Algebraic Expressions and Identities — NCERT Solutions, AI Tutor & Practice

The terms are the parts separated by $+$ or $-$ signs.

• Term $5x^2y$ → coefficient $5$
• Term $-3xy$ → coefficient $-3$
• Term $7y$ → coefficient $7$
• Term $-2$ → this is a constant term (coefficient $-2$)

So the expression has 4 terms with coefficients $5, -3, 7, -2$.

We classify by counting the number of terms.

• (a) $4xy$ → 1 term → Monomial
• (b) $a + b$ → 2 terms → Binomial
• (c) $x^2 + 2x + 1$ → 3 terms → Trinomial
• (d) $-7pqr$ → 1 term → Monomial

Write the expressions and group like terms.

$(3x^2 + 5x - 4) + (-x^2 + 2x + 7)$

Combine like terms:

• $x^2$ terms: $3x^2 + (-x^2) = 2x^2$
• $x$ terms: $5x + 2x = 7x$
• constants: $-4 + 7 = 3$

Result: $2x^2 + 7x + 3$.

We compute $(5a^2 + ab - 4b^2) - (2a^2 - 3ab + b^2)$.

Change the sign of every term being subtracted:

$5a^2 + ab - 4b^2 - 2a^2 + 3ab - b^2$

Group like terms:

• $a^2$: $5a^2 - 2a^2 = 3a^2$
• $ab$: $ab + 3ab = 4ab$
• $b^2$: $-4b^2 - b^2 = -5b^2$

Result: $3a^2 + 4ab - 5b^2$.

Multiply coefficients and variables separately, using the law $x^m \times x^n = x^{m+n}$.

Coefficients: $4 \times (-3) = -12$

$x$ terms: $x^2 \times x = x^{2+1} = x^3$

$y$ terms: $y \times y^3 = y^{1+3} = y^4$

Result: $-12x^3y^4$.

Apply the distributive law: multiply $3p$ by each term.

$3p \times 2p^2 = 6p^3$ $3p \times (-5p) = -15p^2$ $3p \times 4 = 12p$

Result: $6p^3 - 15p^2 + 12p$.

Multiply term by term (each term of the first with each term of the second):

$(x + 4)(x - 3) = x \cdot x + x \cdot (-3) + 4 \cdot x + 4 \cdot (-3)$

$= x^2 - 3x + 4x - 12$

Combine like terms $-3x + 4x = x$:

Result: $x^2 + x - 12$.

Use the identity $(a + b)^2 = a^2 + 2ab + b^2$.

Here $a = x$ and $b = 7$.

$(x + 7)^2 = x^2 + 2(x)(7) + 7^2$ $= x^2 + 14x + 49$

Use the identity $(a - b)^2 = a^2 - 2ab + b^2$.

Here $a = 3a$ and $b = 2b$.

$(3a - 2b)^2 = (3a)^2 - 2(3a)(2b) + (2b)^2$ $= 9a^2 - 12ab + 4b^2$

Write $103 = 100 + 3$ and $97 = 100 - 3$.

Use $(a + b)(a - b) = a^2 - b^2$ with $a = 100$, $b = 3$.

$103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2$ $= 10000 - 9 = 9991$

Compare with $(x + a)(x + b)$: here $x = y$, $a = 8$, $b = 5$.

• $a + b = 8 + 5 = 13$
• $ab = 8 \times 5 = 40$

So, $(y + 8)(y + 5) = y^2 + 13y + 40$

Left Hand Side (LHS):

$(a + b)^2 = (5 + 2)^2 = 7^2 = 49$

$(a - b)^2 = (5 - 2)^2 = 3^2 = 9$

$\text{LHS} = 49 - 9 = 40$

Right Hand Side (RHS):

$4ab = 4 \times 5 \times 2 = 40$

Since $\text{LHS} = \text{RHS} = 40$, the identity is verified.