Simultaneous equation is quite an easy mathematics to solve if you are familiar with the required steps. Note that there are two methods of solving any simultaneous equations either in WAEC, NECO, JAMB or any examination or test you are tried on this topic.
Simultaneous equation is one of the most common topics in mathematics that the West African Examination Council, WAEC test candidates for every year. You will either see it as an objective question or as a theoretical question and sometimes it could appear on both sections.
When simultaneous equation appears under theory section, most likely, you would be mandated to either attempt to solve the question using Graphical method, elimination method or substitution method because the examiners want to test your in-depth understanding on this maths topic. Not to worry, we have broken down the steps on how to solve simultaneous equation substitution method, elimination method as well as the graphical method, ,it would assist candidates or students in a long way in achieving the full score awarded to the question.
For the sake of this post, i will be putting you through the simple and complete breakdown on how to solve simultaneous equations using Substitution method.
Ideally, there are 2 methods of solving simultaneous equations;
- Graphical method
- Algebraic methods (Substitution and Elimination)
I will be citing some examples on how to solve simultaneous equation using The Substitution method.
Example 1:
Let’s take an example ;
6x + 4y = 15___(1)
12x – 3y = 10___(2)
From equation (1), lets make x the subject.
:. 6x = 15 – 4y
:. x= [15y-4y]/6___(3)
Now let’s put x for [15y-4y]/6 in equation (2)
12x – 3y = 10____(2)
:. 12[(15 – 4y)/6] – 3y = 10
:. If we look closely, we discover that 6 can divide 12, which gives 2.
:. 2(15 – 4y) – 3y = 10
Now we expand by removing the bracket.
:. 30 – 8y – 3y = 10
Now we take common terms.
:. – 8y – 3y = 10 – 30
:. – 11y = – 20
Now we divide through by -11.
:. (-11y)/-11 = (-20)/-11
Note: [- ÷ -] = +
:. y = 20/11
Now let’s substitute y for 20/11 in equation (3)
:. x = (15 – 4y)/6____(3)
:. x = 15 – 4[(20/11)]/6
Firstly we multiply through by 11..
x = [{15×11 – 4[(20/11)×11]}/11]/6
:. x = {[165 – 4(20)]/11}/6
Now from here;
x = [(165 – 80)/11]/6
x = (85/11)/6
After further simplification, we arrive at;
:. x = 85/66 = 1.2879 and
y= 20/11 = 1.8182
Did you find this simple to understand? if you have questions on how to best solve simultaneous equation using substitution method in WAEC, NECO, JAMB or any other examination, please reach us via the comment section below and we shall respond accordingly.
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