GSEB Class 8 Maths Solutions Chapter 2 એકચલ સુરેખ સમીકરણ Exercise 2.1

Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 એકચલ સુરેખ સમીકરણ Ex 2.1

નીચેનાં સમીકરણ ઉકેલોઃ

Question 1. x - 2 = 7
Answer:
\( x - 2 = 7 \)
\( \implies x = 7 + 2 \) (Moving -2 to the right side)
\( \implies x = 9 \)
In simple words: To find x, add 2 to 7, which gives 9.

Exam Tip: Remember to change the sign of the number when moving it from one side of the equation to the other.

Question 2. y + 3 = 10
Answer:
\( y + 3 = 10 \)
\( \implies y = 10 - 3 \) (Moving 3 to the right side)
\( \implies y = 7 \)
In simple words: To find y, subtract 3 from 10, which results in 7.

Exam Tip: Always perform the inverse operation when isolating the variable. If a term is added, subtract it from both sides.

Question 3. 6 = z + 2
Answer:
\( 6 = z + 2 \)
\( \implies z + 2 = 6 \) (Swapping both sides)
\( \implies z = 6 - 2 \) (Moving 2 to the right side)
\( \implies z = 4 \)
In simple words: First, switch the sides of the equation. Then, move 2 to the other side by subtracting it from 6 to get 4.

Exam Tip: It's often easier to solve when the variable is on the left side; feel free to swap the entire equation sides.

Question 4. \( \frac {3}{7} + x = \frac {17}{7} \)
Answer:
\( \frac {3}{7} + x = \frac {17}{7} \)
\( \implies x = \frac {17}{7} - \frac {3}{7} \) (Moving \( \frac {3}{7} \) to the right side)
\( \implies x = \frac {17 - 3}{7} \)
\( \implies x = \frac {14}{7} \)
\( \implies x = 2 \)
In simple words: Subtract three-sevenths from seventeen-sevenths. This gives fourteen-sevenths, which simplifies to 2.

Exam Tip: When fractions have the same denominator, you can directly add or subtract their numerators.

Question 5. 6x = 12
Answer:
\( 6x = 12 \)
\( \implies \frac {6x}{6} = \frac {12}{6} \) (Dividing both sides by 6)
\( \implies x = 2 \)
In simple words: To find x, divide 12 by 6, which equals 2.

Exam Tip: Remember that multiplication by a coefficient requires division to isolate the variable.

Question 6. \( \frac{t}{5} = 10 \)
Answer:
\( \frac{t}{5} = 10 \)
\( \implies \frac{t}{5} \times 5 = 10 \times 5 \) (Multiplying both sides by 5)
\( \implies t = 50 \)
In simple words: To find t, multiply 10 by 5, which gives 50.

Exam Tip: The inverse operation of division is multiplication. Use it to remove denominators and isolate the variable.

Question 7. \( \frac{2 x}{3} = 18 \)
Answer:
\( \frac{2 x}{3} = 18 \)
\( \implies \frac{2 x}{3} \times \frac{3}{2} = 18 \times \frac{3}{2} \) (Multiplying both sides by \( \frac{3}{2} \))
\( \implies x = 9 \times 3 \)
\( \implies x = 27 \)
In simple words: To solve for x, multiply 18 by the reciprocal of two-thirds, which is three-halves. This makes x equal to 27.

Exam Tip: When a variable is multiplied by a fraction, multiply both sides by the reciprocal of that fraction to solve for the variable.

Question 8. \( 1.6 = \frac{y}{1.5} \)
Answer:
\( 1.6 = \frac{y}{1.5} \)
\( \implies 1.6 \times 1.5 = \frac{y}{1.5} \times 1.5 \) (Multiplying both sides by 1.5)
\( \implies 2.4 = y \)
\( \implies y = 2.4 \)
In simple words: To find y, multiply 1.6 by 1.5. This calculation gives you 2.4.

Exam Tip: Be careful with decimal multiplication; double-check your calculations to avoid errors.

Question 9. \( 7x - 9 = 16 \)
Answer:
\( 7x - 9 = 16 \)
\( \implies 7x = 16 + 9 \) (Moving -9 to the right side)
\( \implies 7x = 25 \)
\( \implies \frac{7x}{7} = \frac{25}{7} \) (Dividing both sides by 7)
\( \implies x = \frac{25}{7} \)
In simple words: First, add 9 to both sides to get 7x equals 25. Then, divide 25 by 7 to get the value of x.

Exam Tip: Always handle addition/subtraction before multiplication/division when isolating the variable in multi-step equations.

Question 10. \( 14y - 8 = 13 \)
Answer:
\( 14y - 8 = 13 \)
\( \implies 14y = 13 + 8 \) (Moving -8 to the right side)
\( \implies 14y = 21 \)
\( \implies \frac{14y}{14} = \frac{21}{14} \) (Dividing both sides by 14)
\( \implies y = \frac{7 \times 3}{7 \times 2} \)
\( \implies y = \frac{3}{2} \)
In simple words: Add 8 to 13 to get 21. Then, divide 21 by 14 and simplify the fraction by dividing both numbers by 7, which gives three-halves.

Exam Tip: Always simplify fractions to their lowest terms for the final answer.

Question 11. \( 17 + 6p = 9 \)
Answer:
\( 17 + 6p = 9 \)
\( \implies 6p = 9 - 17 \) (Moving 17 to the right side)
\( \implies 6p = -8 \)
\( \implies \frac{6p}{6} = \frac{-8}{6} \) (Dividing both sides by 6)
\( \implies p = \frac{-4}{3} \)
In simple words: First, subtract 17 from 9 to get -8. Then, divide -8 by 6 and simplify to get negative four-thirds.

Exam Tip: Be careful with signs when subtracting numbers, especially when a larger number is subtracted from a smaller one.

Question 12. \( \frac{x}{3} + 1 = \frac {7}{15} \)
Answer:
\( \frac{x}{3} + 1 = \frac {7}{15} \)
\( \implies \frac{x}{3} = \frac {7}{15} - 1 \) (Moving 1 to the right side)
\( \implies \frac{x}{3} = \frac {7 - 15}{15} \) (Taking LCM 15)
\( \implies \frac{x}{3} = \frac{-8}{15} \)
\( \implies \frac{x}{3} \times 3 = \frac{-8 \times 3}{15} \) (Multiplying both sides by 3)
\( \implies x = \frac{-8}{5} \)
In simple words: Subtract 1 from seven-fifteenths to get negative eight-fifteenths. Then, multiply this by 3 to find x, which is negative eight-fifths.

Exam Tip: When combining a fraction and a whole number, always find a common denominator to perform the operation correctly.