Algebraic Methods of Solving a Pair of Linear Equations

There are several algebraic methods to solve a pair of linear equations. These are as follows:

1. Substitution Method

Let us understand the substitution method with an example.

Example: Solve the following pair of linear equations by the substitution method:

• 7x − 15y = 2 …(1)
• x + 2y = 3 …(2)

Solution:

Step 1: We pick one of the equations and write one variable in terms of the other. Take equation (2):

x + 2y = 3

This can be rewritten as x = 3 − 2y …(3)

Step 2: Now substitute the value of x in equation (1). This gives:

7(3 − 2y) − 15y = 2

i.e. 21 − 14y − 15y = 2

i.e. −29y = −19

So y = 19/29

Step 3: Now substitute the value of y in equation (3). This gives:

x = 3 − 2(19/29) = (87 − 38)/29 = 49/29

So x = 49/29, y = 19/29 is the algebraic solution of the given pair of linear equations by the substitution method.

To verify the answer, substitute the values of x and y in equations (1) and (2) separately.

What is the Substitution Method?

We have solved the pair of linear equations by expressing the value of one variable in terms of the other and substituting it into the second equation. Hence this method is called the substitution method.

2. Elimination Method

Apart from the substitution method, another algebraic method to solve a pair of linear equations is the elimination method, in which we eliminate one variable to get a linear equation in the other variable, find its value, and then use it to find the value of the other variable.

Let us understand this with an example.

Example: The ratio of incomes of two persons is 9 : 7, and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.

Solution: Let the monthly incomes of the two persons be ₹9x and ₹7x respectively, and their expenditures be ₹4y and ₹3y respectively. Then the equations formed are:

9x − 4y = 2000 …(1)

and 7x − 3y = 2000 …(2)

Step 1: To make the coefficients of y equal, multiply equation (1) by 3 and equation (2) by 4. This gives:

27x − 12y = 6000 …(3)

28x − 12y = 8000 …(4)

Step 2: Subtract equation (3) from equation (4) to eliminate y, since the coefficients of y are equal:

(28x − 12y) − (27x − 12y) = 8000 − 6000

i.e. x = 2000

Step 3: Substituting x = 2000 in equation (1):

9(2000) − 4y = 2000

i.e. y = 4000

So the solution of the pair of equations is x = 2000, y = 4000. Therefore, the monthly incomes of the two persons are ₹18000 and ₹14000 respectively.

Verification: Their income ratio is 18000 : 14000 = 9 : 7. Also, their expenditure ratio is (18000 − 2000) : (14000 − 2000) = 16000 : 12000 = 4 : 3.

3. Cross-Multiplication Method

Let the given equations be:

a₁x + b₁y + c₁ = 0 …(1)

a₂x + b₂y + c₂ = 0 …(2)

Multiplying equation (1) by b₂ and equation (2) by b₁:

a₁b₂x + b₁b₂y + b₂c₁ = 0 …(3)

a₂b₁x + b₁b₂y + b₁c₂ = 0 …(4)

Subtracting equation (4) from equation (3):

(a₁b₂ − a₂b₁)x + (b₂c₁ − b₁c₂) = 0

x = (b₁c₂ − b₂c₁)/(a₁b₂ − a₂b₁)

Similarly, multiplying equation (1) by a₂ and equation (2) by a₁, and subtracting, we get:

y = (c₁a₂ − c₂a₁)/(a₁b₂ − a₂b₁)

These two results can be combined and remembered easily using the following diagram, which multiplies the values along the arrows and subtracts the products going the other way — this is why the method is called cross-multiplication:

x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)

Before using this method, all terms of both equations are first taken to the left-hand side so that the right-hand side becomes zero, and the coefficients of x, coefficients of y and the constant terms are identified in both equations.

Condition for Solvability

For the system a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, the ratio of the coefficients of the corresponding variables decides the nature of the solution — refer to the coefficient-ratio table above (unique solution, no solution, or infinitely many solutions).

Example — Cross-Multiplication Method

Check whether the pair of equations 2x + y = 35, 3x + 4y = 65 has a unique solution, no solution, or infinitely many solutions. If it has a unique solution, find it.

Solution: 2x + y = 35, 3x + 4y = 65. Taking all terms to the left-hand side:

2x + y − 35 = 0

3x + 4y − 65 = 0

Here a₁ = 2, b₁ = 1, c₁ = −35, a₂ = 3, b₂ = 4, c₂ = −65.

By cross-multiplication:

x / {(1)(−65) − (4)(−35)} = y / {(−35)(3) − (−65)(2)} = 1 / {(2)(4) − (3)(1)}

i.e. x / (−65 + 140) = y / (−105 + 130) = 1 / (8 − 3)

i.e. x/75 = y/25 = 1/5

So x = 15 and y = 5. Since a unique solution exists, the pair of equations is consistent.

A Real-Life Example — Bus Fares

From a bus stand in Bangalore, if we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur, the total cost is ₹46. But if we buy 3 tickets to Malleswaram and 5 tickets to Yeshwanthpur, the total cost is ₹74. Find the fares from the bus stand to Malleswaram and to Yeshwanthpur.

Solution: Let the fare from the bus stand to Malleswaram be ₹x, and to Yeshwanthpur be ₹y. From the given information:

2x + 3y = 46, i.e. 2x + 3y − 46 = 0 …(1)

3x + 5y = 74, i.e. 3x + 5y − 74 = 0 …(2)

Here a₁ = 2, b₁ = 3, c₁ = −46, a₂ = 3, b₂ = 5, c₂ = −74.

By the cross-multiplication method:

x / {(3)(−74) − (5)(−46)} = y / {(−46)(3) − (−74)(2)} = 1 / {(2)(5) − (3)(3)}

i.e. x/(−222 + 230) = y/(−138 + 148) = 1/(10 − 9)

i.e. x/8 = y/10 = 1/1

So x = 8 and y = 10.

Hence, the fare from the Bangalore bus stand to Malleswaram is ₹8, and to Yeshwanthpur is ₹10.

Note: To verify the answer, substitute the values of x and y in equations (1) and (2).

Equations Reducible to a Pair of Linear Equations in Two Variables

We now discuss pairs of equations which are not linear, but which can be reduced to linear equations by suitable substitutions.

Example: Solve the following pair of equations:

• 2/x + 3/y = 13
• 5/x − 4/y = −2

Solution: We can rewrite these equations as:

• 2(1/x) + 3(1/y) = 13 …(1)
• 5(1/x) − 4(1/y) = −2 …(2)

These equations are not in the form ax + by + c = 0. But if we substitute 1/x = p and 1/y = q in equations (1) and (2), we get:

• 2p + 3q = 13 …(3)
• 5p − 4q = −2 …(4)

So we have expressed the equations as a pair of linear equations. Solving these by any method, we get p = 2, q = 3.

Here p = 1/x and q = 1/y

So 1/x = 2 and 1/y = 3

i.e. x = 1/2 and y = 1/3

Verification: Substituting x = 1/2 and y = 1/3 in both equations, we find that both equations are satisfied.

A Real-Life Example — Boat and Stream

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Find the speed of the boat in still water and the speed of the stream.

Solution: Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. Then the speed of the boat downstream = (x + y) km/h, and the speed of the boat upstream = (x − y) km/h.

Also, time = distance/speed.

In the first case, let t₁ be the time (in hours) taken by the boat to go 30 km upstream. Then:

t₁ = 30/(x − y)

Let t₂ be the time (in hours) taken to go 44 km downstream. Then t₂ = 44/(x + y). The total time taken, t₁ + t₂, is 10 hours. So we get the equation:

• 30/(x − y) + 44/(x + y) = 10 …(1)

In the second case, in 13 hours the boat goes 40 km upstream and 55 km downstream. This gives:

• 40/(x − y) + 55/(x + y) = 13 …(2)

To express these equations as a pair of linear equations, substitute:

• 1/(x − y) = u and 1/(x + y) = v …(3)
• 30u + 44v = 10, i.e. 30u + 44v − 10 = 0 …(4)
• 40u + 55v = 13, i.e. 40u + 55v − 13 = 0 …(5)

Solving equations (4) and (5), we get u = 1/5, v = 1/11.

Substituting these values of u, v back in equation (3):

1/(x − y) = 1/5, and 1/(x + y) = 1/11

i.e. x − y = 5 and x + y = 11 …(6)

Solving for x and y:

x = 8, y = 3

So the speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h. To verify, substitute the values of x and y in equations (1) and (2).

Types of Lines Represented by a Pair of Linear Equations

Parallel Lines

Parallel lines are lines in a plane that never meet. This is possible only when the perpendicular distance between them remains constant, i.e. it never changes.

For example, the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Intersecting Lines

Two distinct lines in a plane that have exactly one point in common are called intersecting lines, and this common point is called the point of intersection.

For example, the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are intersecting lines if a₁/a₂ ≠ b₁/b₂.

Coincident Lines

When one or more lines lie exactly on top of another line, they are called coincident lines.

For example, the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincident if a₁/a₂ = b₁/b₂ = c₁/c₂.

A Real-Life Example — Classroom Arrangement

Students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution: Let the number of students in a row be x, and the number of rows be y. Then the total number of students = x × y = xy.

According to the first condition, if there were 3 more students in a row, there would be 1 row less. So the number of students in a row = x + 3, and the number of rows = y − 1.

Total number of students: (x + 3)(y − 1) = xy

Simplifying: x = 3y − 3 …(1)

According to the second condition, if there were 3 fewer students in a row, there would be 2 rows more. So the number of students in a row = x − 3, and the number of rows = y + 2.

Total number of students: (x − 3)(y + 2) = xy

Simplifying: 2x = 3y + 6 …(2)

Solving equations (1) and (2):

x = 9 and y = 4

So the total number of students = xy = 9 × 4 = 36.

Key Points to Remember

• A pair of linear equations in two variables: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
• Graphically, the two lines can intersect (unique solution), be parallel (no solution), or be coincident (infinitely many solutions).
• a₁/a₂ ≠ b₁/b₂ → unique solution (intersecting lines, consistent pair).
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution (parallel lines, inconsistent pair).
• a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many solutions (coincident lines, dependent and consistent pair).
• Algebraic methods to solve a pair of linear equations: substitution method, elimination method, and cross-multiplication method.
• Cross-multiplication formula: x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁).
• Some non-linear equation pairs can be reduced to linear form using a substitution such as 1/x = p, 1/y = q.

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