Maharashtra Board Class 7 Maths Chapter 14 Set 51 Algebraic Formulae Solutions

Question 1. Use the formula to multiply the following:
(i) (x + y)(x - y)
(ii) (3x – 5)(3x + 5)
(iii) (a + 6)(a – 6)
(iv) \(\left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right)\)
Answer:
(i) Here, a = x, b = y
\( (x + y)(x - y) = x^2 - y^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)

(ii) Here, a = 3x, b = 5
\( (3x - 5)(3x + 5) = (3x)^2 - 5^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)
\( = 9x^2 - 25 \)

(iii) Here, A = a, B = 6
\( (a + 6)(a - 6) = a^2 - 6^2 \)
... \( [(A + B)(A - B) = A^2 - B^2] \)
\( = a^2 - 36 \)

(iv) Here, a = \(\frac{x}{5}\), b = 6
\( \left(\frac{x}{5}+6\right)\left(\frac{x}{5}-6\right) = \left(\frac{x}{5}\right)^2 - (6)^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)
\( = \frac{x^2}{25} - 36 \)
In simple words: This question requires applying the algebraic identity \( (a+b)(a-b) = a^2-b^2 \) to simplify given multiplication expressions. For each part, identify 'a' and 'b' and then directly substitute them into the formula to find the expanded form.

🎯 Exam Tip: Remember to correctly identify the 'a' and 'b' terms in each expression, especially when they involve variables or fractions, to avoid calculation errors. Practice recognizing the pattern of the formula to quickly apply it.

Question 2. Use the formula to find the values:
(i) 502 × 498
(ii) 97 × 103
(iii) 54 x 46
(iv) 98 × 102
Answer:
(i) 502 × 498 = (500 + 2) (500 - 2)
Here, a = 500, b = 2

\(\implies (500 + 2) (500 - 2) = 500^2 - 2^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)
\( = 250000 - 4 \)
\( = 249996 \)

\(\implies 502 \times 498 = 249996 \)

(ii) 97 × 103 = (100 - 3) (100 + 3)
Here, a = 100, b = 3

\(\implies (100 - 3) (100 + 3) = 100^2 - 3^2 \)

\(\implies [(a + b)(a - b) = a^2 - b^2] \)
\( = 10000 - 9 \)
\( = 9991 \)

\(\implies 97 \times 103 = 9991 \)

(iii) 54 × 46 = (50 + 4) (50 - 4)
Here, a = 50, b = 4

\(\implies (50 + 4) (50 - 4) = 50^2 - 4^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)
\( = 2500 - 16 = 2484 \)

\(\implies 54 \times 46 = 2484 \)

(iv) 98 × 102 = (100 - 2) (100 + 2)
Here, a = 100, b = 2

\(\implies (100 - 2) (100 + 2) = 100^2 - 2^2 \)
... \( [(a + b)(a - b) = a^2 - b^2] \)
\( = 10000 - 4 \)
\( = 9996 \)

\(\implies 98 \times 102 = 9996 \)
In simple words: This question applies the same \( (a+b)(a-b) = a^2-b^2 \) identity to simplify numerical multiplications by expressing the numbers as sums and differences around a convenient round number. This makes calculations easier and faster.

🎯 Exam Tip: When using the formula for numerical calculations, choose 'a' and 'b' such that 'a' is a round number (like 100, 500) and 'b' is a small integer, simplifying the squaring operation significantly.