Solving Simultaneous Equations with Examples

Simultaneous equations involve finding the variables (unknowns) of two two equations. Usually the unkowns are x & y.

Two common methods of finding the unknowns of simultaneous equations are method of elimination & method of substitution.

Method of Elimination

This method involves eliminating one variable & work the other. Once you find the value of one unknown you then substitute it again in one of the equations to get the other unkown.

Example: Solve the following simultaneous equations
5x + 2y = 10...(1)
4x + 3y = 1...(2)

• Multiply the constant of equation (1) with the constant of equation (2) & the constant of equation (2) with equation (1).
• Subtract & eliminate one variable
• Solve resultant equation

4(5x + 2y = 10)...(1)
5(4x + 3y = 1)...(2)

(20x + 8y =40) -
20x + 15y = 5
Notice the first terms are the same (they will eliminate each other)

8y - 15y = 40-5
-7y = 35
y = 35/-7
y = -5
Now substitute y in one of the equations

5x + 2y = 10
5x + 2(-5) = 10
5x = 10 +10
5x= 20
x = 20/5 = 4

You can first eliminate the y term instead of the x term Example 2:

3x - y = 10...(1)
x + 2y = 28...(2)

1(3x - y =10)
3(x + 2y = 28)
Note that constant of x is 1

3x - y =10-
(3x + 6y = 84)

3x -3x -y - 6y = 10 - 84
-7y = -74
y = -74/ 7
Substitute in (1)

3x - 74/7 = 10
3x = 10+ 74/7
x = (144/7)/3
x = 48/7

Method of Substitution

Method of substitution is rather more complicated than the elimination method. But it is very useful in many circumstances when presented with complex expressions like quadratics. How do you solve this using elimination:

y = x2
x + y = 4

It's rather difficult than using substitution. You can encounter equations like these if you want to find the point of interception between two line.

• Make one variable in either equations the subject.
• Substitute the resultant expression in the equation you didn't make a variable subject.
• Solve the equation

Example 3

Solve
2x - 3y = -2
y + 4x = 24...(2)
Find expressions that are easier to substitute in and will not be difficult to manipulate

Make y subject, y = 24 - 4x

Substitute in equation (1)

2x -3(24 - 4x) =-2
2x - 72 + 12x = -2
Group like terms
14x = 72-2
14x = 70
x = 70/14
x = 5

Now substitute x = 5 in any one equation
(Taking equation (1)
y + 4(5) = 24
y = 24-20
y = 4

If, for example, you make x the subject you end up with complex fractions that are difficult to solve.

y + 4x = 24
Make x the subject
4x = 24 - y
x = (24 - y)/4

substituting this in (1)
2(24 - y)/4 -3y = -2
(24 - y)/2 - 3y = -2
Simplify this you end up with
24 - 7y = -4
7y = 24+4
7y = 28
y = 28/7 = 4

Substitute y = 4 in (1)

2x - 3(4) = -2
2x -12 = -2
2x = 12-2
x = 10/2 = 5

The second substitution is more tedious and prone to errors if your algebra is not strong. Find an easy expression to substitute that doesn't involve fractions.

Example 4

2x - y = 3...(1)
x + y = 2...(2)
Make x the subject in equation (2)
x = 2 - y
Substitute in (1)

2(2-y) - y = 3
4 - 2y - y = 3
4 - 3y = 3
-3y = 3-4 = -1
y = -3/-1 = 3

substitute in (2)
x + 3 = 2
x = 2-3 = -1

Example 5

3x - y = 18...(1)
2x + y = 7...(2)
Make y subject in equation (2)
y = 7 - 2x
Substitute in equation (1)
3x -(7-2x) = 18
the (-) sign in front of y affects the whole expresaopm so you have to put brackets
3x -7 +2x = 18
5x = 18 + 7 = 25
x = 25/5 = 5

Substitute x in any one

2(5) + y = 7
10 + y = 7
y = 7 - 10
y = -3

NB: try to solve example 5 using elimination

You can use either one method to solve simultaneous equations depending on the given expressions. If your algebra is weak then substitution is not ideal for you. Some questions can force you to use either one of the methods.

Substitution is more ideal in higher levels when dealing with simultaneous equations involving quadratics.

Other examples

Sometimes you can be give equations in reverse order, e.g

5x + 4 = y

and

x - 7 = -2y

You can rearrange to the common arrangement.

-y + 5x = -4
2y + x = 7

Then you can solve the equations
Check out Simultaneous equation worked examples overdose