# Learn Basic Mathematics - Solving linear equations

Mathematics is an interesting subject if learned through the basic modules. Linear equations are very useful in finding out the unknowns in the given conditions and in this article we will try to learn how to solve simple linear equations.

### Introduction

Learning Basic Mathematics is fun. Though some people have less interest in Mathematics if small efforts are done to learn it then it becomes a very interesting subject. One way to learn Mathematics is through the gradual learnings through basic lessons. This article is an effort in that direction only to understand how to solve the linear equations.

### What is a linear equation ?

Algebra is the branch of Mathematics where solving various types of equations is taught. The Linear equation is the simplest of them. Once we learn to solve equations, we can find solutions to a variety of Mathematical problems. Linear equations are defined as those equations where multiple powers like square or cube or more powers are not encountered. We have only the simplest form of a variable or unknown. When we are given some data about certain entities and want to find some other things about them then these linear algebraic equations come handy.

For example, if we have bought 50 kg of sugar for Rs 1900 then to find out its rate per kg (say x) we can write an equation of the form -
50x = 1900 which on solving will give x = 1900/50 = Rs 38 per kg.
This is an example of a single variable (only one unknown) linear equation. This one equation is sufficient to find the value of the unknown x. Please remember that at least one equation is required to find the value of one unknown.

For solving a linear equation some basic things are to be kept in mind -
• A linear equation has two parts one is Left Hand Side (LHS) and other is Right Hand Side (RHS) and in between them, there is an equal to (=) sign.
• If we add or subtract anything to one side, we must add or subtract same to other side.
• If we multiply LHS by some number for simplification then we will also have to multiply the RHS with the same number.
• If we move an item from LHS to RHS or vice versa then its sign will change.
• If we change the sign of one item then the sign of all the items will change.

### Solving linear equations with one unknown

We will now see one more example, slightly difficult than the above, which would make the understanding more clear, of solving one linear equation with one unknown.

Example 1 : Ramesh has some chocolates. He gives 4 numbers out of that to Suresh and 2 numbers to Mahesh. He eats 2 chocolates himself. While observing the remaining chocolates he finds that he can divide them into three equal parts and he does that and gives one part to Suresh and one part to Mahesh and keeps rest with him. If Mahesh gets 5 chocolates in total then we have to find out how many chocolates Ramesh had originally.

To solve the above problem we will take help of basic algebra and make some linear equation. So, we would assume that let Ramesh had an unknown number of chocolates as x originally with him.
He gave 4 to Suresh and 2 to Mahesh so what he had left with him would be x-4-2 ==> x-6 (I am using ==> sign for expressing 'that is' or 'means').
Now out of this x-6, he ate 2 numbers himself so he had left with x-6-2 ==> x-8
Out of this x-8, he gave one third to Suresh and another one third to Mahesh.
It means that second time Mahesh (x-8)/3 chocolates.
So a total number of chocolates that Mahesh so far got was 2 + (x-8)/3.
But we have been given that Mahesh got 5 chocolates. This is the point at which a linear equation is formed because we have deduced that total number of chocolates that Mahesh got were 2 + (x-8)/3 and it is given that this quantity is nothing but 5, so equating them, the equation will come in being like this -
2 + (x-8)/3 = 5
In this equation, there is only one variable that is x and we can find its value by solving it. Please note that this equation is having two sides that are one left- hand side (LHS) and one right-hand side (RHS). For solving we will move 2 from LHS to RHS (its sign will change) and get -
(x-8)/3 = 5-2
so (x-8)/3 = 3
Now multiplying both sides by 3 we get -
(x-8) = 9
so taking 8 to RHS (its sign will change) we get -
x = 9 + 8 ==> 17 and this is the solution of the problem. Ramesh had 17 chocolates with him in the beginning. If one wants one can verify the whole sequence starting from the number 17 by going through the distribution of chocolates mentioned in the example.

### Solving linear equations with two unknowns

Now we will go to two -variable linear equations where we have two unknowns and we have to solve the two equations to find them. In this case, we have to get two equations by the information given in the question. So we will go through an example to understand how to solve the two- variable linear equations.

Example 2: Mohan has some fruits in a bag. He says that he has two types of fruits and their total number is 32 and he also says that one type of fruits is three times of other type. Can we find out the number of each type of fruits?

This is a question where two unknowns are there and we will require two equations for solving to find out their individual values. How to get those two equations? Let us assume these two unknowns as x and y where x is the number of fruit 1 and y is the number of fruit 2. So, it is given that the total number of fruits is 32 which gives us an equation like -
x + y = 32 …………….. (1)
This equation is a single equation and we can not solve it to get the values of these two unknowns and hence we search for another equation. We were told that one type of fruits are three times of the other one which means -
x = 3y ………………… (2)

So we have now two equations with us (1) and (2). How to solve them? We will use the method of substitution here which says that from one of the equation take the value of a variable (unknown) and put it in other equation. We have two options with us now as we can take the value of one of the unknowns from either of the equation and put it in the other one. The one which seems to be the easiest is to put the value of x from equation (2) in equation (1). One can do it in other ways also (can you just think of those other ways also?). So, if we put the value of x from equation (2) to equation (1) then equation (1) would become like -
3y + y = 32
which seems easy to go further -
4y = 32
Now dividing both sides by 4 it gives y = 8

So, we got the result that one of the fruits is 8 numbers in quantity. Now put this value of y in either of the equation and that will give us the value of x. I put this value of y (that is 8) in equation (1) and get -
x + 8 = 32
Now I take 8 from LHS to RHS, which changes its sign, and I get -
x = 32 - 8
which means -
x =24
With this, we have solved two linear equations to get the value of two unknowns. One type of fruit is 24 in numbers while the other is only 8.

### What if there are more unknowns

When unknowns are more, we require more equations to solve for them, at least equal to the number of unknowns. Let me illustrate it with a simple example.

Example 3: There are three items in a bag say copybooks, pencils, and sharpeners. It is told that copybooks and pencils together are 30 numbers, copybooks and sharpeners together are 28 numbers, and pencils and sharpeners together are 18 numbers. Please note that now variables are three so let us say x (copybooks), y (pencils), and z (sharpeners) are these unknowns which we have to find out.

So from the information given in this question we can form three equations and what we have now with us to solve is -
x + y = 30 …………… (1)
x + z = 28 …………… (2)
y + z = 18 …………… (3)

Now, the process of solving them will be slightly longer but our approach and logic would remain the same as in the earlier examples. We will first put the value of x from (1) in the equation number (2) and that would give us a new equation altogether between y and z. Let us do it.
From equation (1) we get -
x = 30 - y and putting this in equation (2) gives us -
30 - y + z = 28 which can be written as -
y - z = 2 ……………… (4)

Now, equation (3) and (4) can be solved with the same method that we used in the earlier example 2. Solving them would give us the values of y and z. Try yourself and you would get y=10 and z=8. If we use these values then from equation (1) or (2) we will get x = 20. That leads us to the answer of Example 3 as - copybooks are 20, pencils are 10, and sharpeners are 8.

In Mathematics, there are many methods to solve a particular problem and sometimes some tricks are also there which make our work easier if we know those tricks. For example in the case of example 3, one can solve these three equations by adding all of them and getting 2x + 2y + 2z = 30 + 28 + 18 ==> x + y + z = 36. Remember, we have to add LHS to LHS and RHS to RHS. From this new equation so obtained if we subtract the equations (1), (2), and (3) one by one (again LHS to LHS and RHS to RHS) we get the result for x, y and z quickly. Once the basics are learned one can learn these tricks also for quick computations.

### Conclusion

Linear algebraic equations are very useful in finding out the unknowns in any problem which can be transformed into linear equations. The only requirement is that number of equations should be at least equal to the number of unknowns. Even the problems which look quite complex on the face of it sometimes can be solved easily using linear equations.

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