Primary School Students‎ > ‎Algebra‎ > ‎

Linear Equation in two variables

Simultaneous Linear Equations

Linear equations are those equations which are linear in its variables i.e. the power of the variables are 1. A single variable linear equation looks like this: ax = b where x is the variable. A double variable linear equation looks like this: ax + by = c and a triple variable linear equation looks like this: ax + by + cz = d. The number of variables can extend to any number of variables.

The solution of a linear requires as much equations as there are variables in it. The number of variables is equal to the number of equations. When we find the solution of a set of linear equations then the solutions can be unique or dependent or no solution. The solution of a linear equation can be found with the help of substitution, elimination or with the help of determinants. Let us look at the solutions of some linear equations.

Single variable linear equation

A single variable linear equation ax = b has solution x = b/a. When we plot it on a graph then we get a straight line parallel to y-axis. As the value of y is not present in the equation therefore the solution is independent of y.

Double variable linear equation


A double variable linear equation can be solved either by substitution method or elimination method. The solution on a graph is the point of intersection of the lines represented by the two equations.

The equations are 2x − y = 3 and 4x + y = −6

Let us solve the above two equations by substitution method and elimination method.

Substitution method


In substitution method we express one variable in terms of the other and substitute it to get value of one variable. Then substitute it in any equation to get value of other variable.

Step 1:

Express any one equation in terms of one variable.
2x−y = 3 ⇒ y = 2x − 3

Step 2:

Substitute it in next equation.
4x + y = −6
⇒ 4x + 2x − 3 = −6
⇒ 6x = −3
⇒ x = −(1/2) = −.5

Step 3:

Substitute value in the first equation.
y = 2x − 3 = 2(−.5) − 3 = −4
So solution is x = −.5 and y = −4

Elimination method


The elimination method is based on the fact that if one variable is removed from the equation then the reduced equation contains only one variable. After we can find the solution. Substitute one solution in one equation to get the other solution.

Step 1:

Eliminate one variable by making coefficient of that variable equal in two equations.
2x − y = 3
4x + y = −6
Add the two equations
6x = −3
x = −(1/2) = −.5
Substitute in anyone
2(−.5) − y = 3
y = − 4

the linear equations of multi variables is found by Gauss elimination method. We will discuss about such methods in some other posts.

Linear Equation in two variable

Let us first look at the form of linear equations in two variables.

ax + by = c

Such equations arise when we have two things changing at the same time. The simplest is to watch it on a graph. The x-axis forms one variable and the y-axis forms the other variable. A line has infinite number of solutions but we have to get a unique solution. And this is possible only when the number of lines is two and they intersect or in other words we have two linear equations in two variable. Hence to solve a system of two variable linear equation the required equations are

ax+ by = c
dx + ey = f
A system of two variable linear equation is solved either by Substitution method or by Elimination method.

Substitution method


In substitution method we find the value of one variable in terms of the other and substitute it in the other equation.

ax + by = c ----(i)
dx + ey = f ----(ii)
From (i) x = (c - by)/a

Substituting in second we get
d(c - by)/a + ey = f
dc/a - dby/a + ey= f
dc/a - (db - ea)y/a = f
dc - (db - ea)y = fa
(db - ea)y = (dc - fa)
y = (dc - fa)/(db - ea)

Substituting in (i) we get
ax + b(dc - fa)/(db - ea) = c
ax = c - b(dc - fa)/(db - ea)
ax = [c(db - ea) - b(dc - fa)]/(db - ea)
x = [cdb - cea - bdc + bfa]/a(db - ea)
x = (bf - ce)a/a(db - ea)
x = (bf - ce)/(db - ea)

x = (bf - ce)/(db - ea) = (ce - bf)/(ea - db)
y = (dc - fa)/(db - ea) = (fa - dc)/(ea - db)

Elimination method


In elimination method we eliminate one variable by equaling the other variable in both the equations. Then we substitute the value of first variable to get the value of other variable.

ax + by = c ----(i)
dx + ey = f ----(ii)

Multiplying (i) by e and (ii) by b, we get
aex + bey = ce ----(iii)
dbx + bey = fb ----(iv)

Subtracting (iv) from (iii) we get
(ae - db)x = (ce - fb)
x = (ce - fb)/(ae - db)
As substituted above in substitution method we get,
y = (fa - dc)/(ea - db)

x = (ce - bf)/(ea - db)
y = (fa - dc)/(ea - db)


Comments