Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.6

Question 1. Fill in the blanks:
(i) The value of x in the equation x + 5 = 12 is ___________.
Answer:
(i) Given the equation, \( x + 5 = 12 \).
To find x, we move 5 to the other side of the equation. This is called transposition.
\( x = 12 - 5 \)
\( x = 7 \)
The solution for x is 7, meaning 7 makes the equation true.
In simple words: To find x, take 5 away from both sides of the equation. So, x equals 7.

🎯 Exam Tip: Always show the transposition step clearly, indicating when a number changes sides and its sign changes.

 

(ii) The value of y in the equation y – 9 = (-5) + 7 is ___________.
Answer:
(ii) Given the equation, \( y - 9 = (-5) + 7 \).
First, simplify the right side of the equation:
\( y - 9 = 2 \)
Now, transpose -9 to the right side by adding 9 to both sides:
\( y = 2 + 9 \)
\( y = 11 \)
This process of moving terms helps isolate the variable to find its value.
In simple words: First, solve the numbers on the right side. Then, move the -9 to the other side by adding it, and you will get y = 11.

🎯 Exam Tip: Remember that when transposing a number, its sign always flips (plus becomes minus, minus becomes plus).

 

(iii) The value of m in the equation 8m = 56 is __________.
Answer:
(iii) Given the equation, \( 8m = 56 \).
Here, 8 is multiplied by m. To find m, we need to divide both sides by 8.
\( \frac{8m}{8} = \frac{56}{8} \)
\( m = 7 \)
Dividing both sides by the same non-zero number keeps the equation balanced.
In simple words: To find m, divide both sides of the equation by 8. So, m equals 7.

🎯 Exam Tip: Always perform the same operation (addition, subtraction, multiplication, division) on both sides of an equation to keep it balanced.

 

(iv) The value of p in the equation \( \frac{2p}{3} = 10 \) is __________.
Answer:
(iv) Given the equation, \( \frac{2p}{3} = 10 \).
First, multiply both sides by 3 to remove the denominator:
\( \frac{2p}{3} \times 3 = 10 \times 3 \)
\( 2p = 30 \)
Next, divide both sides by 2 to find the value of p:
\( \frac{2p}{2} = \frac{30}{2} \)
\( p = 15 \)
This method helps simplify the equation step-by-step to isolate the variable.
In simple words: First, multiply both sides by 3. Then, divide both sides by 2 to get p = 15.

🎯 Exam Tip: To solve equations with fractions, multiply by the denominator to clear the fraction before isolating the variable.

 

(v) The linear equation in one variable has ___________ solution.
Answer:
(v) A linear equation in one variable has **one** solution.
For example, in \( 2x + 3 = 7 \), there is only one value of x that makes the equation true (which is \( x = 2 \)).
In simple words: A linear equation with only one unknown letter will always have just one answer for that letter.

🎯 Exam Tip: Understand that the "one" in "one variable" refers to the type of variable (e.g., only 'x' or only 'y'), and it also means there's a unique solution for that variable.

 

Question 2. Say True or False.
(i) The shifting of a number from one side of an equation to other is called transposition.
Answer:
(i) **True**
Transposition is indeed the method of moving a term from one side of an equation to the other by changing its sign. This is a fundamental step in solving linear equations.
In simple words: Yes, when you move a number across the equals sign in an equation, it is called transposition.

🎯 Exam Tip: Always remember that the sign of the term must be changed when it is transposed to the other side of the equation.

 

(ii) Linear equation in one variable has only one variable with power 2.
Answer:
(ii) **False**
A linear equation in one variable has only one variable with a power of one. If the variable has a power of 2, it is a quadratic equation, not a linear one. For example, \( x^2 + 2x = 0 \) is quadratic.
In simple words: This is false. A linear equation means the variable has a power of 1, not 2. If it has a power of 2, it's a different kind of equation.

🎯 Exam Tip: The highest power of the variable defines the type of equation: power 1 for linear, power 2 for quadratic, and so on.

 

Question 3. Match the following

(A) (i), (ii), (iv), (iii), (v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii), (i), (iv), (v), (ii)
(D) (iii), (i), (v), (iv), (ii)
Answer: (C) (iii), (i), (iv), (v), (ii)
For each equation, let's find the value of x:
a. \( \frac{x}{2} = 10 \)
Multiply both sides by 2:
\( \frac{x}{2} \times 2 = 10 \times 2 \)
\( x = 20 \) (Matches iii)

b. \( 20 = 6x - 4 \)
Transpose -4 to the left side:
\( 20 + 4 = 6x \)
\( 24 = 6x \)
Divide both sides by 6:
\( \frac{24}{6} = \frac{6x}{6} \)
\( x = 4 \) (Matches i)

c. \( 2x - 5 = 3 - x \)
Transpose -x to the left side and -5 to the right side:
\( 2x + x = 3 + 5 \)
\( 3x = 8 \)
Divide both sides by 3:
\( \frac{3x}{3} = \frac{8}{3} \)
\( x = \frac{8}{3} \) (Matches iv)

d. \( 7x - 4 - 8x = 20 \)
Combine like terms on the left side:
\( (7x - 8x) - 4 = 20 \)
\( -x - 4 = 20 \)
Transpose -4 to the right side:
\( -x = 20 + 4 \)
\( -x = 24 \)
Multiply both sides by -1:
\( x = -24 \) (Matches v)

e. \( \frac{4}{11} - x = \frac{-7}{11} \)
Transpose \( \frac{4}{11} \) to the right side:
\( -x = \frac{-7}{11} - \frac{4}{11} \)
\( -x = \frac{-7 - 4}{11} \)
\( -x = \frac{-11}{11} \)
\( -x = -1 \)
Multiply both sides by -1:
\( x = 1 \) (Matches ii)
The correct matching is a-(iii), b-(i), c-(iv), d-(v), e-(ii).
In simple words: For each equation, solve for x by moving numbers and letters around, remembering to change signs. Then, match the answer you get for x to the given options.

🎯 Exam Tip: When matching, solve each equation completely before trying to find its pair to avoid confusion, especially with negative signs and fractions.

 

Question 4. Find x:
(i) \( \frac{2x}{3} - 4 = \frac{10}{3} \)
Answer:
(i) Given the equation, \( \frac{2x}{3} - 4 = \frac{10}{3} \).
First, transpose -4 to the right side of the equation:
\( \frac{2x}{3} = \frac{10}{3} + 4 \)
To add the terms on the right side, find a common denominator for 4 (which is \( \frac{4}{1} \)):
\( \frac{2x}{3} = \frac{10}{3} + \frac{4 \times 3}{1 \times 3} \)
\( \frac{2x}{3} = \frac{10}{3} + \frac{12}{3} \)
\( \frac{2x}{3} = \frac{10 + 12}{3} \)
\( \frac{2x}{3} = \frac{22}{3} \)
Now, multiply both sides by 3 to clear the denominators:
\( \frac{2x}{3} \times 3 = \frac{22}{3} \times 3 \)
\( 2x = 22 \)
Finally, divide both sides by 2 to solve for x:
\( \frac{2x}{2} = \frac{22}{2} \)
\( x = 11 \)
Following these steps helps solve equations with fractions accurately.
In simple words: Move the -4 to the other side. Add the fractions on the right side. Then, multiply both sides by 3, and finally divide by 2 to find x = 11.

🎯 Exam Tip: When combining fractions with whole numbers, always convert the whole number to a fraction with the same denominator before adding or subtracting.

 

(ii) \( y + \frac{1}{6} - 3y = \frac{2}{3} \)
Answer:
(ii) Given the equation, \( y + \frac{1}{6} - 3y = \frac{2}{3} \).
First, combine the y terms on the left side:
\( (y - 3y) + \frac{1}{6} = \frac{2}{3} \)
\( -2y + \frac{1}{6} = \frac{2}{3} \)
Next, transpose \( \frac{1}{6} \) to the right side:
\( -2y = \frac{2}{3} - \frac{1}{6} \)
To subtract the fractions on the right side, find a common denominator, which is 6:
\( -2y = \frac{2 \times 2}{3 \times 2} - \frac{1}{6} \)
\( -2y = \frac{4}{6} - \frac{1}{6} \)
\( -2y = \frac{4 - 1}{6} \)
\( -2y = \frac{3}{6} \)
Simplify the fraction on the right:
\( -2y = \frac{1}{2} \)
Finally, divide both sides by -2 to solve for y. Dividing by -2 is the same as multiplying by \( -\frac{1}{2} \):
\( y = \frac{1}{2} \times \left(-\frac{1}{2}\right) \)
\( y = -\frac{1}{4} \)
Careful handling of signs and fractions is key to solving such equations.
In simple words: Combine the 'y' terms first. Then, move the fraction to the other side and subtract. Finally, divide by -2 to get the value of y as \( -\frac{1}{4} \).

🎯 Exam Tip: Remember to combine like terms (like 'y' terms) on each side of the equation before attempting to transpose terms.

 

(iii) \( \frac{1}{3} - \frac{x}{3} = \frac{7x}{12} + \frac{5}{4} \)
Answer:
(iii) Given the equation, \( \frac{1}{3} - \frac{x}{3} = \frac{7x}{12} + \frac{5}{4} \).
To simplify, find the LCM of all denominators (3, 3, 12, 4), which is 12.
Multiply the entire equation by 12 to clear the denominators:
\( 12 \left( \frac{1}{3} - \frac{x}{3} \right) = 12 \left( \frac{7x}{12} + \frac{5}{4} \right) \)
\( 12 \times \frac{1}{3} - 12 \times \frac{x}{3} = 12 \times \frac{7x}{12} + 12 \times \frac{5}{4} \)
\( 4 - 4x = 7x + 15 \)
Now, group the x terms on one side and constant terms on the other. Transpose -4x to the right and 15 to the left:
\( 4 - 15 = 7x + 4x \)
\( -11 = 11x \)
Finally, divide both sides by 11 to find x:
\( \frac{-11}{11} = \frac{11x}{11} \)
\( x = -1 \)
Multiplying by the LCM at the start often makes solving equations with multiple fractions much easier.
In simple words: Find the smallest number that all denominators can divide into (LCM), which is 12. Multiply the whole equation by 12 to get rid of fractions. Then, move all 'x' terms to one side and numbers to the other, and solve for x.

🎯 Exam Tip: Multiplying the entire equation by the Least Common Multiple (LCM) of all denominators is a very efficient way to eliminate fractions early on.

 

Question 5. Find x
(i) \( -3(4x + 9) = 21 \)
Answer:
(i) Given the equation, \( -3(4x + 9) = 21 \).
First, expand the bracket on the left side by multiplying -3 with each term inside:
\( (-3 \times 4x) + (-3 \times 9) = 21 \)
\( -12x - 27 = 21 \)
Next, transpose -27 to the right side by adding 27 to both sides:
\( -12x = 21 + 27 \)
\( -12x = 48 \)
Finally, divide both sides by -12 to solve for x:
\( \frac{-12x}{-12} = \frac{48}{-12} \)
\( x = -4 \)
Always be careful with the signs when expanding brackets and dividing.
In simple words: Multiply -3 by everything inside the bracket. Then, move the -27 to the other side by adding it. Finally, divide by -12 to find x = -4.

🎯 Exam Tip: Remember to distribute the number outside the bracket to *all* terms inside, paying close attention to negative signs.

 

(ii) \( 20 - 2(5 - p) = 8 \)
Answer:
(ii) Given the equation, \( 20 - 2(5 - p) = 8 \).
First, expand the bracket on the left side by multiplying -2 with each term inside:
\( 20 - (2 \times 5) - (2 \times -p) = 8 \)
\( 20 - 10 + 2p = 8 \)
Combine the constant terms on the left side:
\( 10 + 2p = 8 \)
Next, transpose 10 to the right side by subtracting 10 from both sides:
\( 2p = 8 - 10 \)
\( 2p = -2 \)
Finally, divide both sides by 2 to solve for p:
\( \frac{2p}{2} = \frac{-2}{2} \)
\( p = -1 \)
Simplifying constants early on can make the equation easier to handle.
In simple words: Multiply -2 by everything in the bracket. Then, add the numbers on the left. Move the single number to the right side, and finally divide by 2 to find p = -1.

🎯 Exam Tip: Be careful with the minus sign in front of the 2: treat it as multiplying by -2, not just 2.

 

(iii) \( (7x - 5) - 4(2 + 5x) = 10(2 - x) \)
Answer:
(iii) Given the equation, \( (7x - 5) - 4(2 + 5x) = 10(2 - x) \).
First, expand all brackets:
Left side: \( 7x - 5 - (4 \times 2) - (4 \times 5x) = 7x - 5 - 8 - 20x \)
Right side: \( (10 \times 2) - (10 \times x) = 20 - 10x \)
So the equation becomes:
\( 7x - 5 - 8 - 20x = 20 - 10x \)
Combine like terms on the left side:
\( (7x - 20x) + (-5 - 8) = 20 - 10x \)
\( -13x - 13 = 20 - 10x \)
Now, transpose -10x to the left side and -13 to the right side:
\( -13x + 10x = 20 + 13 \)
\( -3x = 33 \)
Finally, divide both sides by -3 to solve for x:
\( \frac{-3x}{-3} = \frac{33}{-3} \)
\( x = -11 \)
This type of equation requires careful expansion and organization of terms.
In simple words: Expand all brackets by multiplying the numbers outside. Then, gather all 'x' terms on one side and all numbers on the other. Add or subtract them, and then divide to find x = -11.

🎯 Exam Tip: For equations with multiple brackets, expand one bracket at a time, then combine like terms before transposing.

 

Question 6. Find x and m:
(i) \( \frac{3x - 2}{4} - \frac{x - 3}{5} = -1 \)
Answer:
(i) Given the equation, \( \frac{3x - 2}{4} - \frac{x - 3}{5} = -1 \).
First, find the Least Common Multiple (LCM) of the denominators 4 and 5, which is 20.
Multiply the entire equation by 20 to clear the denominators:
\( 20 \left( \frac{3x - 2}{4} \right) - 20 \left( \frac{x - 3}{5} \right) = 20 \times (-1) \)
\( 5(3x - 2) - 4(x - 3) = -20 \)
Now, expand the brackets:
\( (5 \times 3x) - (5 \times 2) - (4 \times x) - (4 \times -3) = -20 \)
\( 15x - 10 - 4x + 12 = -20 \)
Combine like terms on the left side:
\( (15x - 4x) + (-10 + 12) = -20 \)
\( 11x + 2 = -20 \)
Next, transpose 2 to the right side:
\( 11x = -20 - 2 \)
\( 11x = -22 \)
Finally, divide both sides by 11 to solve for x:
\( \frac{11x}{11} = \frac{-22}{11} \)
\( x = -2 \)
Multiplying by the LCM at the start simplifies the equation significantly.
In simple words: Find the LCM of 4 and 5, which is 20. Multiply the whole equation by 20 to remove fractions. Expand the brackets. Combine 'x' terms and numbers. Move numbers to one side, and finally divide to find x = -2.

🎯 Exam Tip: When multiplying by the LCM, remember to apply it to *every* term in the equation, including the constant on the right side.

 

(ii) \( \frac{m + 9}{3m + 15} = \frac{5}{3} \)
Answer:
(ii) Given the equation, \( \frac{m + 9}{3m + 15} = \frac{5}{3} \).
To solve this, we can use cross-multiplication:
\( 3(m + 9) = 5(3m + 15) \)
Now, expand both sides:
\( (3 \times m) + (3 \times 9) = (5 \times 3m) + (5 \times 15) \)
\( 3m + 27 = 15m + 75 \)
Next, group the 'm' terms on one side and constant terms on the other. Transpose 3m to the right and 75 to the left:
\( 27 - 75 = 15m - 3m \)
\( -48 = 12m \)
Finally, divide both sides by 12 to solve for m:
\( \frac{-48}{12} = \frac{12m}{12} \)
\( m = -4 \)
Cross-multiplication is very effective for equations where a single fraction equals another single fraction.
In simple words: Multiply the top of one fraction by the bottom of the other (cross-multiplication). Expand both sides. Move 'm' terms to one side and numbers to the other. Then divide to get m = -4.

🎯 Exam Tip: Cross-multiplication is a shortcut for clearing denominators when you have a fraction equal to another fraction. Be sure to distribute correctly when expanding.