Simultaneous Linear Equations up to Three Variables

The System of Equations: Suppose you went to the market and bought trousers and a pair of shirts. Your friend asks you the cost of each of the item. You tell him that the cost of the trousers is three hundred more than the cost of one shirt. The total amount you paid to the shopkeeper was one thousand and fifty rupees.

Obviously, your friend can calculate the prices of each of the item. How does he do so? He simply translated the statements into a system of equations and got the answer. But what will happen if we have more items and the number of the simultaneous equations is also more than two? In this section, we are going to learn about the system of equations and the methods to solve them.

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The System of Simultaneous Linear Equations

An equation is a mathematical statement showing the relationship of equality. It is a general form of showing the relationship by using a combination of letters, numbers, and symbols. The constant values have the fixed values like 12, 5, −4 etc. and the variables denote the unknowns. They are represented by English letters like a, b, c, x, y, z etc.

Browse more Topics under Business Mathematics

• Ratios
• Proportions
• Laws of Indices, Exponents
• Logarithms and Anti-Logarithms
• Quadratic and Cubic Equations in One Variable
• Types and Algebra of Matrices
• Determinant of a Matrix
• Inverse of a Matrix
• Solving System of Equations by Cramer’s Rule
• Linear Inequalities

I. System of Equations in Two Variables

A relationship between two unknown values is shown by an equation. A set of values for the two unknowns satisfy the equality statement. The general form of an equation is ax + by = c. Here, a, and b are non−zero coefficients and c are the constants and x, and y are variables. Two such equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ are a pair of simultaneous equations in x, and y.

Translating Statement into Mathematical Equations

To solve real-life problems, we need to convert them into mathematical form.

1. Note down the facts from the problem.
2. Use Variables for denoting unknown quantities.
3. Try to establish a relationship between known and unknown quantities.
4. Form an equation by this relationship.
5. Solve it to get the desired result.

Methods of Solving a System of Equations in Two Variables

A value of each of the unknown or variable satisfies both the equation simultaneously. These values of the variables are the roots of the equations.

A. Elimination Method

In this method, we eliminate one of the variables (say x) from the equation. In other words, two linear equations are reduced to form only one linear equation by eliminating one of the unknowns. Suppose we have,

x = 16 – y… (a)
and, x = 4 + y … (b)

Since the L.H.S. of both the equations is same. We have, 16 – y = 4 + y or, 12 = 2y or, y = 6 and solving for x, we have x = 10.

B. Cross Multiplication Method

In this method we cross – multiply the equations with respective coefficients. Subtracting the two equations thus formed gives the required answer. If the two equations are

a₁ x + b₁ y + c₁ = 0, and
a₂ x + b₂ y + c₂ = 0,

in more general form, we have

Assume we have,

x + y = 16 … (i)
x – y = 4 … (ii)

Multiplying (i) by (−1) and (ii) by (1), and then subtracting the new equations we have,

−x – x – y + y = −16 – 4

or, −2x = −20

or, x = 10 and solving for y, we have y = 6.

II. System of Equations in Three Variables

A relationship between three variables shown in the form of a system of three equations is a triplet of simultaneous equations. The general form of equations in this form is ax + by + cz = d. Here, a, b, and c are non – zero coefficients, d is a constant. Here, x, y, and z are unknown variables.

Methods of Solving a System of Equations in Three Variables

A value of each of the unknown or variable satisfying all the equation simultaneously gives the roots of the equations.

A. Elimination Method

This method is similar to that in two variables. One of the variables is eliminated to form a linear equation and the equations are then solved. Suppose we have,

2x – y + 3z = 10 …(a)
x + 3y – 2z = 5 …(b)
and, 3x – 2y + 4z = 12 … (c)

Here, we eliminate y by multiplying (a) by 3 and then solving the first two equations. We proceed in a similar way and solving the equations, we have x = 2, y = 3, and z = 3.

B. Cross Multiplication Method

In this method, we find the value of x and y just as we find the values for equations in two variables. From the known values of x and y and using any of the equations we calculate the value of z. Suppose we have,

2x – 2y + 3z = 20 …(a)
x + 3y – z = 12 …(b)
and, 3x – y + 4z = 22 … (c)

Using the first two equations and putting the values in the formula for x and y, we have

x = [(– 2 × (– z – 12)) – 3 × (3z – 20)] ÷ [(2 × 3) – (1 × – 2)]. Here, (3z – 20) and (– z – 12) are the respective c₁ and c₂.

or, x = (– 7z + 84) ⁄ 8. Similarly, y = (5z + 4) ⁄ 8.

Substituting these values of x, and y in (c) we have

x = 21, y = – 7, and z = – 12.

Solved Example for You

Problem: Which of the following is the correct values of x, and y for the equations

2x – 3y – 12 = 0, and
x – 6y – 15 = 0.

The options are:

1. x = – 2, y = 3
2. y = – 2, x = – 3
3. x = 3, y = – 2
4. y = – 3, x = – 2

Solution: Solving the above equations by the method of cross – multiplication, we have

x = [(–3 × –15) – (–6 × –12)] ÷ (2 × –6) – (1 × –3)] = 3.

and, similarly, y = – 2. The correct choice is (c) i.e., x = 3 and y = –2