Chapter 2- Linear Equations in One Variable Units

Exercise 2.1

1. Solve the following equations and check your results

1. 3x=2x+18

Solution:
3x=2x+18
3x-2x=18 (transposing 2x to L.H.S )
x=18
Now, to check the result we put the value of x=18 in L.H.S and R.H.S of the equation.
L.H.S = 3x = 3 x 18 = 54
R.H.S = 2x + 18 = 2 x 18 +18 = 36 + 18 = 54
L.H.S = R.H.S

2. 5t-3=3t-5

Solution:
5t – 3= 3t – 5
5t – 3t= -5 + 3 (transposing 3t to L.H.S & -3 to R.H.S)
2t = – 2
t = -2/2
t = -1
Now, to check the result we put the value of t=-1 in L.H.S and R.H.S of the equation.
L.H.S = 5t-3 = 5 x (-1) – 3 = – 5 – 3 = -8
R.H.S = 3t-5 = 3 x (-1) – 5 = – 3 – 5 = – 8
L.H.S = R.H.S

3. 5x+9=5+3x

Solution:
5x+9=5+3x
5x-3x=5-9 (transposing 3x to L.H.S & 9 to R.H.S)
2x=-4
x=-4/2
x=-2
Now, to check the result we put the value of x=-2 in L.H.S and R.H.S   of the equation.
L.H.S = 5x+9 = 5 x (-2) + 9 = – 10 + 9 = -1
R.H.S = 5+3x = 5 + 3 x (-2)  = 5 + (-6) = – 1
L.H.S = R.H.S

4. 4z+3=6+2z

Solution:
4z+3=6+2z
4z-2z=6-3 (transposing 2z to L.H.S and 3 to R.H.S)
2z=3

Now, to check the result we put the value of z=3/2 in L.H.S and R.H.S of the equation.
L.H.S = 4z+3 = 4×3/2+3=6+3=9
R.H.S = 6+2z=6+2×3/2=6+6=9
L.H.S = R.H.S

5. 2x-1=14-x

Solution:
2x-1=14-x
2x-x=14-1                (transposing -x to L.H.S and -1 to R.H.S )
3x = 15
x=15/3
x=5
Now, to check the result we put the value of x=5 in L.H.S and R.H.S   of the equation.
L.H.S =2x-1=2(5)-1=10-1=9
R.H.S = 14-x=14-5=9
L.H.S = R.H.S

6. 8x+4=3(x-1)+7

Solution:
8x+4=3(x-1)+7
8x+4=3x-3+7
8x+4=3x+4
8x+3x=4-4 (transposing 3x to L.H.S and 4 to R.H.S )
5x=0
x=0
Now, to check the result we put the value of x=0 in L.H.S and R.H.S   of the equation.
L.H.S = 8x+4=8×0+4=0+4=4
R.H.S = 3(x-1)+7=3(0-1)+7=3(-1)+7=-3+7=4
L.H.S = R.H.S

Solution:

5x=4(x+10)
5x=4x+40
5x-4x=40 (transposing 4x to L.H.S)
x=40

Now, to check the result we put the value of x=40 in L.H.S and R.H.S of the equation.
L.H.S = x=40
R.H.S = 4/5 (x+10)=4/5 (40+10)=4/5×50=4×10=40
L.H.S = R.H.S

Solution:

Now, to check the result we put the value of x=10 in L.H.S and R.H.S of the equation.
L.H.S = 2x/3+1=2(10)/3+1=20/3+1=(20+3)/3=23/3
R.H.S = 7x/15+3=7(10)/15+3=70/15+3=14/3+3=23/3
L.H.S.=R.H.S

Solution:

Now, to check the result we put the value of y=7/3 in L.H.S and R.H.S   of the equation.

Solution:

Now, to check the result we put the value of in L.H.S and R.H.S   of the equation.

Exercise 2.2

2. Solve the following linear equations.

Solution :

we take LCM of all the denominators i.e (2,5,3,4)
here, the LCM is 60 then multiply each term by the LCM

Solution:

We take L.C.M of all the denominators i.e (2,4,6)
Here, the L.C.M is 12 then multiply each term by the L.C.M

Solution:

We take  L.C.M of all the denominators i.e. (3,6,2)
Here, the L.C.M is 6 then multiply each term by the L.C.M

Solution:

Solution:

We take L.C.M of all the denominators i.e (3 , 4)
Here, the L.C.M is 12 then multiply each term by the L.C.M

Solution:

We take L.C.M of all the denominators i.e (2,3)
Here, the L.C.M is 6 then multiply each term by the L.C.M

Simplify and solve the following linear equation

Solution:

Solution:

Solution:

Solution:

Download Chapter 2- Linear Equations in One Variable 
NCERT 
PDF

Leave a Reply

Your email address will not be published. Required fields are marked *

Request a Free Trial Class

Book Demo Blog Archive Pages (#19)

NCERT Solutions for Class 8

Maths Chapters

       

Get Amazon Voucher $25

Refer a friend and earn Amazon Voucher Worth $25(or equivalent to your local currency)

Refer a Friend
Book Your 
 IB 
 IGCSE 
 AS/A Level 
 SAT Exam Preparation 
Demo Class

Personalized Learning

Online Class

Book Demo Blog Archive Pages (#19)