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Algebra Homework Help : Equations

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Preparation Just what you need to know !

What is an Equation?

Algebra is a branch of mathematics where certain kinds of problems are solved using equations. An equation can be looked upon as a weighing scale or beam balance (see figure).

 

Instead of two pans separated by a fulcrum, an equation has two sides separated by the equal (=) sign. Rather than use objects and weights, an equation uses constants, variables and mathematical symbols.

 

A constant is a known number whose value does not change, e.g., 15. A variable is an unknown quantity whose value needs to be found. It is represented by any letter of the alphabet (usually, x, y or z). The mathematical symbols are +, −, x, / and ( ). The first four symbols denote addition, subtraction, multiplication and division. The brackets (parentheses) are grouping symbols and require that what is within them be calculated first.

 

To understand the concept of an equation, consider a simple example.

 

____ + 15 = 25

 

The blank above stands for an unknown number. It is easy to find out that the blank equals 10 (or 25 − 15). In algebra, the unknown number is called the variable and represented by a letter of the alphabet (say, x). Thus, the problem could be written as
x + 15 = 25

 

and the answer would then be
x = 10

 

Here, x is the variable, = is the equal sign, and + is the mathematical symbol for addition. The numbers 10, 15 and 25 are the constants.

 

To understand the above example better, imagine a weighing scale with 25 marbles in the right pan. In the left pan, there is a bag (of negligible weight) with an unknown number of marbles and some extra marbles outside the bag. The scale is in perfect balance. All the marbles are identical. Your task is to find out the number of marbles in the bag without opening it.

 

Suppose we remove all the extra marbles outside the bag in the left pan and find them to be 15 in number. If we remove the same number of marbles (in this example, 15) from the right pan, then the scale will continue to be in perfect balance. Since only the bag is in the left pan and 10 marbles remain in the right pan, we know (without opening the bag) that there are 10 marbles in the bag.

 

From the above example, it is important to note that the scale is in perfect balance so long as the same operation (addition, subtraction, etc.) is performed on both sides.

 

Similarly, in algebra, an equation remains balanced so long as the same operation (addition, subtraction, multiplication, and division) is performed on both sides.

 

This basic idea can be used to systematically solve equations as shown in the next section.

 

How to Solve an Equation?

If the same quantity is added to or subtracted from both sides of an equation, the two sides of the equation remain in balance.

 

If both sides of an equation are multiplied or divided by the same number, the two sides of the equation remain in balance.

 

The answer is obtained when only the variable exists on one side. The answer may be checked by substituting in the left-hand side (LHS) and the right-hand side (RHS) of the equation.
Examples are given below.

 

Example

 

Equation : x + 27 = 71

 

Subtract 27 from both sides: x + 27 − 27 = 71 − 27

 

Answer : x = 44

 

Check: LHS = x + 27 = 44 + 27 = 71 and RHS = 71

 

Note that the multiplication sign is often not written explicitly in algebra. Thus, 3 multiplied by y is written as 3y rather than 3 x y. Here, 3 is called the coefficient. The coefficient is the number by which the variable is being multiplied. If only y is written, then the coefficient is 1. If −2y is written, then the coefficient is −2. If −y is written, then the coefficient is −1.

 

Example

 

Equation : 3y + 34 = 2y + 89

 

Subtract 34 from both sides : 3y + 34 − 34 = 2y + 89 − 34

 

Simplification gives : 3y = 2y + 55

 

Subtract 2y from both sides: 3y − 2y = 2y − 2y + 55

 

Answer : y = 55

 

Check : LHS = 3y + 34 = 3(55) + 34 = 165 + 34 = 199

and       RHS = 2y + 89 = 2(55) + 89 = 110 + 89 = 199

 

Example

 

Equation : 5z + 97 = 3z + 33

 

Subtract 97 from both sides : 5z + 97 − 97 = 3z + 33 − 97

 

Simplification gives : 5z = 3z − 64

 

Subtract 3z from both sides : 5z − 3z = 3z − 3z − 64

 

Simplification gives : 2z = −64

 

Dividing both sides by the coefficient 2: 2z / 2 = −64 / 2

 

Answer : z = −32

 

Check : LHS = 5z + 97 = 5(−32) + 97 = −160 + 97 = −63
and       RHS = 3z + 33 = 3(−32) + 33 = −96 + 33 = −63


Practice Exercise for Algebra Module on Simple Equations

 

 

Practice Exercise for Algebra Module on Linear Equations I

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