# CLASS-7ALGEBRAIC EQUATIONS

EQUATIONS

Mathematical Sentence –

We would like to consider the mathematical sentence, 5 + 9 = 14. We would like to consider that, what we know that is true because, the sum of 5 & 9 is 14, now consider the mathematical sentence, X + 5 = 12, it is to be considered or may not be considered or may not be true depending on the value of X, similarly, Y – 8 = 14 is true only if Y = 10, such mathematical sentence which are true for only particular values of the literals involved are called open mathematical sentence.

Linear equation in one variable –

An open mathematical sentence showing the equality of two quantities or algebraic expression in terms of one or more variables (literals) is called an equations. If the number of variables used is one then the equation is in ONE VARIABLE, and if the variable, say x appears in the first degree only, that is x ( not x², xetc.) then the equation is to be considered as a LINEAR EQUATIONS in x, that is, the degree of the equation is 1.

As we all know that, the form of a linear equation in one variable x is ax + b = cx + d, where a, b, c, & d are constants.

Examples –

Equations          Number of Variables        Linear Equation or Not

4x + 8y = 9                Two                          Yes

8y – 6 = 10                 One                         Yes

9x + 5 = 0                  One                         Yes

4x²+ x = 10                One                          No

9x + 3y – 6z = 12         Three                        Yes

7X⁴ = 5y²- 17z + 5       Three                        No

Solution of an Equations

A value of the variable which makes the left-hand side of an equation equal to the right-hand side is called the root or solutions of the equations.

Example

1) The equation 3A = 90 has the solution A = 30, because 3 X 30 = 90

2) The equation 5B = 140 has the solution B = 28, because 5 X 28 = 140

3) The equation 6C = 120 has the solution C = 20, because 6 X 20 = 120

Laws of equality –

The following laws of equality are used to solve an equations.

If the same number or quantity is added to both sides of an equations, the two sides remain equal, i.e, if A = B then A + K = B + K

Example.1)  If, ( X + 5 ) = 12 , then 3 ( X + 5 ) – 15  = ?

Ans.)        3 ( X + 5 ) – 15

=  ( 3 X 12)  - 15     ,  where ( X + 5 ) =

=   36 - 15

=   21               (Ans.)

Example.2)  If, ( X² + 3X – 8 ) = 15 , then  4 ( X² + 3 X – 8 ) + 25  =  ?

Ans.)     4 ( X² + 3 X – 8 ) + 25

=  4 ( X² + 3X – 8 ) + 25

=  4 ( 15 ) + 25  ,  where ( X² + 3X – 8 ) = 15

=   ( 4  X 15 ) + 25

=   60 + 25

=   85                   (Ans.)

Rules of Transposition

Any term from one side of an equation can be transposed or shifted to the other side by changing its sign. this is the same as subtracting the term from both sides of the equations,

Example – if, 5x + 3 = x – 8 then

a)  5x + 3 + 8 = x,  b)  5x – x = - 8 – 3,  c) 5x + 3 – x + 8 = 0, d)  0 = - 5x – 3 + x - 8

an equation does not change if the left-hand side is written on the right-hand side and the right-hand side on the left-hand side. For example 5x + 3 = x – 8 can be written as x – 8 = 5x + 3

Method of Solving Equations –

Step.1) Simplify both sides and separate the terms with the variable and the constant terms.

Step.2) Bring all the terms containing the variable to the left-hand side and the constant terms to the right-hand side using the rule of transposition.

Step.3) Divide both sides of the equations by the resulting coefficient of the variable.

Verification of solution –

Substitute the value of the variable ( you have obtained ) on both sides of the equations and check whether the values of both sides are equal. if they are the solution is correct, and we say that the value of the variable satisfies the equations.

Find x if 10 ( x – 2 ) = 5x – 8

Ans.)     10 ( x – 2 ) = 5x – 8

=>   10 x - 20 = 5x - 8

=>   10x – 5x = 20 – 8

=>     5x =  12

x = 12/5                (Ans.)

Verification –

Substituting 12/5 for x on the left-hand side (LHS)

10 ( x – 2 )  =  10 ( 12/5 – 2 ), where x = 12/5

=  10 X 2/5  =  4

Substituting 12/5 for x on the right-hand side (RHS)

5x – 8  =  ( 5 X  12/5 )  - 8

=  12 – 8

= 4

Therefore , LHS = RHS = 4, so the solution is correct.       (Ans.)