NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Introduction
In the preceding class, we have learned about algebraic expressions and equations. Here, we shall confine ourselves to the study of linear equations in one variable.
An equation essentially contains a sign of equality (=). It is a statement of equality involving an unknown quantity.
→ Also, the expressions we use to form a linear equation are linear only, i.e., the highest power of the variable occurring in the expression is 1, and that too only in one variable. A linear equation may have linear expressions on both sides of the sign of equality.
The expression on the left of the sign of equality is called the Left Hand Side whereas the expression on the right of the sign of equality is called the Right Hand Side.
Read and Learn More NCERT Solutions For Class 8 Maths
→ The value of the variable for which LHS = RHS is called a solution of the linear equation.
To find the solution of a linear equation in one variable, we assume that the two sides of the equation are balanced.
→ We are free to perform the same operation (suitable) on both sides of the equation such as we can add to or subtract from both sides of the equation the same quantity (number).
→ Also, we can multiply or divide both sides of the equation by the non-zero quantity (number).
Solving Equations Having The Variable On Both Sides
We transpose the terms so that the terms containing the variables are on the LHS and constant numbers on the RHS.
Then, we can solve the equation by simplifying both sides and dividing by a suitable number (if required).
NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Exercise 2.1
Solve the following equations and check your results
Question 1. 3x = 2x + 18
Solution:
3x = 2x + 18
3x- 2x = 18
Transposing 2x from RHS to LHS
x = 18
This is the required solution
Check:
LHS = 3x = 3 (18)
= 54
RHS = 2x + 18 = 2 (18)+18
= 36 + 18
= 54
∴ x = 18
∴ LHS = RHS
2. 5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
5t – 3t, = -5 + 3
Transposing 3t from RHS to LHS- 3 from LHS to RHS
2t =-2
t = \(\frac{2}{2}\)
t = -1
Dividing both sides by 2
This is the required solution
.
Check:
LHS = 5t – 3 = 5(- 1)- 3
= -5 – 3 = -8
∴ t = -1
RHS = 3t – 5 = 3(- 1)- 5
= -3 – 5
= -8
∴ LHS = RHS
Question 3. 5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
5x – 3x = 5 – 9
Transposing 3 from RMS to LHS and 9 from LHS to RMS
2 x =-4
x = \(\frac{-4}{2}\)
x = – 2
Dividing both sides by 2
This is the required solution
Check:
LHS = 5x + 9
= 5(-2)+9 = -10+9
= -1
RHS = 5 + 3x
= 5+3(-2) = 5 -6
= -1
∴ LHS = RHS
Question 4. 4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
4z-2z = 6-3
Transposing 2z from RHS to LHS 3 from LHS to RHS
2z = 3
z= \(\frac{3}{2}\)
Dividing both sides by 2
This is the required solution
Check:
LHS = 4z + 3
= 4 (\(\frac{3}{2}\)) +3
= (\(\frac{12}{2}\)) +3 = 6+3
= 9
RHS = 6 + 2z
= 6+ 2(\(\frac{3}{2}\))
= 6+ \(\frac{6}{2}\) = 6+3
= 9
∴ LHS = RHS
Question 5. 2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
2x + x = 14+ 1
Transposing- x from RHS to LHS and -1 from LHS to RHS
3x = 15
x= \(\frac{15}{3}\)
x = 5
Dividing both sides by 3
This is the required solution
Check:
LHS = 2x – 1 = 2(5) – 1
= 10 – 1
= 9
RHS = 14 – x
=14 – 5
= 9
∴ LHS = RHS
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 from RHS to LHS and 4 from LHS to RHS
5x = 0
x = \(\frac{0}{5}\)
= 0
Dividing both sides by 5
This is the required solution
Check:
LHS = 8x + 4 = 8(0)+ 4
= 4
RHS = 3(x – 1) + 7= 3(0-1)+7
= -3+7
= 4
∴ LHS = RHS
Question 7. \(x=\frac{4}{5}(x+10)\)
Solution:
x = \(\frac{4}{5}(x+10)\)
5x = 4(x + 10)
Multiplying both sides by 5
5x = 4x + 40
5x-4x = 40
Transposing 4x from RHS to LHS
x = 40
This is the required solution
Check:
LHS = 5x = 5(40)
= 200
RHS = 4(x + 10) = 4 (40+10)
= 4(50)
= 200
∴ LHS = RHS
8. \(\frac{2 x}{3}+1=\frac{7 x}{15}+3 \)
Solution:
⇒ \(\frac{2 x}{3}+1=\frac{7 x}{15}+3 \)
⇒ \(\frac{2 x}{3}-\frac{7 x}{15}=3-1\)
Transposing \(\frac{7 x}{15}\) from RHS to LHS and 1 from LHS to RHS
⇒ \(\frac{2 x}{3}-\frac{7 x}{15}=2\)
⇒ \(15\left(\frac{2 x}{3}-\frac{7 x}{15}\right)=2 \times 15\)
Multiplying both sides by 15
∴ LCM (3, 15) = 15
10x – 7x = 30
3x = 30
x = \(\frac{30}{3}\)
x = 10
This is the required solution
∴ LHS = RHS
Question 9. \(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:
⇒ \(2 y+\frac{5}{3}=\frac{26}{3}-y\)
⇒ \(2 y+y=\frac{26}{3}-\frac{5}{3}\)
Transposing – y from RHS to LHS \(\frac{5}{3}\) and from LHS to RHS
3 y= \(\frac{26-5}{3}\)
3 y= \(\frac{21}{3}\)
3y = 7
y = \(\frac{7}{3}\)
∴ LHS = RHS
10. 3m-5m-\(\frac{8}{5}\)
Solution:
3m-5m= –\(\frac{8}{5}\)
3m-5m-\(\frac{8}{5}\)
Transposing 5m from RHS to LHS
= -2m = –\(\frac{8}{5}\)
m = \(\frac{-8}{5 \times(-2)}=\frac{4}{5}\)
This is the required solution
∴ LHS = RHS
Reducing Equations To Simpler Form
We multiply both sides of the equation by the LCM of the denominators of the terms in the expressions occurring in the given equation.
→ We transpose properly so that all the variable terms come on LHS and constant terms on RHS.
→ Then, combining like terms on both sides of the equation and dividing both sides by a suitable number (if required), we can find out the required solution.
Finally, we check this solution for its validity. Reject if the solution is invalid
NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Exercise 2.2
Solve the following linear equations:
Question 1. \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:
⇒ \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
We have, \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
It is a linear equation since it involves linear expressions only
⇒ \(\frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\)
Transposing \(\frac{x}{3}\) from RHS to LHS and \(\frac{-1}{5}\) from LHS to RHS
⇒ \( \frac{3 x-2 x}{6}= \frac{5+4}{20}\)
⇒ \(\frac{x}{6}= \frac{9}{20}\)
x = \(\frac{9}{20} \times 6=\frac{9}{10} \times 3\)
Multiplying both sides by 6
x = \(\frac{27}{10}\)
This is the required solution
Check:
LHS = \(\frac{x}{2}-\frac{1}{5}=\frac{1}{2} \times \frac{27}{10}-\frac{1}{5} \)
= \(\frac{27}{20}-\frac{1}{5}=\frac{27-4}{20} \)
= \(\frac{23}{20}\)
RHS = \(\frac{x}{3}+\frac{1}{4}=\frac{1}{3} \times \frac{27}{10}+\frac{1}{4} \)
=\(\frac{9}{10}+\frac{1}{4}=\frac{18+5}{20}\)
∴ LCM(10,4)= 20
= \(\frac{23}{20}\)
∴ LHS = RHS
Question 2. \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)
Solution:
⇒ \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)
It is a linear equation since it involves linear expressions only
⇒ \(\frac{6 n-9 n+10 n}{12}=21\)
⇒ \(\frac{7 n}{12}\) =21
n = \(21 \times \frac{12}{7}=36\)
Multiplying both sides by \(\frac{12}{7}\)
This is the required solution
Check:
LHS = \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\)
= \(\frac{36}{2}-\frac{3}{4} \times 36+\frac{5}{6} \times 36\)
18 – 27 + 30 = 21
RHS = 21
∴ LHS = RHS
Question 3. \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:
⇒ \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
It is linear equation since it involves linear expressions only.
x- \(\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7\)
Transposing \(\frac{5 x}{2}\) from RHS to LHS and 7 from LHS to RHS
⇒ \(\frac{6 x-16 x+15 x}{6}=\frac{17-42}{6}\)
∴ LCM (3,2) = 6
⇒ \(\frac{5 x}{6}=\frac{-25}{6} \)
⇒ \(\frac{-25}{6} \times \frac{6}{5}\)
Multiplying both sides by \(\frac{6}{5}\)
x =-5
This is the required solution
∴ LHS = RHS
Question 4. \(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:
⇒ \(\frac{x-5}{3}=\frac{x-3}{5}\)
It is a linear equation since it involves linear expressions only.
⇒ \(\frac{x}{3}-\frac{5}{3}=\frac{x}{5}-\frac{3}{5} \)
⇒ \(\frac{x}{3}-\frac{x}{5}=\frac{5}{3}-\frac{3}{5}\)
Transposing \(\frac{x}{5}\) From RHS to LHS and-\(\frac{x}{5}\)from LHS to RHs
⇒ \(\frac{5 x-3 x}{15}=\frac{25-9}{15}\)
⇒ \(\frac{2 x}{15}=\frac{16}{15} \)
=\(\frac{16}{15} \times \frac{15}{2}=8\)
Multiply both sides by \(\frac{15}{2}\)
This is the required solution
∴ LHS = RHS
Question 5. \(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:
⇒ \(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
It is a linear equation since it involves linear expressions only.
⇒ \(\frac{3}{4} t-\frac{2}{4}-\frac{2}{3} t-\frac{3}{3}=\frac{2}{3}-t\)
⇒ \(\frac{3}{4} t-\frac{1}{2}-\frac{2}{3} t-1=\frac{2}{3}-t\)
⇒ \(\frac{3}{4} t-\frac{2}{3} t+t=\frac{2}{3}+\frac{1}{2}+1\)
Transposing – t from RHS to LHS and – \(\frac{1}{2}\) and -1 from LHs to RHS
⇒ \(\frac{9 t-8 t+12 t}{12}\)
= \(\frac{4+3+6}{6}\)
LCM (4, 3) = 12; LCM (3, 2) = 6
⇒ \(\frac{13 t}{12}=\frac{13}{6} \)
= \(\frac{13}{6} \times \frac{12}{13}=2\)
Multiplying both sides by \(\frac{12}{3}\)
This is the required solution
∴ LHS = RHS
Question 6. \(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
It is a linear equation since it involves linear expressions only
m- \(\frac{m}{2}+\frac{1}{2}=1-\frac{m}{3}+\frac{2}{3} \)
m- \(\frac{m}{2}+\frac{m}{3}=1+\frac{2}{3}-\frac{1}{2}\)
Transposing – \(\frac{m}{3}\) from RHS to LHS and \(\frac{1}{2}\) from LHS to RHS
⇒ \(\frac{6 m-3 m+2 m}{6}=\frac{6+4-3}{6}\)
⇒ \(\frac{5 m}{6}=\frac{7}{6} \)
m = \(\frac{7}{6} \times \frac{6}{5}=\frac{7}{5}\)
Multiplying both sides by \(\frac{6}{5}\)
This is the required solution
∴ LHS = RHS
Question 7. 3(t – 3) = 5 (2t + 1)
Solution:
3(t – 3) = 5(2t + 1)
3t – 9 = 10t + 5
3t – 10t = 5 + 9
-7t = 14
t =- \( \frac{14}{7}\) = -2
t =-2
Dividing both sides by – 7
This is the required solution.
∴ LHS = RHS
Question 8. 15 (y -4)- 2(y – 9) + 5(y + 6) = 0
Solution:
15(y- 4)- 2(y- 9) + 5(y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
15y – 2y- 5y -60 +18 +30 = 0
18y – 12 = 0
18y = 12
Transposing- 12 from LHS to RHS
y =\(\frac{12}{18} \)
y =\(\frac{12 \div 6}{18 \div 6}\)
=\(\frac{2}{3}\)
This is the required solution.
Check:
LHS = \(15(y-4)-2(y-9)+5(y+6) \)
= \(15\left(\frac{2}{3}-4\right)-2\left(\frac{2}{3}-9\right)+5\left(\frac{2}{3}+6\right) \)
= \(15\left(\frac{2-12}{3}\right)-2\left(\frac{2-27}{3}\right)+5\left(\frac{2+18}{3}\right) \)
= \(15\left(-\frac{10}{3}\right)-2\left(-\frac{25}{3}\right)+5\left(\frac{20}{3}\right) \)
= \(-50+\frac{50}{3}+\frac{100}{3} \)
= \(\frac{-150+50+100}{3}=\frac{0}{3}=0\)
RHS = 0
∴ LHS = RHS
Question 9. 3(5z – 7) -2 (9z – 11) = 4 (8z – 13) – 17
Solution:
3(5z – 7) -2 (9z – 11) = 4 (8z – 13) – 17
15 z-21-18z+22 = 32z -52-17
-3z+1 = 32z- 69
– 3z- 32z = – 69-1
35z = – 70
z = \(\frac{-70}{-35}=\frac{70}{35}\)
= 2
This is the required solution.
∴ LHS = RHS
Question 10. 0.25 (4f – 3) = 0.05(10f – 9)
Solution:
0.25(4f- 3) = 0.05(10f- 9)
f- 0.75 = 0.5f- 0.45
f- 0.5f = – 0.45 + 0.75
Transposing 0.5f from RHS to LHS and – 0.75 from LHS to RHS
0.5f = 0.30
f = \(\frac{0.30}{0.5}\)
= 0.6
Dividing both sides by 0.5
This is the required solution
∴ LHS = RHS
NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Multiple-Choice Questions
Question 1. The standard form of a linear equation is 7. The root, of the equation 3y + 4 = by – 4 in one variable x is
- ax + 6=0
- ax2 + bx + c = 0
- ox3 + bx2 + cx + d = 0
- ax4 + bx3 + cx² + dx + e = 0.
Solution: 1. ax + 6=0
Question 2. Of the following, the linear equation in one variable x, is
- \(\frac{4}{x}=\frac{x}{4}\)
- \(\frac{1}{x}+\frac{1}{x-1}=1\)
- \(\frac{x}{2}+\frac{x}{3}=\frac{1}{4}\)
- 2 + 2x + 3 = 0
Solution: 3. \(\frac{x}{2}+\frac{x}{3}=\frac{1}{4}\)
Question 3. The degree of the equation x² – 2x + 1 = x² – 3 is
- 1
- 2
- 0
- 3
Solution: 1. 1
x² – 2x + 1 = x² – 3
x² – 2x + 1 – x² + 3= 0
– 2x + 4 = 0
-2x = – 4
2x = 4
Question 4. The root of the equation 2x+3 = 2(x- 4) is
- 2
- 4
- 0
- Does not exist
Solution: 4. Does not exist
2x +3 = 2 (x- 4)
2x + 3 = 2x – 8
3 = – 8 which is impossible.
Question 5.The value of x in \(\frac{3}{4x}\) = 7-x is
- 4
- 3
- \(\frac{7}{3}\)
- 7
Solution: 1. 4
⇒ \(\frac{3}{4} x=7-x \)
⇒ \(\frac{3}{4} x+x=7\)
⇒ \(\frac{7}{4} x=7 \)
x = 4
Question 6. The root of the equation 2y = 5(3 + y) is
- 5
- \(\frac{-1}{5}\)
- -5
- -1/5
Solution: 3. -5
2y = 5 (3 + y)
⇒ 2y= 15 + 5y
⇒ by – 2y = – 15
⇒ 3 y =-15
y =-\(\frac{15}{3}\)
= -5
Question 7. The root of the equation 3y + 4 = 5y – 4
- 1
- 3
- 4
- 2
Solution: 4. 2
3y+ 4 = 5y- 4
5y – by = 4 + 4
y= \(\frac{8}{2}\)
= 4
Question 8. The root of the equation 9z- 15 – 9- 3z is
- 1
- 2
- 3
- 4
Solution: 2. 2
2y = 8
92-15 = 9 – 32
⇒ 92 + 32 = 9+15
12 z = 24
z = \(\frac{24}{12}\)
= 2
Question 9. The root of the equation 3x= \(\frac{20}{7}-x\) is
- \(\frac{7}{5}\)
- \(\frac{5}{7}\)
- \(-\frac{7}{5}\)
- \(-\frac{5}{7}\)
Solution: 2. \(\frac{5}{7}\)
3 x= \(\frac{20}{7}-x \)
3 x+x= \(\frac{20}{7}\)
4 x= \(\frac{20}{7}\)
x = \(\frac{20}{7 \times 4}=\frac{5}{7}\)
Question 10. The root of the equation 2y = 5 (7- y) is
- 5
- -5
- 3
- – 3.
Solution: 1. 5
2y = 5 (7 – y) => 2y = 35 – 5y 2y + 5y = 35
7y = 35
y = \(\frac{35}{7}\)
= 5
Question 11. The root of the equation (2x – 1) + (x – 1) = x + 2 is
- 1
- 2
- -1
- -2
Solution: 2. 2
(2x – 1) +(x – 1) = x + 2
2x – 2 + x-1 = x+2
2x + x -x = 2+2
2x = 4
x= \(\frac{4}{2}\)
= 2
Question 12. The root of the equation, 13x – 14 = 9x + 10 is
- 1
- 2
- 3
- 6
Solution: 4. 6
13x – 14 = 9x + 10
13x- 9x = 10+14
4x = 24
x = \(\frac{24}{4}\)
= 6
Question 13. The root of the equation,11x – 5- x + 6 = 2x + 17 is
- 1
- 2
- 3
- 4
Solution: 2. 2
11x – 5 – x + 6 = 2x+ 17
8x = 16
x = 2
NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable True-False
State whether the following are True or False statements
1. A polynomial of degree 1 is called a linear polynomial – True
2. The equations 2.v + 1 = 5 and 4* + 2 = 10 are identical – True
3. If x is an odd number, then the next even number is x + 2 –False
4. Ifa and b are positive integers, then the solution of the linear equation ax = b is always positive – True
5. Two different equations can never have the same solution – False
NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Fill In The Blanks
1. In a linear equation, the highest power ofthe variable appearing in the equation is → One
2. The other name of a solution of an equation is → Root
3. If x is an even number, then the next even number is → x+2
4. A term of the equation can be transposed from LHS to RHS by changing its→ Sign
5. Find the value of x for which the expressions 3x+ 1 and 2x + 9 become equal → 8
6. Is ax²+ bx + c = 0 a linear equation → No