

Sol.

Write constants on right side:

\(2x + y = 5\)

\(3x - 2y = 4\)

Compute determinant \(\Delta\).

\(\Delta = \begin{vmatrix} 2 & 1 \\[4pt] 3 & -2 \end{vmatrix} = 2(-2) - 3(1) = -4 - 3 = -7\)

Compute \(\Delta_x\).

\(\Delta_x = \begin{vmatrix} 5 & 1 \\[4pt] 4 & -2 \end{vmatrix} = 5(-2) - 4(1) = -10 - 4 = -14\)

Compute \(\Delta_y\).

\(\Delta_y = \begin{vmatrix} 2 & 5 \\[4pt] 3 & 4 \end{vmatrix} = 2(4) - 3(5) = 8 - 15 = -7\)

Since $\Delta \neq 0$, find \(x\) and \(y\).

\(x = \dfrac{\Delta_x}{\Delta} = \dfrac{-14}{-7} = 2\)

\(y = \dfrac{\Delta_y}{\Delta} = \dfrac{-7}{-7} = 1\)

Answer: \((2,1)\).

Graphical Method 

Each linear equation represents a straight line in the plane. The nature of the solution set corresponds to how the two lines relate.

• If the two lines intersect at one point, the system has a unique solution: the coordinates of the intersection point.
• If the two lines are parallel and distinct, they have no common point and the system has no solution (the system is inconsistent).
• If the two lines coincide (are the same line), every point on the line satisfies both equations and the system has infinitely many solutions (the system is dependent and consistent).

Consistency: Conditions and Tests

Consider the pair:

The following three situations may arise.

(i) If

then the system is consistent and has a unique solution.

(ii) If

then the system is inconsistent and has no solution.

(iii) If

then the system is dependent and consistent (infinitely many solutions).