NCERT Maths Class 8th Chapter 2 Linear Equation In One Variable Exercise 2.3 Problems

Linear equation is an equation having variables with power for example: ax+b=0 where x is the variable, a and b are real numbers. The statement of equality of two algebraic expressions involving one or more variables is known as linear equation. Linear equations are used to solve any arithmetic equations to know the exact value or root of the variable. So here we are going to discuss a few problems of chapter linear equations in one variable.

Everyone thinks that maths is a very difficult subject but no that’s wrong actually it’s a very easy subject if we understand. Whatever subject if we study in everything maths is included and each and every topic is related with maths. It’s a very easy subject, no need to fear at all about this subject.
1. 3x=2x+18
Sol: 3x-2x=18
1x=18
x=18
LHS= 3x
3(18)
54
RHS= 2x+18
2(18)+18
54
Therefore, LHS=RHS

So now we give some problems try to solve them very easy problems

2. 5t-3=3t-5
Sol: 5t-3=3t-5
5t-3t=-5+3
2t=-2
t=-1
Putting the value of t in RHS and LHS we get, 5x(-1)-3=3x(-1)-5
-5-3=-3-5
-8=-8
Therefore LHS=RHS

3. 5x+9=5+3x
Sol: 5x+9=5+3x
5x-3x=5-9
2x=-4
x=-2
Putting the value of x in RHS and LHS we get, 5x(-2)+9=5+3x(-2)
-10+9=5+(-6)
-1=-1
Therefore LHS=RHS

4. 4z+3=6+2z
Sol: 4z+3=6+2z
4z-2z=6-3
2z=3
z=3/2
Putting the value of z in RHS and LHS we get
(4×3/2)+3=6+(2×3/2)
6+3=6+3
9=9
Therefore LHS=RHS

5. 2x-1=14-x
Sol: 2x-1=14-x
2x+x=14+1
3x=15
x=15
Putting the value of x in RHS and LHS we get
(2×5)-1=14-5
10-1=9
9=9
Therefore LHS=RHS

6. 8x+4=3(x-1)+7
Sol: 8x+4=3(x-1)+7
8x+4=3x-3+7
8x+4=3x+4
8x-3x=4-4
5x=0
x=0
Putting the value of x in RHS and LHS we get
(8×0)+4=3(0-1)+7
0+4=0-3+7
4=4
Therefore LHS=RHS

7. x=4/5(x+10)
Sol: X=4/5(x+10)
x=4x/5+40/5
x-(4x/5)=8
(5x-4x)/5=8
x=8×5
x=40
Putting the value of x in RHS and LHS we get
40=4/5(40+10)
40=4/5×50
40=200/5
40=40
Therefore LHS=RHS

8. 2x/3+1=7x/15+3
Sol: 2x/3+1=7x/15+3
2x/3-7x/15=2
3x=2×15
3x=30
x=30/3
x=10
Putting the value of x in RHS and LHS we get

9. 2y+5/3=26/3-y
Sol: 2y+5/3=26/3-y
2y+y=26/3-5/3
3y=(26-5)/3
3y=(26-5)/3
3y=21/3
3y=7
y=7/3
Putting the value of y in RHS and LHS we get
(2×7/3)+5/3=26/3-7/3
14/3+5/3=26/3-7/3
(14+5)/3=(26-7)/3
19/3=19/3
Therefore LHS=RHS

10. 3m=5m-8/5
Sol: 3m=5m-8/5
5m-3m=8/5
2m=8/5
2mx5=8
10m=8
m=8/10
m=4/5
Putting the value of m in RHS and LHS we get
3x(4/5)=(5×4/5)-8/5
12/5=4-(8/5)
12/5=(20-8)/5
12/5=12/5
Therefore LHS=RHS

Conclusion:

These are some of the problems which are very easy to solve all problems. These are very important problems. It helps us to gain knowledge and it is useful for all kinds of competitive exams. Maths is a very easy and interesting subject.

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