E x a mp l e 1 . 10
Sketch the graphs of the following equations.
3
2
2
6
x
y
x
y
S o l u t i o n
As shown in the opposite graph
.
Ac Activity x.x
Spend at least 20 -30 minutes on this activity.
State whether the following system of linear equations are coincident, parallel or
intersect. Explain your answers.
a)
2
5
4
2
10
x
y
x
y
b)
2
3
6
4
6
0
x
y
x
y
c)
2
5
4
x
y
x
y
3
3
x
y
Activity 1.3

Unit 1: Basic Algebra and Linear Equations
1.15
1.4
Solving systems of linear
equations
There are many techniques of solving systems of linear equations. These include
graphical methods, algebraic methods and elementary row operation. However in
this course, we will be dealing with algebraic methods only.
Algebraic methods are where algebraic operations are performed on linear equations
to yield an equation with only one variable and then solve the new equation. The
rules for obtaining any two equivalent equations are shown below.
Equation Operations Producing Equivalent Systems
Equivalent systems of equations are systems that have the same solution set
.
There are two algebraic methods to study and these are
Substitution
and
Elimination Methods
.
Substitution Method
E x a mp l e 1 . 11
2
3
3
(1)
Solve
4
(2)
x
y
x
y
Summary of Steps
1.
Make one variable the subject, for one of the equations.
2.
Substitute the resultant equation in the remaining equations.
3.
If we obtain an equation with only one variable, solve it. Otherwise, repeat steps
1 and 2 until such an equation is obtained.
4.
Find the values of the remaining variables by back-substitution.
5.
Check the solution.
Theorem:
A system of linear equations is transformed into an equivalent system if:
(1)
Two equations are interchanged.
(2)
An equation is multiplied by a nonzero constant.
(3)
A constant multiple of one equation is added to another equation.

Unit 1: Basic Algebra and Linear Equations
1.16
S o l u t i o n
Step 1:
Making
the subject of equation (2) we have
4.
y
y
x
Step 2:
We substitute this result in equation (1) and simplify, we have
2
3(
4)
3
2
3
12
3
x
x
x
x
Step 3:
Since now we have only one variable (
), we solve for
5
15
3.
x
x
x
x
Step 4:
Knowing
3, we can find
by substituting the value of
into either (1) or (2).
Note that both give the same answers.
x
y
x
(This technique is also known as back substitution.)
In here we are using equation (2).
4
4
3
4
1.
x
y
y
x
2(3)
3( 1)
3
Step 5:
Check
( 1)
3
4
E x a mp l e 1 . 12
Solve the following system by substitution method.
36
9
(1)
3
12
(2)
3
y
x
y
x
S o l u t i o n
Substitute (1) into (2), we have
1
3
(36
9 )
12
3
3
12
3
12
12
12
x
x
x
x
This result does not give a value of either
y
nor
x
. However it gives a true statement.

#### You've reached the end of your free preview.

Want to read all 24 pages?

- Fall '17
- Arvind Patel
- Linear Equations, Equations, Elementary algebra