Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3
Solve the following equations and check your results :
Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
∴ 3x – 2x = 18 (Transposing 2x to LHS)
∴ x = 18
Check:
LHS = 3x = 3 × 18 = 54
RHS = 2x + 18
= 2(18) + 18
= 36 + 18 = 54
LHS = RHS
Thus, the answer is correct.
Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
∴ 5t – 3t – 3 = – 5 (Transposing 3t to LHS)
∴ 2t – 3 = -5
∴ 2t = – 5 + 3 (Transposing -3 to RHS)
∴ 2t = – 2
∴ \(\frac{2 t}{2}=\frac{-2}{2}\) (Dividing both the sides by 2)
∴ t = – 1
Check:
LHS = 5t – 3
= 5 (- 1) – 3
= – 5 – 3 = -8
RHS = 3t – 5
= 3 (- 1) – 5
= – 3 – 5 = – 8
∴ LHS = RHS
Thus, the answer is correct.
Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
∴ 5x + 9 – 3x = 5 (Transposing 3x to LHS)
∴ 2x + 9 = 5
∴ 2x = 5 – 9 (Transposing 9 to RHS)
∴ 2x = -4
∴ \(\frac{2 x}{2}=\frac{-4}{2}\) (Dividing both the sides by 2)
x = – 2
Check:
LHS = 5x + 9
= 5 (-2) + 9
= – 10 + 9 = -1
RHS = 5 + 3x
= 5 + 3 (-2)
= 5 – 6 = -1
∴ LHS = RHS
Thus, the answer is correct.
Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
∴ 4z + 3 – 2z = 6 (Transposing 2z to LHS)
∴ 2z + 3 = 6
∴ 2z = 6 – 3 (Transposing 3 to RHS)
∴ 2z = 3
∴ \(\frac{2 z}{2}=\frac{3}{2}\) (Dividing both the sides by 2)
∴ z = \(\frac {3}{2}\)
Check:
LHS = 4z + 3
=4 (\(\frac {3}{2}\)) + 3
= 6 + 3 = 9
RHS = 6 + 2z
= 6 + 2(\(\frac {3}{2}\))
= 6 + 3 = 9
LHS = RHS
Thus, the answer is correct.
Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14-x
∴ 2x – 1 + x = 14 (Transposing -x to LHS)
∴ 3x – 1 = 14
∴ 3x = 14 + 1 (Transposing -1 to RHS)
∴ 3x = 15
∴ \(\frac{3 x}{3}=\frac{15}{3}\) (Dividing both the sides by 3)
x = 5
Check:
LHS = 2x – 1
= 2 (5) – 1
= 10 – 1 = 9
RHS = 14 – x
= 14 – 5 = 9
LHS = RHS
Thus, the answer is correct.
Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
8x + 4 = 3(x – 1) + 7
∴ 8x + 4 = 3x – 3 + 7
∴ 8x + 4 = 3x + 4
∴ 8x + 4 – 3x = 4 (Transposing 3x to LHS)
∴ 5x + 4 = 4
∴ 5x = 4 – 4 (Transposing 4 to RHS)
∴ 5x = 0
∴ \(\frac{5 x}{5}=\frac{0}{5}\) (Dividing both the sides by 5)
∴ x = 0
Check:
LHS = 8x + 4
= 8 (0) + 4
= 0 + 4 = 4
RHS = 3 (x – 1) + 7
= 3(0 – 1) + 7
= 3 (-1) + 7
= – 3 + 7 = 4
∴ LHS = RHS
Thus, the answer is correct.
Question 7.
x = \(\frac {4}{5}\) (x + 10)
Solution:
x = \(\frac {4}{5}\) (x + 10)
∴ x = \(\frac{4 x}{5}+10 \times \frac{4}{5}\)
∴ x = \(\frac{4 x}{5}+8\)
∴ x – \(\frac{4 x}{5}\) = 8 (Transposing \(\frac{4 x}{5}\) to LHS)
∴ \(\frac{5 x-4 x}{5}\) = 8 (LCM = 5)
∴ \(\frac{x}{5}\) = 8
∴ \(\frac{x}{5}\) × 5 = 8 × 5 (Multiplying both the sides by 5)
∴ x = 40
Check:
LHS = x = 40
RHS = \(\frac {4}{5}\) (x + 10)
= \(\frac {4}{5}\) (40 + 10)
= \(\frac {4}{5}\) (50)
= 4 × 10 = 40
∴ LHS = RHS
Thus, the answer is correct.
Question 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 1 to RHS)
∴ \(\frac{2 x}{3}\) = \(\frac{7 x}{15}\) + 2
∴ \(\frac{2 x}{3}-\frac{7 x}{15}\) = 2 (Transposing \(\frac{7 x}{15}\) to LHS)
∴ \(\frac{2 x \times 5-7 x}{15}\) (LCM = 15)
∴ \(\frac{10 x-7 x}{15}\) = 2
∴ \(\frac{3 x}{15}\) = 2
∴ \(\frac{x}{5}\) = 2
∴ \(\frac{x}{5}\) × 5 = 2 × 5 (Multiplying both the sides by 5)
∴ x = 10
Check:
LHS = \(\frac{2 x}{3}\) + 1
= \(\frac{2(10)}{3}\) + 1
= \(\frac{20}{3}\) + 1 = \(\frac{23}{3}\)
RHS = \(\frac{7 x}{15}\) + 3
= \(\frac{7(10)}{15}\) + 3
= \(\frac {14}{3}\) + 3
= \(\frac {23}{3}\)
∴ LHS = RHS
Thus, the answer is correct.
Question 9.
2y + \(\frac {5}{3}\) = \(\frac {26}{3}\) – y
Solution:
2y + \(\frac {5}{3}\) = \(\frac {26}{3}\) – y
∴ 2y + y + \(\frac {5}{3}\) = \(\frac {26}{3}\) (Transposing -y to LHS)
∴ 3y + \(\frac {5}{3}\) = \(\frac {26}{3}\)
∴ 3y = \(\frac{26}{3}-\frac{5}{3}\) (Transposing \(\frac {5}{3}\) to RHS)
∴ 3y = \(\frac{26-5}{3}\)
∴ 3y = \(\frac {21}{3}\)
∴ 3y = 7
∴ \(\frac{3 y}{3}=\frac{7}{3}\) (Dividing both the sides by 3)
∴ y = \(\frac {7}{3}\)
Check:
LHS = 2y + \(\frac {5}{3}\)
= 2(\(\frac {7}{3}\) ) + \(\frac {5}{3}\)
= \(\frac{14}{3}+\frac{5}{3}\)
= \(\frac{14+5}{3}\)
= \(\frac {19}{3}\)
RHS = \(\frac {26}{3}\) – y
= \(\frac{26}{3}-\frac{7}{3}\)
= \(\frac{26-7}{3}\)
= \(\frac {19}{3}\)
Question 10.
3m = 5m – \(\frac {8}{5}\)
Solution:
3m = 5m – \(\frac {8}{5}\)
∴ 3m – 5m = –\(\frac {8}{5}\) (Transposing 5m to LHS)
∴ -2m = –\(\frac {8}{5}\)
∴ 2m = \(\frac {8}{5}\) [Multiplying both the sides by (-1)]
∴ \(\frac{2 m}{2}=\frac{8}{5} \times \frac{1}{2}\) (Dividing both the sides by 2)
∴ m = \(\frac {4}{5}\)
Check:
LHS = 3m
= 3(\(\frac {4}{5}\))
= \(\frac {12}{5}\)
RHS = 5m – \(\frac {8}{5}\)
= 5(\(\frac {4}{5}\)) – \(\frac {8}{5}\)
= 4 – \(\frac {8}{5}\)
= \(\frac{20-8}{5}\)
= \(\frac {12}{5}\)
∴ LHS = RHS
Thus, the answer is correct.