Solving Simultaneous Linear Equations
At the end of this section you should be familiar with:
• How to solve two simultaneous equations in two unknowns
using:
(a) Algebra: i) Substitution Method and ii) Elimination Method.
(b) Graphical methods
• Determining when two equations in two unknowns have:
(a) A unique solution
(b) No solution
(c) Infinitely many solutions
• Solving three simultaneous equations in three unknowns.

A solution of an equation in an unknown, say x, is simply the value
for x for which the left-hand side (LHS) of the equation is equal to
the right-hand side (RHS).
For example, consider the equation x + 4 = 6
The solution of this equation is x = 2
x = 2 is the only value of x for which the LHS = RHS.
The statement 'x = 2 satisfies the equation' is another way of saying
that x = 2 is a solution.

Two equations in two unknowns
Example: Given the simultaneous equations
x + 3y = 4
-x + 2y = 6
(a) Solve for x and y algebraically.
(b) Solve for x and y graphically.
Solution: First we will solve by using Substitution method. We number equations as
x + 3y = 4 ………………(1)
-x + 2y = 6 ………………(2)
Now we find
x = 4 – 3y …………….(3). Now we substitute this value of x into equation (2).
- (4 – 3y) + 2y = 6 or, -4 + 3y +2y = 6 or, 5y = 6+ 4 or, 5y = 10 or, y =10/5 = 2. Now we
substitute this value of y = 2 into equation (3). x= 4 – 3(2) = 4 – 6 = -2.
So the solution is: (x, y) = (-2, 2). Now you can check.

Graphical Method
In graphical method, first we will draw two straight lines in the same
graph. The point where the two lines cross each other, that point is
the solution point.
For drawing the straight lines we we have to select some points for
both the lines separately:
x + 3y = 4
1+3y=4
3y=4-1=3
Y =3/3=1 x+3y=4
-2 +3y=4
3y=4+2=6 y=6/3=2
x y (x, y)
1 1 (1,1)
-2 2 (-2,2)

Put x=0,
-x + 2y = 6
0+2y=6
y=6/2=3
Put x=2, -2+2y =6 2y=6+2=8 y=8/2=4.
Draw the graph by drawing two straight lines. The two lines cross at the solution point (-2,2)
Graph by hand.
x y (x,y)
0 3 (0,3)
2 4 (2,4)

Method of Elimination
Example: Given the simultaneous equations
x + 3y = 4 ……………(1)
-x + 2y = 6 …………..(2)
Solve for x and y algebraically.
Solution: Eliminate x from the system of equations by adding equations (1) and
(2). The two equations reduce to a single equation in which the only unknown
is y. Solve for y, then substitute the value of y into either of the original
equations and solve for x:
x + 3y = 4 ………. (1)
-x + 2y = 6 …………(2)
----------------------------- Adding
0+ 5y =10 so y = 10/5 = 2.
Solve for x by substituting y = 2 into either equation (1) or equation (2):

-x + 2(2) = 6 substituting y = 2 into equation (2)
-x = 6 – 4=2
x = -2
So (x, y) = (-2, 2).
Example 2: Given the simultaneous equations [Solve by Elimination Method].
2x + 3y=12.5 ……………(1)
-x + 2y = 6 ……………..(2)
In this example, neither the x nor the y terms are the same. However, all terms
on both sides of any equation may be multiplied by a constant without
affecting the solution of the equation. So, if equation (2) is multiplied by 2, the
x terms in both equations will be the same with opposite signs. Then, proceed
as in previous Example.

Eliminate x from the system of equations by multiplying equation (2) by 2
and then add the equations; 2x and -2x cancel to leave a single equation in one
unknown, y:
2x + 3 y = 12.5 ………. (1) as given
-2x + 4y = 12 ……….. (2) x 2
-----------------------------------------
0 + 7y = 24.5 adding
Y=24.5/7 = 3.5
Solve for the value of x by substituting y = 3.5 into either equation (1) or
equation (2).
-x + 2(3.5) = 6 substituting y = 3.5 into (2) to find the value of x.
-x = 6 – 7 = -1.
So x=1 (x, y) =(1, 3.5)

Example: Given the simultaneous equations
2x + 3y = 2 …………(1)
5x + 2y = 6 ………….(2)
In these two equations, neither the x nor the y terms are the same. If equation (1) is
multiplied by 2 and equation (2) is multiplied by -3, the y terms in both equations will be the
same with opposite signs. Then proceed as in previous Example above.
Eliminate y terms from the system of equations:
4x + 6y = 4 …….(1) x2
-15x – 6y = -18 …….(2) x -3
-------------------------------------------Adding
-ll x + 0 = -14
x=14 /11= 1.2727.
Solve for y by substituting x = 1.2727 into either equation (1) or equation (2):
2(1.2727) + 3y = 2 substituting x = 1.2727 into equation (1)
3y = 2 – 2.5454 y=-.5454/3=- .1818 (x, y) =(1.2727, -0.1818).

Unique, infinitely many and no solutions of simultaneous
equations
A set of simultaneous equations may have
• A unique solution
• No solution
• Infinitely many solutions.
Unique solution :This occurs when a set of equations has one set of
values which satisfy all equations.
No solution: This occurs when a set of equations has no set of
values which satisfy all equations.
Infinitely many solutions: A set of equations has infinitely many
solutions when there is an infinite number of sets of values that
satisfy all equations.

Example: Given the simultaneous equations
y = 1 + x
y= 2 + x
(a) Solve for x and y algebraically.
(b) Solve for x and y graphically.
Solution: y = 1 + x
y= 2 + x
___________Subtracting equations
0 = -1. 0 = — 1 is not possible, therefore, there is no solution.
A false statement (or a contraction) like 0 = — 1 indicates a set of equations
with no solution.
The two equations are plotted in Figure 3.3. The lines will never meet since
they are parallel and thus will never have a point (solution) in common. Online
page#114.

SIMULTANEOUS EQUATIONS WITH INFINITELY MANY
SOLUTIONS
Example: Given the simultaneous equations
y = 2 — x …………….(1)
2y = 4- 2x ……………… (2)
Solve algebraically and graphically.
Solution: When equation (1) is multiplied by -2,
-2y = -4 +2x
2y = 4 – 2x
------------------------------Adding
0 =0 Zero indeed equals zero. In this situation we say, there is unlimited number of
solutions for these simultaneous equations.
Graph, page # 115.