4. Linear - linear come from the Latin word ‘linearis’ which means created by
lines.
it consists of one variable and no exponent.
For ex - y=25x+100
Equation – A statement in which states two or more algebraic expression
are equal.
What is a linear equation in one variable
The equation involving only one variable in first order is called linear
equation.
For ex – 2x-3=9

5. we know that an algebraic
expression is an equation involving
Variables.It has an equality
sign.the expression on the left of the
Equality sign is the left hand
side (LHS).the expression on the
Right of the equality sign is the right
hand side (RHS).

6. In an equation the values of the expression on the LHS and
RHS equal. This happens to true only for certain values of the
Variables these values are the solution of the equation.
How to find an solution of an equation
we the two sides of the equation are balanced.
We perform the same mathematical operation on both
Sides of equation ,so that the balance is not disturbed.
A few such steps give the solution.

7. 1.To solve an equation of the form x+a=b E.x.: Solve x+4=10
Solution: x+4=10 => x+4-4=10-4 (subtracting 4 from both the sides)
=> x=6
2.To solve an equation of the form x-a=b E.g.: Solve y-6=5 equal.
Solution: y-6=5 => y-6+6=5+6 (adding 6 to both sides) => y=11 1.
3)example 1: 2x-3=7
Solution:
Adding 3 to both sides.
2x-3+3=7+3 (the balance is not disturbed)
2x=10
Next divide both sides by 2:
2x/2=10/2 (required solution)
X=5

8. a. In an equation, an added term is transposed (taken) from
one side to the other, it is subtracted. i.e.,2y+9=4
Solution: transposing 9 to RHS
2y =4 – 9
2y = -5 (solution)
Dividing both sides by 2 , y =-5/2
To check the answer :LHS=2(-5/2)+9=-5+9=4=RHS (as required)
b. In an equation, a term in multiplication is
transposed to the other side, it is divided. i.e., 3x=12
=> x=12/3=4 (3 is transposed).
c.In an equation a term in division is taken to
the other side it is multiplied. i.e., y/4=6
=> y=6×4=24 (4 is transposed)
GREAT JOB!!!GREAT JOB!!!

9. A. sum of two number is 74 .one number is
10 more than the other .what are the number
ans.let first number be x
2nd number be x+10
ATQ: x+(x+10)=74
2x + 10 = 74
2x = 74-10
X= 32 this one is the number.
The other number is x+10=32+10=42

10. b) Banish has three as many two rupee coins as he has five rupee coin he has an
all a sum of RS 77 how many coins of each denomination does he have?
Ans –let the number of 5 rupee coins that banish has be x. then the number of
two rupee coin he has is three times x or 3x.
The amount banish has:
From 5 rupee coins,Rs 5*x=5*
From 2 rupee coins,Rs 2*3x=6x
Hence the total money ha has =11x
ATQ: 11X = 77
X = 77/11=7
Thus the number of 5 rupee coin
=x=7
Number of two rupee coin =3x=3x7=21

11. C .The difference between two number is 66,the ratio of
the two number is 5:2 what are the two numbers.
Ans. Since the ratio of two number is 2:5 we
May take one number be 2x and the other be 5x.
ATQ:5x-2=66
3x = 66
x=66/3
x=22
so the number are 5x =2*22 or44 and
5*22 or 110 the difference between them = 110-44=66 as
desired

12. We know that in the equation 2x-3=7,the two
Expression is 2x-3and7. In most examples we come to know that
RHS is just a number. But this is not needed always be so; both
sides could have expression with variable. For example, the
equation 2x-3=x+2 has expression with variable on both sides; the
expression on the LHS ism (2x-3) and the expression on the RHS
is(x+2).

13. a) Solve
2x-3=x+2
Ans or 2x=x+2+3
or 2x=x+5
or 2x-x=x+5-x
Here we subtracted from both side of equation not a number
constant, but a term involving the variable.
We can do this as variable are also number.also,note that
subtracting x from both sides amount to transposing x to
LHS.

14. (b)5t – 3= 3t – 5
Or 5t – 3t= -5 +3
Or 2t=-2
Or t=-2/2
Or t =-1
(c)5x+9=5+3x
Or 5x-3x=5-9
Or 2x = -4
Or x = -4/2
Or x = -2

15. (a) arjun is twice as old as shiv .five years ago his age was
three times shiv age .find their present ages.
Ans :let us take shiv present age to be x years
Then arjun present age would be 2x years
Shiv age five years ago was (x-5) years
Arjun age five years ago was (2x-5)years
ATQ, 2X-5=3(X-5)
2X-5=3X-I15
15-5=3x-2x
10=x
Shiv present age =x=10years
Arjun present age = 2x=2*10=20 years

16. Means reducing it to calculate is easily
(a)solve:
X/2-1/5=x/3+1/4
or x/2-x/3=1/4+1/5
or 3x-2x/6=5+4/20
or x/6=9/20
or x=9*6/20
or x=54/20
or x27/10 = 2.7