RBSE Solutions Class 6 Maths Chapter 12 Algebra Exercise 12.3

Algebra Ex 12.3

 

Question 1. State which of the following are equations (with a variable). Identify the variable from the equations with a variable.
(i) \( 5x = 0 \)
(ii) \( t - 7 > 5 \)
(iii) \( 4 \div 2 = 2 \)
(iv) \( 2x - 1 < 5 \)
(v) \( 7 = 14 \times 2 + q \)
(vi) \( 15000 = 2t + 3500 \)
Answer: The table below shows which statements are equations with a variable and identifies the variable for those that are. Equations always have an equal sign (=) and variables are letters that stand for unknown numbers.

In simple words: Look for an equal sign and a letter in the statement. If both are there, it's an equation with a variable, and the letter is the variable.

🎯 Exam Tip: Remember that equations always use an equals sign (=), while inequalities use signs like \( > \) (greater than) or \( < \) (less than).

 

Question 2. For the equation, \( 10y = 50 \), pick out the solution which satisfies the equation from the values \( y = 10 \), \( y = 8 \) and \( y = 5 \).
Answer: The given equation is \( 10y = 50 \). We need to find which value of \( y \) makes the equation true. Finding the solution to an equation means finding the specific value of the variable that makes both sides of the equation equal.
Let's test each value:
If we put \( y = 10 \):
\( 10 \times 10 = 100 \)
Since \( 100 \neq 50 \), \( y = 10 \) is not the solution.
If we put \( y = 8 \):
\( 10 \times 8 = 80 \)
Since \( 80 \neq 50 \), \( y = 8 \) is not the solution.
If we put \( y = 5 \):
\( 10 \times 5 = 50 \)
Since \( 50 = 50 \), this value satisfies the equation.
Therefore, \( y = 5 \) is the correct solution.In simple words: We tested each number. Only \( y=5 \) worked because when you multiply 10 by 5, you get 50, which is what the equation asked for.

🎯 Exam Tip: When checking potential solutions, always substitute the value into the original equation and calculate both sides to see if they are equal.

 

Question 3. A possible solution is given with each of the equations given below. Put the value of the variable in the equations and show that the value satisfy/do not satisfy the equation.
(i) \( 3x - 7 = 5 \), \( x = 5 \)
(ii) \( 3p + 2 = 8 \), \( p = 2 \)
Answer: When a value satisfies an equation, it means that if you plug that value into the equation, both sides will have the same numerical result.
(i) Given equation: \( 3x - 7 = 5 \)
Given value: \( x = 5 \)
Let's substitute \( x = 5 \) into the left side of the equation:
\( 3 \times 5 - 7 \)
\( = 15 - 7 \)
\( = 8 \)
Since \( 8 \neq 5 \), the value \( x = 5 \) does not satisfy the equation.
(ii) Given equation: \( 3p + 2 = 8 \)
Given value: \( p = 2 \)
Let's substitute \( p = 2 \) into the left side of the equation:
\( 3 \times 2 + 2 \)
\( = 6 + 2 \)
\( = 8 \)
Since \( 8 = 8 \), the value \( p = 2 \) satisfies the equation.In simple words: For the first problem, \( x=5 \) did not work because when we put 5 into the equation, we got 8, not 5. For the second problem, \( p=2 \) did work because when we put 2 into the equation, we got 8, which matched the right side.

🎯 Exam Tip: Always perform calculations on one side of the equation at a time when checking for satisfaction, and then compare the final result with the other side.

 

Question 4. Complete the table and by inspection of the table find the solution to the equation :
(i) \( 3x = 15 \)
Answer: The task is to complete the table for the equation \( 3x = 15 \) and find the solution. The table shows values of \( x \) and corresponding values of \( 3x \). We look for the row where \( 3x \) equals 15. Solving by inspection means looking at the pattern or values in the table directly to find the answer without doing complex calculations.

From the table, when \( 3x = 15 \), the value of \( x \) is 5. Therefore, the solution to the equation \( 3x = 15 \) is \( x = 5 \).In simple words: We looked at the table to find when '3 times x' gives 15. The table shows that this happens when \( x \) is 5.

🎯 Exam Tip: When filling out a table, calculate each value of the expression carefully for the given input variable, and double-check your arithmetic.

 

Question 4. Complete the table and by inspection of the table find the solution to the equation :
(ii) \( \frac{p}{3} = 4 \)
Answer: We need to complete the table for \( \frac{p}{3} = 4 \) and find the solution. The table shows different values of \( p \) and what \( \frac{p}{3} \) equals for each. We are looking for the point where \( \frac{p}{3} \) is 4. This method helps visualize how different input values change the output, making it easier to spot the value that satisfies the equation.

Looking at the table, when \( \frac{P}{3} = 4 \), the value of \( P \) is 12. So, the solution to the equation \( \frac{P}{3} = 4 \) is \( P = 12 \).In simple words: We looked for where dividing 'P' by 3 gives 4. The table showed this happens when \( P \) is 12.

🎯 Exam Tip: It's helpful to simplify fractions in your head or on scratch paper when inspecting the table, e.g., \( \frac{6}{3} = 2 \) and \( \frac{9}{3} = 3 \).

 

Question 4. Complete the table and by inspection of the table find the solution to the equation :
(iii) \( x - 3 = 5 \)
Answer: The goal is to complete the table for the equation \( x - 3 = 5 \) and find its solution. The table shows various values of \( x \) and the result of \( x - 3 \). We need to find when \( x - 3 \) equals 5. Using a table for simple equations allows us to systematically test values and clearly see when the equation holds true.

By inspecting the table, we see that when \( x - 3 = 5 \), the corresponding value of \( x \) is 8. Thus, the solution to the equation \( x - 3 = 5 \) is \( x = 8 \).In simple words: We looked at the table to find when 'x minus 3' gives 5. The table showed that this happens when \( x \) is 8.

🎯 Exam Tip: Pay close attention to negative results when subtracting, especially when the number being subtracted is larger than the starting number.