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 Algebra - II « PreviousNext » Indices (Exponents) 
Author: S.LAL
Created: 21 Aug, 2010; Last Modified: 29 Apr, 2018

Algebra II - 01

Simultaneous Linear Equations

An equation which has linear polynomials of two variables is a linear equation in two variables or a simultaneous linear equation. Some examples are:

(i) 7 x + 2 y = 27 (ii) 2 3 x − 3 5 z = 4
(iii) 3 x + 2 y − 5 2 = 0  

A simultaneous linear equation has the form:

a x + b y + c = 0 { a , b , c ∈ R ; a   ≠   0 , b   ≠   0 }
.......(1)

Each problem involving simultaneous linear equations contains a set of two equations which relate the two unknown quantities or variables. These equations have to be considered together to find the solutions.

Solving simultaneous equations

Two common methods for solving simultaneous equations are the method of elimination by substitution and the method of elimination by equating coefficients.

Method of elimination by substitution

The steps are:

  1. Use any one equation and express one unknown variable in terms of the other by making it the subject of the formula.
  2. Substitute this value for the unknown variable into the other equation.
  3. Solve the resulting other equation.

Solve { 2 x − 7 y = − 8 3 x + 2 y = 13 . (Schultze, Elements Algebra, p. 133 Art. 177)

Consider the 1st equation. Solving for x we get:

2 x − 7 y = − 8 ⇒ 2 x = 7 y − 8 ⇒ x = 7 y − 8 2

Substituting this value of x in the second equation gives:

3 x + 2 y = 13 ⇒ 3 ( 7 y − 8 2 ) + 2 y = 13 ⇒ 3 ( 7 y − 8 ) + 4 y = 26 (Multiplying with 2 throughout) ⇒ 21 y − 24 + 4 y = 26 ⇒ 25 y = 50 ⇒ y = 2

Thus, y = 2. Putting this value for y in any of the equations gives x = 3.

Method of elimination by equating coefficients

The steps are:

  1. Multiply, if necessary, the equations by such numbers as will make the coefficients one of the unknown variables equal (the targeted coefficient is the LCM of the original coefficients for the unknown variable).
  2. If the signs of these coefficients are the same, subtract the equations; if opposite, add them. This will eliminate one of the unknown variables.
  3. Solve the resulting equation for the other unknown variable. The solution can be substituted in any of the original equations to get the value of the eliminated unknown variable.

Solve { 3 x + 2 y = 13 2 x − 7 y = − 8 . (Schultze, Elements Algebra, p. 130 Art. 175 Ex. 1)

Consider eliminating x from the equations.

On inspection, if we multiply the 1st equation with 2, and the 2nd equation with 3, the coefficients of x in both equations become the same, i.e. 6, which is the targeted LCM for the original coefficients of 3 and 2 for variable x. (Remember that multiplication of both sides of an equation by the same non-zero number does not change the equality).

Thus, we have

{ 2 × ( 3 x + 2 y = 13 ) 3 × ( 2 x − 7 y = − 8 ) ⇔ { 6 x + 4 y = 26 6 x − 21 y = − 24 ...(1) ...(2)

Subtracting (2) from (1) gives;

25 y = 50 ⇒ y = 2

With y = 2, substituting in any of the original equations gives x = 3.

A shop sells two kinds of cycles – bicycles and tricycles. There are 7 cycles in the shop having a total of 19 wheels. Determine how many bicycles and tricycles are there in the shop, considering that a bicycle has two wheels and a tricycle has three. (Adapted from: FHSST, Mathematics 10 - 12, p. 102)

Let b be the number of bicycles, and t be the number of tricycles in the shop.

Since there are 7 cycles in the shop, therefore

b + t = 7 ...(1)

Also, if we consider the total of the wheels being 19,

2 b + 3 t = 19 ...(2)

(1) and (2) form a pair of simultaneous equations. From (1) we get

b = 7 − t

Substituting the value of b above into (2), we have,

2   ( 7 − t ) + 3 t = 19 ⇒ 14 − 2 t + 3 t = 19 ⇒ t = 19 − 14 = 5

Substituting the value t = 5 obtained above into (1), we get b = 2. Thus, the shop has 2 bicycles and 5 tricycles.

Solve { 7 x + 8 y = 2 2 x + 12 y = 20 . ICSE

In this case, it will be easier to solve the equations if we let

a = 1 x and b = 1 y

Thus, the original equations can be written in the form:

{ 7 a + 8 b = 2 2 a + 12 b = 20

Using the method of "elimination by equating coefficients" to first eliminate a, it can be seen that the targeted coefficient for a is 14 (LCM of 7 and 2). Thus,

{ 2 × ( 7 a + 8 b = 2 ) 7 × ( 2 a + 12 b = 20 ) ⇔ { 14 a + 16 b = 4 14 a + 84 b = 140 ...(1) ...(2)

Subtracting (2) from (1) gives:

− 68 b = − 136 ⇒ b = 2

Thus, with b = 2, substituting in any of the equations gives a = −2.

This, in turn, implies that x = − 1 2 and y = 1 2


Of course, these kinds of problems can also be solved in the normal way without involving a and b, in which 1 x and 1 y themselves are considered as unknown quantities, rather that x and y.

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List of References

Bansal, RK, Concise Mathematics I.C.S.E., Part I – Class IX, New Delhi: Selina Publishers, 2005.
FHSST, Mathematics Grades 10 - 12, ver 0, viewed 10 March, 2009, <http://www.fhsst.org>, 2008.
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.

Bibliography

Bansal, RK, Concise Mathematics I.C.S.E., Part I – Class IX, New Delhi: Selina Publishers, 2005.
Gupta, SD & Banerjee, A, ICSE Mathematics for Class 9, Patna, India: Bharati Bhawan, 2003.
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.

 Algebra - II « PreviousNext » Indices (Exponents) 
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