 Hello and welcome to the session. In this session we will discuss linear equations in one variable. Equations with linear expressions in one variable only are known as linear equations in one variable. Then we have an algebraic equation is an equality involving variables. Like if you consider the equation 2x minus 3 equal to 5, this is an algebraic equation. And as you can see there is an equality sign there. And the expression to the left side of this equality sign that is this expression is the LHS that is the left hand side. And the expression on the right side of this equality sign that is this expression is the RHS that is the right hand side. Then next is in an equation the values of the expressions on the LHS, RHS are equal only for certain values of the variable. And these values are the solutions of the equation. Like for this equation 2x minus 3 equal to 5, the value of the LHS would be equal to the value of the RHS for some values for the variable x. And these values of the variable x are called the solutions of this equation. While finding the solution of an equation we assume that the equation is balanced and then we perform some mathematical operations on both the sides of the equation so that the balance is not disturbed. Consider the equation 2x minus 3 equal to 5, we assume that this equation is balanced and then we perform mathematical operations on both sides of this equation so that the balance is not disturbed. Like we add 3 on both the sides of this equation so this gives us 2x equal to 8. Now we need to find the value of the variable x so we divide both sides of the equation by 2 so this would give us x equal to 4 is the solution of the equation. Now we can also check whether x equal to 4 is the correct solution or not. Let's see how we can check this. We take x equal to 4 and put that value in the LHS and the RHS of the given equation. Now since in the given equation we have that the variable is just on the left-hand side so we substitute x equal to 4 in the left-hand side itself. So consider the LHS it is 2x minus 3 now we put x equal to 4 in this we get 8 minus 3 that is equal to 5 and as you can see the RHS is also 5 so we get that LHS is equal to the RHS hence x equal to 4 is the correct solution. This is how we solve the equations which have linear expressions on one side and the numbers on the other side. Next let's try solving equations having the variable on both sides and equation is the equality of the values of 2 expressions. Consider this equation 3x equal to 2x plus 18 it has variables on both sides as you can see. Let's try and solve this equation. Now we assume that the both sides of this equation are balanced and now we can perform some mathematical operations so as to get the value of the variable x. Now first we transpose this 2x from the right-hand side to the left-hand side so we get 3x minus 2x is equal to 18 that is we get x is equal to 18. So we have got the solution of the given equation. Next we check whether this x equal to 18 is the correct value for the given equation or not. Consider the LHS it is 3x now we substitute x equal to 18 in this so we have 3 into 18 so this gives us 54. Next we consider the RHS it is 2x plus 18 we take x equal to 18 in this so we get 2 into 18 plus 18 and this is equal to 54. Now as you can see that the value of the LHS is equal to the value of the RHS hence x equal to 18 is the correct solution of the equation. Next we have reducing equations to simpler form. Sometimes the expressions forming the equations have to be simplified before we can solve them using the usual methods. Let's see how we can simplify an equation. Consider the equation x minus 5 upon 3 equal to x minus 3 upon 5. Now this is not in a simple form first we will reduce this to a simpler form and then solve it by usual methods. To do this we multiply both sides of the equation by the LCM of the denominators of the both sides that is 3 and 5 LCM of 3 and 5 is 15. So we multiply both sides of the equation by 15 and we have 3 5 times as 15 5 3 times as 15 so we have 5 into x minus 5 is equal to 3 into x minus 3. Now we open the brackets on both the sides so we get 5x minus 5 into 5 that is 25 equal to 3 into x is 3x minus 3 into 3 is 9. Now we have reduced the equation to a simpler form now we can solve it by usual methods. To do this we transpose this 3x to the left hand side so we get 5x minus 3x minus 25 is equal to minus 9 and then we transpose this minus 25 to the right hand side. So we get 5x minus 3x is equal to 25 minus 9 that is we get 2x is equal to 16. Now to get the value of the variable x we divide both the sides by 2 so we get x is equal to 8 is the solution of the given equation. We can also check if x equal to 8 is the correct solution or not. Consider the equation x minus 5 is equal to x minus 3 upon 5. Now let's take the LHS first it is x minus 5 upon 3 we substitute x equal to 8 in this so we get the value of the LHS as 1. Then next we consider the RHS x minus 3 upon 5 and we substitute x equal to 8 in this so this gives us 1. Thus we see that LHS is equal to the RHS hence the solution x equal to 8 is correct. There we have equations reducible to linear form sometimes the given equations are not in a linear form. So we can make them or we can reduce them to a linear form by multiplying both sides of the equation by a suitable expression. Consider the equation 8x minus 3 upon 3x is equal to 2 this is not in a linear form. Let's reduce this equation to a linear form by multiplying both sides of this equation by the expression 3x. So we have 3x into 8x minus 3 upon 3x equal to 3x into 2 so this gives us 8x minus 3 is equal to 6x. Now we can easily solve this equation. First we transpose the 6x to the left hand side so we get 8x minus 3 minus 6x equal to 0. Then we transpose this minus 3 to the right hand side so we get 8x minus 6x equal to 3 this gives us 2x is equal to 3. Now to get the value for the variable x we divide both the sides by 2. So we get 2x upon 2 is equal to 3 upon 2 which gives us x equal to 3 upon 2 is the solution of the given equation. Let's check if x equal to 3 upon 2 is the correct value or not. Consider the equation let's first take the LHS of this equation that is 8x minus 3 upon 3x. Now we substitute x equal to 3 upon 2 in this so we get 8 into 3 upon 2 minus 3 upon 3 into 3 upon 2. So this gives us 9 into 2 upon 9 that is equal to 2. Now let's consider RHS as you can see that RHS is 2 thus we see that the value of the LHS is equal to the value of the RHS hence the given solution x equal to 3 upon 2 is correct. Now let's discuss some applications of linear equations in one variable. We can solve different problems on numbers, ages, perimeters, combination of currency nodes and so on using the linear equations. Let's consider a problem which says that if we subtract 1 upon 2 from a number and multiply the result by 1 upon 2 we get 1 upon 8. What is the number? Let's try and solve this problem using linear equations. Since we need to find the numbers so let the number be equal to x. According to the equation we have that if we subtract 1 upon 2 from the number that is x minus 1 upon 2 and we multiply the result by 1 upon 2 that is 1 upon 2 into x minus 1 upon 2 we get 1 upon 8. Now we multiply both sides of this equation by 2 so we get 2 into 1 upon 2 into x minus 1 upon 2 equal to 2 into 1 upon 8. So this gives us x minus 1 upon 2 equal to 1 upon 4. Now to get the value for the variable x we transpose this minus 1 upon 2 to the right hand side. So we get x equal to 1 upon 2 plus 1 upon 4 which gives us x equal to 3 upon 4. Hence we say the required number is 3 upon 4. Let's check if this is correct or not. The number that we got is 3 upon 4. Now according to the equation we have that if we subtract 1 upon 2 from this number that is 3 upon 4 minus 1 upon 2 this gives us 1 upon 4 and then if we multiply this result by 1 upon 2 we should get 1 upon 8. So 1 upon 2 into 1 upon 4 we get 1 upon 8. Hence this number 3 upon 4 is the correct number. This is how we use linear equations to solve different problems. So this completes the session. Hope you have understood the concept of linear equations as one variable.