Equations With One Variable - Part I

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An equation relates quantities written on the left hand side to the quantities written on the right hand side. Sometimes quantities are unknown. In that case we have to use equations to find the unknown quantity. To make the discussion a bit more clear, say you are designing a square flowerbed. You are told that the area of the square cannot be larger than 10 m2. You know from your basic geometry that a square is a figure whose sides are equal. Let us assume that the side is x meters.  
x2  = 10 m2  

The last equation shows an equation with one variable.
A variable is generally indicated by an alphabet.
Variable is a quantity that varies under different conditions given in the equations. The constant numbers attached to a variable is called its coefficient. In the above example, the coefficient of x2 is 1.  

The quantity on the right hand side was known : 10m2.
Hence it is very easy to solve the equation. From the chapter of square roots, we can see that x would be a square root of 10 m2, which is 3.16m.

Instead of a square if the flower bed had to be made out as a rectangle, then we would have had  the following equation

x  *  y   = 10m2

Then x and y can take a whole range of values that satisfy the above equation. With just one equation, the values of x and y will not be unique.

This is an equation that has two variables. We will discuss such equations in later chapters.

What we will study in this chapter :

1. Equations with a variable on one side
2.  Equations with a variable on both the sides

1. Equations with a variable on one side  
Notice the following equations.

1.  5x + 2 = 7  

In this equation, the coefficient of variable x is 5. There is a constant term 2 on the left hand side and 7 on the right hand side.  
The following is the method for solving this equation and determining the value of x :  

5x + 2 = 7

Add - 2 on both the sides 
5x + 2 ñ 2 = 7 ñ 2  
5x + 0 = 5   
5x = 5

Divide both sides by 5.  
x = 1

To check if the answer is correct or not, substitute the value of x = 1 back into the original equation 
5 x + 2 = 7

LHS = 5 (1) + 2 = 5 + 2 = 7
RHS = 7

LHS = RHS  
Therefore our answer x = 1 is correct as this value satisfies the given equation.  

2. (y/3)  -  3 = 0 

In this equation, the coefficient of y is 1/3. The LHS has a constant - 3 and the RHS has a constant 0.

Do the same procedure as shown for equation (1). Add + 3 on both the sides.

     y/3  -   3 + 3   = 0   +  3 
     y/3  -    0   = 3  
  y = 9

If you put this value of y, you will see that the LHS = RHS will be satisfied.  

3.  7z = 49

In this equation, the coefficient of z is 7. There is no constant on the LHS. The RHS has a constant 49. The equation is simple. Divide both the sides by 7 and we get

z = 7

Substitute this value of z in the original equation (3) 7z = 49.
LHS = 7 * 7 = 49  
RHS = 49

LHS = RHS and thus the equation is satisfied.

To summarize the procedure for finding values of a variable in a single variable equation, we should first get rid of the constants on the side of the equation that has the variable. Then we should try to see how to have the coefficient of the variable 1 : multiply or divide by a suitable number to get the coefficient of the variable as 1.  

 

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