Fundamental Concepts of Algebra

A symbol having a fixed numerical value in all situations is a constant, whereas a symbol whose value changes with the situation is a variable. Also, every combination of a constant and a variable is a variable.

Constants = 6, $\mathbf{\frac{5}{4}}$ and 0

Variables = $\bm{4y, -3x, \frac{4}{5}xy, az, 7p, \frac{9x}{y}, \frac{3}{4x}}$ and $\bm{-\frac{xz}{3y}}$

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{4x, -x}$ and $\bm{\frac{2}{3}x}; \bm{-3y, \frac{4}{5}y}$ and $\bm{y}$.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{\frac{2}{3}xy, -4yx}$ and $\bm{yx}; \bm{2yz, \frac{-2}{3}yz}$ and $\bm{\frac{zy}{3}}$.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{-ab^2, 7b^2a}$ and $\bm{2ab^2}; \bm{b^2a^2}$ and $\bm{-3a^2b^2}$.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{5ax, 7xa}$ and $\bm{\frac{2ax}{3}}; \bm{-5by}$ and $\bm{\frac{by}{7}}$.

(i) True, as the value of $16y$ changes with the value of $y$, so every combination of a constant and a variable is a variable.

(ii) False, as $5x$ is a single term. The sign of multiplication does not separate the terms.

(iii) True, as terms are separated by the signs + and - only, so $5 + x$ has two terms 5 and $x$.

(iv) False, as $2x^2 + x$ has two unlike terms, so it is a binomial.

(v) True, as $ax^2 + bx + c$ has three unlike terms.

(vi) False, as $8 \times ab = 8ab$ has only one term, so it is a monomial.

(vii) True, as $8 + ab$ has two unlike terms.

(viii) True, as in $x^3 - 5xy + 6x + 7$ the sum of the powers of the variables in each term is a whole number.

(ix) True, as $x^3 - 5xy + 6x + 7$ has more than two terms.

(x) False, as $5x = 5 \times x$, so the coefficient of $x$ in $5x$ is 5.

(xi) True, as $-ab = (-1) \times ab$, so the coefficient of $ab$ in $-ab$ is -1.

(xii) False, as $-3xy = (-3x) \times y$, so the coefficient of $y$ in $-3xy$ is $-3x$.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$2a - b$ has the terms $2a$ and $-b$ → 2 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$3 \times x + \frac{a}{2} = 3x + \frac{a}{2}$, which has the terms $3x$ and $\frac{a}{2}$ → 2 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$3x - \frac{x}{p}$ has the terms $3x$ and $-\frac{x}{p}$ → 2 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$a \div x \times b + c = \frac{ab}{x} + c$, which has the terms $\frac{ab}{x}$ and $c$ → 2 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$3x \div 2 + y + 4 = \frac{3x}{2} + y + 4$, which has the terms $\frac{3x}{2}, y$ and 4 → 3 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$xy \div 2 = \frac{xy}{2}$, which has only one term → 1 term.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$x + y \div a = x + \frac{y}{a}$, which has the terms $x$ and $\frac{y}{a}$ → 2 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$2x + y + 8 \div y = 2x + y + \frac{8}{y}$, which has the terms $2x, y$ and $\frac{8}{y}$ → 3 terms.

Terms are separated by the signs +(plus) and -(minus) only. The signs ×(multiplication) and ÷(division) do not separate the terms.

$2 \times a + 3 \div b + 4 = 2a + \frac{3}{b} + 4$, which has the terms $2a, \frac{3}{b}$ and 4 → 3 terms.

The terms having the same literal coefficients are like terms, and the terms that do not have the same literal coefficients are unlike terms.

(i) True, as the literal coefficient of each of $xy$ and $-yx$ is $xy$.

(ii) False, as the literal coefficients $x^2y$ and $y^2x$ are different, so they are unlike terms.

(iii) True, as the literal coefficient of each of $a$ and $-a$ is $a$.

(iv) False, as the literal coefficient of each of $-ba$ and $2ab$ is $ab$, so they are like terms.

(v) False, as 5 has no literal coefficient while the literal coefficient of $5x$ is $x$, so they are unlike terms.

(vi) True, as the literal coefficients $xy$ and $xyz$ are different.

An expression with only one term is a monomial, with two unlike terms is a binomial and with three unlike terms is a trinomial.

(i) $xy$ has 1 term → Monomial.

(ii) $xy + x$ has 2 terms → Binomial.

(iii) $2x \div y = \frac{2x}{y}$, which has 1 term → Monomial.

(iv) $-a$ has 1 term → Monomial.

(v) $ax^2 - x + 5$ has 3 terms → Trinomial.

(vi) $-3bc + d$ has 2 terms → Binomial.

(vii) $1 + x + y$ has 3 terms → Trinomial.

(viii) $1 + x \div y = 1 + \frac{x}{y}$, which has 2 terms → Binomial.

(ix) $x + xy - y^2$ has 3 terms → Trinomial.

Any factor or group of factors of a product is the coefficient of the remaining factors.

(i) $x = 1 \times x$, so the coefficient of $x$ is 1.

(ii) $-x = (-1) \times x$, so the coefficient of $x$ is −1.

(iii) $-3x = (-3) \times x$, so the coefficient of $x$ is −3.

(iv) $-5ax = (-5a) \times x$, so the coefficient of $x$ is $\bm{-5a}$.

(v) $\frac{3}{2}xy = \left(\frac{3}{2}y\right) \times x$, so the coefficient of $x$ is $\bm{\frac{3}{2}y}$.

(vi) $\frac{ax}{y} = \left(\frac{a}{y}\right) \times x$, so the coefficient of $x$ is $\bm{\frac{a}{y}}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-3xy^2 = (-3y^2) \times x$, so the coefficient of $x$ is $\bm{-3y^2}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-ax = (-a) \times x$, so the coefficient of $x$ is $\bm{-a}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-y = (-1) \times y$, so the coefficient of $y$ is −1.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$\frac{2}{a}y = \left(\frac{2}{a}\right) \times y$, so the coefficient of $y$ is $\bm{\frac{2}{a}}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-2xyz = (-2z) \times xy$, so the coefficient of $xy$ is $\bm{-2z}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-axy^2 = (-y^2) \times ax$, so the coefficient of $ax$ is $\bm{-y^2}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-3ax^2y = (-3a) \times x^2y$, so the coefficient of $x^2y$ is $\bm{-3a}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$5axy^2 = (5a) \times xy^2$, so the coefficient of $xy^2$ is $\bm{5a}$.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

The numeral coefficient of $5xy$ is 5.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

$abc = 1 \times abc$, so the numeral coefficient of $abc$ is 1.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

The numeral coefficient of $5pqr$ is 5.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

$\frac{-2x}{y} = (-2) \times \frac{x}{y}$, so the numeral coefficient is −2.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

The numeral coefficient of $\frac{2}{3}xy^2$ is $\mathbf{\frac{2}{3}}$.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

$\frac{-15xy}{2z} = \left(\frac{-15}{2}\right) \times \frac{xy}{z}$, so the numeral coefficient is $\mathbf{\frac{-15}{2}}$.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

$-7x \div y = \frac{-7x}{y} = (-7) \times \frac{x}{y}$, so the numeral coefficient is −7.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

$-3x \div (2y) = \frac{-3x}{2y} = \left(\frac{-3}{2}\right) \times \frac{x}{y}$, so the numeral coefficient is $\mathbf{\frac{-3}{2}}$.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 1 and 2. The greatest is 2.

Hence, the degree of the polynomial = 2.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 2, 1 and 0. The greatest is 2.

Hence, the degree of the polynomial = 2.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 3, 8 and 10. The greatest is 10.

Hence, the degree of the polynomial = 10.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 0 and 20. The greatest is 20.

Hence, the degree of the polynomial = 20.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 0, 1 and 3. The greatest is 3.

Hence, the degree of the polynomial = 3.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms. For a term containing more than one variable, the power of that term is the sum of the powers of all the variables in it.

Degrees of the terms are (2 + 1) = 3, 2 and (2 + 5) = 7. The greatest is 7.

Hence, the degree of the polynomial = 7.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms. For a term containing more than one variable, the power of that term is the sum of the powers of all the variables in it.

Degrees of the terms are 3, (2 + 3) = 5 and (1 + 5) = 6. The greatest is 6.

Hence, the degree of the polynomial = 6.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms. For a term containing more than one variable, the power of that term is the sum of the powers of all the variables in it.

Degrees of the terms are 2, 3 and (2 + 7) = 9. The greatest is 9.

Hence, the degree of the polynomial = 9.

For the addition or subtraction of like terms, the rules are the same as those for the addition or subtraction of integers. Unlike terms cannot be added or subtracted to get a single term.

(i) The sum of -2 and -5 = -2 - 5 = −7 and the sum of $-2x$ and $-5x = -2x - 5x = (-2 - 5)x = \bm{-7x}$.

(ii) The sum of 8 and -3 = 8 - 3 = 5 and the sum of $8ab$ and $-3ab = 8ab - 3ab = (8 - 3)ab = \bm{5ab}$.

(iii) The sum of -15 and -4 = -15 - 4 = −19 and the sum of $-15x$ and $-4y = \bm{-15x - 4y}$, as $-15x$ and $-4y$ are unlike terms.

(iv) 15 + 8 + 3 = 26 and $15x + 8y + 3x = (15 + 3)x + 8y = \bm{18x + 8y}$.

(v) 12 - 9 + 15 = 18 and $12ab - 9ab + 15ba = (12 - 9 + 15)ab = \bm{18ab}$.

(vi) 25 - 7 - 9 = 9 and $25xy - 7xy - 9yx = (25 - 7 - 9)xy = \bm{9xy}$.

Solving,

$$\begin{aligned}\Rightarrow 8xy + 3xy \\[1em] \Rightarrow (8 + 3)xy \\[1em] \Rightarrow 11xy\end{aligned}$$

Hence, the required sum = $\bm{11xy}$.

Solving,

$$\begin{aligned}\Rightarrow 2xyz + xyz + 6xyz \\[1em] \Rightarrow (2 + 1 + 6)xyz \\[1em] \Rightarrow 9xyz\end{aligned}$$

Hence, the required sum = $\bm{9xyz}$.

Solving,

$$\begin{aligned}\Rightarrow 2a + 3a + 4b \\[1em] \Rightarrow (2 + 3)a + 4b \\[1em] \Rightarrow 5a + 4b\end{aligned}$$

Hence, the required sum = $\bm{5a + 4b}$.

Solving,

$\Rightarrow 3x + 2y$

As $3x$ and $2y$ are unlike terms, they cannot be added to get a single term.

Hence, the required sum = $\bm{3x + 2y}$.

Solving,

$\Rightarrow 5m + 3n + 4p$

As $5m, 3n$ and $4p$ are unlike terms, they cannot be added to get a single term.

Hence, the required sum = $\bm{5m + 3n + 4p}$.

Solving,

$$\begin{aligned}\Rightarrow 6a + 3a + 9ab \\[1em] \Rightarrow (6 + 3)a + 9ab \\[1em] \Rightarrow 9a + 9ab\end{aligned}$$

Hence, the required sum = $\bm{9a + 9ab}$.

Solving,

$$\begin{aligned}\Rightarrow 3p + 4q + 9q \\[1em] \Rightarrow 3p + (4 + 9)q \\[1em] \Rightarrow 3p + 13q\end{aligned}$$

Hence, the required sum = $\bm{3p + 13q}$.

Solving,

$$\begin{aligned}\Rightarrow 5ab + 4ba + 6b \\[1em] \Rightarrow (5 + 4)ab + 6b \\[1em] \Rightarrow 9ab + 6b\end{aligned}$$

Hence, the required sum = $\bm{9ab + 6b}$.

Solving,

$$\begin{aligned}\Rightarrow 50pq + 30pq + 10pr \\[1em] \Rightarrow (50 + 30)pq + 10pr \\[1em] \Rightarrow 80pq + 10pr\end{aligned}$$

Hence, the required sum = $\bm{80pq + 10pr}$.

Solving,

$$\begin{aligned}\Rightarrow -2y + (-y) + (-3y) \\[1em] \Rightarrow (-2 - 1 - 3)y \\[1em] \Rightarrow -6y\end{aligned}$$

Hence, the required sum = $\bm{-6y}$.

Solving,

$$\begin{aligned}\Rightarrow -3b + (-b) \\[1em] \Rightarrow (-3 - 1)b \\[1em] \Rightarrow -4b\end{aligned}$$

Hence, the required sum = $\bm{-4b}$.

Solving,

$$\begin{aligned}\Rightarrow -2c + (-c) + (-5c) \\[1em] \Rightarrow (-2 - 1 - 5)c \\[1em] \Rightarrow -8c\end{aligned}$$

Hence, the required sum = $\bm{-8c}$.

Solving,

$$\begin{aligned}\Rightarrow 6a - a - 5a - 2a \\[1em] \Rightarrow (6 - 1 - 5 - 2)a \\[1em] \Rightarrow -2a\end{aligned}$$

Hence, the value of the given expression = $\bm{-2a}$.

Solving,

$$\begin{aligned}\Rightarrow 2b - 3b - b + 4b \\[1em] \Rightarrow (2 - 3 - 1 + 4)b \\[1em] \Rightarrow 2b\end{aligned}$$

Hence, the value of the given expression = $\bm{2b}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 2x - 4x + 7x \\[1em] \Rightarrow (3 - 2 - 4 + 7)x \\[1em] \Rightarrow 4x\end{aligned}$$

Hence, the value of the given expression = $\bm{4x}$.

Solving,

$$\begin{aligned}\Rightarrow 5ab + 2ab - 6ab + ab \\[1em] \Rightarrow (5 + 2 - 6 + 1)ab \\[1em] \Rightarrow 2ab\end{aligned}$$

Hence, the value of the given expression = $\bm{2ab}$.

Solving,

$$\begin{aligned}\Rightarrow 8x - 5y - 3x + 10y \\[1em] \Rightarrow (8 - 3)x + (-5 + 10)y \\[1em] \Rightarrow 5x + 5y\end{aligned}$$

Hence, the value of the given expression = $\bm{5x + 5y}$.

Solving,

$$\begin{aligned}\Rightarrow -7x + 9x + 2x - 2x \\[1em] \Rightarrow (-7 + 9 + 2 - 2)x \\[1em] \Rightarrow 2x\end{aligned}$$

Hence, the value of the given expression = $\bm{2x}$.

Solving,

$$\begin{aligned}\Rightarrow 5ab - 2ab - 8ab + 6ab \\[1em] \Rightarrow (5 - 2 - 8 + 6)ab \\[1em] \Rightarrow ab\end{aligned}$$

Hence, the value of the given expression = $\bm{ab}$.

Solving,

$$\begin{aligned}\Rightarrow -8a - 3a + 12a + 13a - 6a \\[1em] \Rightarrow (-8 - 3 + 12 + 13 - 6)a \\[1em] \Rightarrow 8a\end{aligned}$$

Hence, the value of the given expression = $\bm{8a}$.

Solving,

$$\begin{aligned}\Rightarrow 6ba - 4ab \\[1em] \Rightarrow (6 - 4)ab \\[1em] \Rightarrow 2ab\end{aligned}$$

Hence, the required difference = $\bm{2ab}$.

Solving,

$$\begin{aligned}\Rightarrow 6.8b - 4.8b \\[1em] \Rightarrow (6.8 - 4.8)b \\[1em] \Rightarrow 2b\end{aligned}$$

Hence, the required difference = $\bm{2b}$.

Solving,

$$\begin{aligned}\Rightarrow 10.5abc - 3.5abc \\[1em] \Rightarrow (10.5 - 3.5)abc \\[1em] \Rightarrow 7abc\end{aligned}$$

Hence, the required difference = $\bm{7abc}$.

Solving,

$$\begin{aligned}\Rightarrow 8\frac{1}{2}nm - 3\frac{1}{2}mn \\[1em] \Rightarrow \frac{17}{2}mn - \frac{7}{2}mn \\[1em] \Rightarrow \left(\frac{17 - 7}{2}\right)mn \\[1em] \Rightarrow \frac{10}{2}mn \\[1em] \Rightarrow 5mn\end{aligned}$$

Hence, the required difference = $\bm{5mn}$.

Solving,

$$\begin{aligned}\Rightarrow 2a^2b^2 + 5ab^2 + 8a^2b^2 - 3ab^2 \\[1em] \Rightarrow (2 + 8)a^2b^2 + (5 - 3)ab^2 \\[1em] \Rightarrow 10a^2b^2 + 2ab^2\end{aligned}$$

Hence, the simplified expression = $\bm{10a^2b^2 + 2ab^2}$.

Solving,

$$\begin{aligned}\Rightarrow 4a + 3b - 2a - b \\[1em] \Rightarrow (4 - 2)a + (3 - 1)b \\[1em] \Rightarrow 2a + 2b\end{aligned}$$

Hence, the simplified expression = $\bm{2a + 2b}$.

Solving,

$$\begin{aligned}\Rightarrow 2xy + 4yz + 5xy + 3yz - 6xy \\[1em] \Rightarrow (2 + 5 - 6)xy + (4 + 3)yz \\[1em] \Rightarrow xy + 7yz\end{aligned}$$

Hence, the simplified expression = $\bm{xy + 7yz}$.

Solving,

$$\begin{aligned}\Rightarrow ab + 15ab - 11ab - 2ab \\[1em] \Rightarrow (1 + 15 - 11 - 2)ab \\[1em] \Rightarrow 3ab\end{aligned}$$

Hence, the simplified expression = $\bm{3ab}$.

Solving,

$$\begin{aligned}\Rightarrow 6a^2 - 3b^2 + 2a^2 + 5b^2 - 4a^2 \\[1em] \Rightarrow (6 + 2 - 4)a^2 + (-3 + 5)b^2 \\[1em] \Rightarrow 4a^2 + 2b^2\end{aligned}$$

Hence, the simplified expression = $\bm{4a^2 + 2b^2}$.

Solving,

$$\begin{aligned}\Rightarrow 8abc + 2ab - 4abc + ab \\[1em] \Rightarrow (8 - 4)abc + (2 + 1)ab \\[1em] \Rightarrow 4abc + 3ab\end{aligned}$$

Hence, the simplified expression = $\bm{4abc + 3ab}$.

Solving,

$$\begin{aligned}\Rightarrow 9xyz + 15yxz - 10zyx - 2zxy \\[1em] \Rightarrow (9 + 15 - 10 - 2)xyz \\[1em] \Rightarrow 12xyz\end{aligned}$$

Hence, the simplified expression = $\bm{12xyz}$.

Solving,

$$\begin{aligned}\Rightarrow 13pqr + 2p + 4q - 6pqr + 5pqr \\[1em] \Rightarrow (13 - 6 + 5)pqr + 2p + 4q \\[1em] \Rightarrow 12pqr + 2p + 4q\end{aligned}$$

Hence, the simplified expression = $\bm{12pqr + 2p + 4q}$.

Solving,

$$\begin{aligned}\Rightarrow 4ab + 0 - 2ba \\[1em] \Rightarrow (4 - 2)ab \\[1em] \Rightarrow 2ab\end{aligned}$$

Hence, the simplified expression = $\bm{2ab}$.

Solving,

$$\begin{array}{rrrr} &3a &+ 4b &+ 7c \\ &-5a &+ 3b &- 6c \\ + &4a &- 2b &- 4c \\ \hline &2a &+ 5b &- 3c \end{array}$$

Hence, the required sum = $\bm{2a + 5b - 3c}$.

Solving,

$$\begin{array}{rrrr} &2x^2 &+ xy &- y^2 \\ &-x^2 &+ 2xy &+ 3y^2 \\ + &3x^2 &- 10xy &+ 4y^2 \\ \hline &4x^2 &- 7xy &+ 6y^2 \end{array}$$

Hence, the required sum = $\bm{4x^2 - 7xy + 6y^2}$.

Solving,

$$\begin{array}{rrrr} &x^2 &- x &+ 1 \\ &-5x^2 &+ 2x &- 2 \\ + &3x^2 &- 3x &+ 1 \\ \hline &-x^2 &- 2x &+ 0 \end{array}$$

Hence, the required sum = $\bm{-x^2 - 2x}$.

Solving,

$$\begin{array}{rrrr} &a^2 &- ab &+ bc \\ &-2a^2 &+ 2ab &+ bc \\ + &3a^2 &+ ab &- 3bc \\ \hline &2a^2 &+ 2ab &- bc \end{array}$$

Hence, the required sum = $\bm{2a^2 + 2ab - bc}$.

Solving,

$$\begin{array}{rrrr} &4x^2 &- 3x &+ 7 \\ &-x^2 &+ 4x &+ 8 \\ + &-2x^2 &+ 5x &- 10 \\ \hline &x^2 &+ 6x &+ 5 \end{array}$$

Hence, the required sum = $\bm{x^2 + 6x + 5}$.

Solving,

$$\begin{array}{rrrr} &3x &+ 4xy &- y^2 \\ &-4x &+ xy &+ 2y^2 \\ + &6x &- xy &+ 3y^2 \\ \hline &5x &+ 4xy &+ 4y^2 \end{array}$$

Hence, the required sum = $\bm{5x + 4xy + 4y^2}$.

Solving,

$$\begin{array}{rrrr} &-17x^2 &- 2xy &+ 23y^2 \\ &15x^2 &+ 7xy &- 9y^2 \\ + &13x^2 &- 4xy &+ 3y^2 \\ \hline &11x^2 &+ xy &+ 17y^2 \end{array}$$

Hence, the required sum = $\bm{11x^2 + xy + 17y^2}$.

Solving,

$$\begin{array}{rrrrr} &-x^2 &- 3xy &+ 3y^2 &+ 8 \\ &3x^2 &+ 4xy &- 5y^2 &- 3 \\ + &2x^2 &- 6xy &+ y^2 &- 2 \\ \hline &4x^2 &- 5xy &- y^2 &+ 3 \end{array}$$

Hence, the required sum = $\bm{4x^2 - 5xy - y^2 + 3}$.

Solving,

$$\begin{array}{rrrrr} &a^3 &- 2b^3 &+ a & \\ &-2a^3 &+ b^3 & &+ b \\ + &4a^3 &+ 2b^3 &- 5a &- 2b \\ \hline &3a^3 &+ b^3 &- 4a &- b \end{array}$$

Hence, the required sum = $\bm{3a^3 + b^3 - 4a - b}$.

Solving,

$$\begin{array}{rrr} &+3a & \\ &+a &+ 2b \\ &- &- \\ \hline &+2a &- 2b \end{array}$$

Hence, the value of the given expression = $\bm{2a - 2b}$.

Solving,

$$\begin{array}{rrr} &+5x &- 3y \\ &+x &+ y \\ &- &- \\ \hline &+4x &- 4y \end{array}$$

Hence, the value of the given expression = $\bm{4x - 4y}$.

Solving,

$$\begin{array}{rrr} &+8a &+ 15b \\ &-7a &+ 3b \\ &+ &- \\ \hline &+15a &+ 12b \end{array}$$

Hence, the value of the given expression = $\bm{15a + 12b}$.

Solving,

$$\begin{array}{rrr} &+8x &+ 7y \\ &-3x &+ 4y \\ &+ &- \\ \hline &+11x &+ 3y \end{array}$$

Hence, the value of the given expression = $\bm{11x + 3y}$.

Solving,

$$\begin{aligned}\Rightarrow 7 - (4a - 5) \\[1em] \Rightarrow 7 - 4a + 5 \\[1em] \Rightarrow 12 - 4a.\end{aligned}$$

Hence, the value of the given expression = $\bm{12 - 4a}$.

Solving,

$$\begin{array}{rrrr} &+a &- 4b &- 2c \\ &+5a &- 3b &+ 2c \\ &- &+ &- \\ \hline &-4a &- b &- 4c \end{array}$$

Hence, the required difference = $\bm{-4a - b - 4c}$.

Solving,

$$\begin{array}{rrrr} &+12x &+ 7y &- 21z \\ &+4x &- 6y &+ 3z \\ &- &+ &- \\ \hline &+8x &+ 13y &- 24z \end{array}$$

Hence, the required difference = $\bm{8x + 13y - 24z}$.

Solving,

$$\begin{array}{rrrrr} &+5a &- 7b &+ 2c & \\ &-a &- 4b &+ 4c &+ 5 \\ &+ &+ &- &- \\ \hline &+6a &- 3b &- 2c &- 5 \end{array}$$

Hence, the required difference = $\bm{6a - 3b - 2c - 5}$.

Solving,

$$\begin{array}{rrrr} &+x &- y &- z \\ &-8x &- 12y &+ 17z \\ &+ &+ &- \\ \hline &+9x &+ 11y &- 18z \end{array}$$

Hence, the required difference = $\bm{9x + 11y - 18z}$.

Solving,

$$\begin{array}{rrrr} &+ab &- 2cd &+ 2ac &+ bd \\ &+2ab &+ cd &- ac &- 2bd \\ &- &- &+ &+ \\ \hline &-ab &- 3cd &+ 3ac &+ 3bd \end{array}$$

Hence, the required difference = $\bm{-ab - 3cd + 3ac + 3bd}$.

Solving,

$$\begin{array}{rrrr} &+ab &+ bc &- ca \\ &-ab &+ bc &- ca \\ &+ &- &+ \\ \hline &+2ab & & \end{array}$$

Hence, the required difference = $\bm{2ab}$.

Solving,

$$\begin{array}{rrrr} &+3x &+ 5y &- 4z \\ &+5x &+ 6y &- 3z \\ &- &- &+ \\ \hline &-2x &- y &- z \end{array}$$

Hence, the required difference = $\bm{-2x - y - z}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{1}{2}p - \frac{1}{3}q - \frac{3}{2}r\right) - \left(-\frac{3}{2}p + q - r\right) \\[1em] \Rightarrow \frac{1}{2}p - \frac{1}{3}q - \frac{3}{2}r + \frac{3}{2}p - q + r \\[1em] \Rightarrow \frac{1}{2}p + \frac{3}{2}p - \frac{1}{3}q - q - \frac{3}{2}r + r \\[1em] \Rightarrow \left(\frac{1}{2} + \frac{3}{2}\right)p + \left(-\frac{1}{3} - 1\right)q + \left(-\frac{3}{2} + 1\right)r \\[1em] \Rightarrow \frac{1 + 3}{2}p + \frac{-1 - 3}{3}q + \frac{-3 + 2}{2}r \\[1em] \Rightarrow \frac{4}{2}p - \frac{4}{3}q - \frac{1}{2}r \\[1em] \Rightarrow 2p - \frac{4}{3}q - \frac{1}{2}r.\end{aligned}$$

Hence, the required difference = $\bm{2p - \frac{4}{3}q - \frac{1}{2}r}$.

Solving,

$$\begin{array}{rrrr} &+a^2 &+ a &+ 1 \\ &+a^2 &- a &+ 1 \\ &- &+ &- \\ \hline & &+ 2a & \end{array}$$

Hence, the required difference = $\bm{2a}$.

Sum of $x + y - 2z$ and $2x - y + z$,

$$\begin{array}{rrrr} &x &+ y &- 2z \\ + &2x &- y &+ z \\ \hline &3x &0 &- z \end{array}$$

Now, subtracting $x + y + z$ from this sum,

$$\begin{array}{rrrr} &+3x &0 &- z \\ &+x &+ y &+ z \\ &- &- &- \\ \hline &+2x &- y &- 2z \end{array}$$

Hence, the required difference = $\bm{2x - y - 2z}$.

Sum of $3a - 2b + 4c$ and $3b - 2c$,

$$\begin{array}{rrrr} &3a &- 2b &+ 4c \\ + & &+ 3b &- 2c \\ \hline &3a &+ b &+ 2c \end{array}$$

Now, subtracting $a - b - c$ from this sum,

$$\begin{array}{rrrr} &+3a &+ b &+ 2c \\ &+a &- b &- c \\ &- &+ &+ \\ \hline &+2a &+ 2b &+ 3c \end{array}$$

Hence, the required difference = $\bm{2a + 2b + 3c}$.

Sum of $3x - y + z$ and $x + y - 3z$,

$$\begin{array}{rrrr} &3x &- y &+ z \\ + &x &+ y &- 3z \\ \hline &4x &0 &- 2z \end{array}$$

Now, subtracting $x - 2y - z$ from this sum,

$$\begin{array}{rrrr} &+4x &0 &- 2z \\ &+x &- 2y &- z \\ &- &+ &+ \\ \hline &+3x &+ 2y &- z \end{array}$$

Hence, the required difference = $\bm{3x + 2y - z}$.

Sum of $x - 2z$ and $x + y + z$,

$$\begin{array}{rrrr} &x & &- 2z \\ + &x &+ y &+ z \\ \hline &2x &+ y &- z \end{array}$$

Sum of $x + y$ and $x - z$,

$$\begin{array}{rrrr} &x &+ y & \\ + &x & &- z \\ \hline &2x &+ y &- z \end{array}$$

Now, subtracting the second sum from the first sum,

$$\begin{array}{rrrr} &+2x &+ y &- z \\ &+2x &+ y &- z \\ &- &- &+ \\ \hline &0 &0 &0 \end{array}$$

Hence, the required difference = 0.

The product of the given monomials = (product of their numeral coefficients) × (product of their literals), and in multiplication the powers of like factors are added.

(i) $6 \times 3 =$ 18 and $6x \times 3x = (6 \times 3)(x \times x) = \bm{18x^2}$.

(ii) $6 \times 3 =$ 18 and $6x^2 \times 3x^3 = (6 \times 3)(x^{2+3}) = \bm{18x^5}$.

(iii) $5 \times 4 =$ 20 and $5x \times 4y = (5 \times 4)(x \times y) = \bm{20xy}$.

(iv) $4 \times 7 =$ 28 and $4ax \times 7x = (4 \times 7)(a \times x^{1+1}) = \bm{28ax^2}$.

(v) $6 \times 2 =$ 12 and $6xy \times 2xy = (6 \times 2)(x^{1+1} \times y^{1+1}) = \bm{12x^2y^2}$.

(vi) $12 \times 4 =$ 48 and $12ax^2 \times 4ax = (12 \times 4)(a^{1+1} \times x^{2+1}) = \bm{48a^2x^3}$.

(vii) $1 \times 8 =$ 8 and $a^2xy^2 \times 8a^3x^2y = (1 \times 8)(a^{2+3} \times x^{1+2} \times y^{2+1}) = \bm{8a^5x^3y^3}$.

(viii) $15 \times 3 =$ 45 and $15x \times 3x^5y^2 = (15 \times 3)(x^{1+5} \times y^2) = \bm{45x^6y^2}$.

The product of the given monomials = (product of their numeral coefficients) × (product of their literals), and in multiplication the powers of like factors are added.

(i) $4x \times 6x \times 2 = (4 \times 6 \times 2)(x^{1+1}) = \bm{48x^2}$.

(ii) $3ab \times 6ax = (3 \times 6)(a^{1+1} \times b \times x) = \bm{18a^2bx}$.

(iii) $x \times 2x^2 \times 3x^3 = (1 \times 2 \times 3)(x^{1+2+3}) = \bm{6x^6}$.

(iv) $5 \times 5a^3 = (5 \times 5)a^3 = \bm{25a^3}$.

(v) $6 \times 6x^2 \times 6x^2y^2 = (6 \times 6 \times 6)(x^{2+2} \times y^2) = \bm{216x^4y^2}$.

(vi) $-8x \times -3x = (-8 \times -3)(x^{1+1}) = \bm{24x^2}$.

(vii) $-5 \times -3x \times 5x^2 = (-5 \times -3 \times 5)(x^{1+2}) = \bm{75x^3}$.

(viii) $8 \times -4xy^2 \times 3x^3y^2 = (8 \times -4 \times 3)(x^{1+3} \times y^{2+2}) = \bm{-96x^4y^4}$.

(ix) $-4x \times 5xy \times 3z = (-4 \times 5 \times 3)(x^{1+1} \times y \times z) = \bm{-60x^2yz}$.

(x) $5x \times 2x^2y \times -7y^3 \times 2x^3y^2 = (5 \times 2 \times -7 \times 2)(x^{1+2+3} \times y^{1+3+2}) = \bm{-140x^6y^6}$.

Solving,

$$\begin{aligned}\Rightarrow 3x^3 \times 5x^4 \\[1em] \Rightarrow (3 \times 5)(x^{3+4}) \\[1em] \Rightarrow 15x^7\end{aligned}$$

Hence, $\bm{3x^3 \times 5x^4 = 15x^7}$.

Solving,

$$\begin{aligned}\Rightarrow 5a^2 \times 7a^7 \\[1em] \Rightarrow (5 \times 7)(a^{2+7}) \\[1em] \Rightarrow 35a^9\end{aligned}$$

Hence, $\bm{5a^2 \times 7a^7 = 35a^9}$.

Solving,

$$\begin{aligned}\Rightarrow 3abc \times 6ac^3 \\[1em] \Rightarrow (3 \times 6)(a^{1+1} \times b \times c^{1+3}) \\[1em] \Rightarrow 18a^2bc^4\end{aligned}$$

Hence, $\bm{3abc \times 6ac^3 = 18a^2bc^4}$.

Solving,

$$\begin{aligned}\Rightarrow a^2b^2 \times 5a^3b^4 \\[1em] \Rightarrow (1 \times 5)(a^{2+3} \times b^{2+4}) \\[1em] \Rightarrow 5a^5b^6\end{aligned}$$

Hence, $\bm{a^2b^2 \times 5a^3b^4 = 5a^5b^6}$.

Solving,

$$\begin{aligned}\Rightarrow 2x^2y^3 \times 5x^3y^4 \\[1em] \Rightarrow (2 \times 5)(x^{2+3} \times y^{3+4}) \\[1em] \Rightarrow 10x^5y^7\end{aligned}$$

Hence, $\bm{2x^2y^3 \times 5x^3y^4 = 10x^5y^7}$.

Solving,

$$\begin{aligned}\Rightarrow abc \times bcd \\[1em] \Rightarrow a \times b^{1+1} \times c^{1+1} \times d \\[1em] \Rightarrow ab^2c^2d\end{aligned}$$

Hence, $\bm{abc \times bcd = ab^2c^2d}$.

Solving,

$$\begin{array}{rrrr} &a &+& b \\ \times & &&ab \\ \hline &a^2b &+& ab^2 \end{array}$$

Hence, $\bm{(a + b) \times ab = a^2b + ab^2}$.

Solving,

$$\begin{array}{rrrr} &3ab &-& 4b \\ \times & &&3ab \\ \hline &9a^2b^2 &-& 12ab^2 \end{array}$$

Hence, $\bm{(3ab - 4b) \times 3ab = 9a^2b^2 - 12ab^2}$.

Solving,

$$\begin{array}{rrrr} &2xy &-& 5by \\ \times & &&4bx \\ \hline &8bx^2y &-& 20b^2xy \end{array}$$

Hence, $\bm{(2xy - 5by) \times 4bx = 8bx^2y - 20b^2xy}$.

Solving,

$$\begin{array}{rrrr} &4x &+& 2y \\ \times & &&3xy \\ \hline &12x^2y &+& 6xy^2 \end{array}$$

Hence, $\bm{(4x + 2y) \times 3xy = 12x^2y + 6xy^2}$.

Solving,

$$\begin{array}{rrrr} &x^2 &-& x \\ \times & &&2x \\ \hline &2x^3 &-& 2x^2 \end{array}$$

Hence, $\bm{(x^2 - x) \times 2x = 2x^3 - 2x^2}$.

Solving,

$$\begin{array}{rrrr} &1 &+& 4x \\ \times & &&x \\ \hline &x &+& 4x^2 \end{array}$$

Hence, $\bm{(1 + 4x) \times x = x + 4x^2}$.

Solving,

$$\begin{array}{rrrr} &9xy^2 &+& 3x^2y \\ \times & &&5xy \\ \hline &45x^2y^3 &+& 15x^3y^2 \end{array}$$

Hence, $\bm{(9xy^2 + 3x^2y) \times 5xy = 45x^2y^3 + 15x^3y^2}$.

Solving,

$$\begin{array}{rrrr} &6x &-& 5y \\ \times & &&3axy \\ \hline &18ax^2y &-& 15axy^2 \end{array}$$

Hence, $\bm{(6x - 5y) \times 3axy = 18ax^2y - 15axy^2}$.

Solving,

$$\begin{array}{rrrrrr} &-x &+& y &-& z \\ \times & &&&&-2x \\ \hline &2x^2 &-& 2xy &+& 2xz \end{array}$$

Hence, $\bm{(-x + y - z) \times (-2x) = 2x^2 - 2xy + 2xz}$.

Solving,

$$\begin{array}{rrrr} &xy &-& yz \\ \times & &&x^2yz^2 \\ \hline &x^3y^2z^2 &-& x^2y^2z^3 \end{array}$$

Hence, $\bm{(xy - yz) \times x^2yz^2 = x^3y^2z^2 - x^2y^2z^3}$.

Solving,

$$\begin{array}{rrrr} &2xyz &+& 3xy \\ \times & &&-2y^2z \\ \hline &-4xy^3z^2 &-& 6xy^3z \end{array}$$

Hence, $\bm{(2xyz + 3xy) \times (-2y^2z) = -4xy^3z^2 - 6xy^3z}$.

Solving,

$$\begin{array}{rrrr} &-3xy^2 &+& 4x^2y \\ \times & &&-xy \\ \hline &3x^2y^3 &-& 4x^3y^2 \end{array}$$

Hence, $\bm{(-3xy^2 + 4x^2y) \times (-xy) = 3x^2y^3 - 4x^3y^2}$.

Solving,

$$\begin{array}{rrrr} &-x^2y &-& 3x^2y^2 \\ \times & &&4xy \\ \hline &-4x^3y^2 &-& 12x^3y^3 \end{array}$$

Hence, $\bm{4xy \times (-x^2y - 3x^2y^2) = -4x^3y^2 - 12x^3y^3}$.

Solving,

$$\begin{array}{rrrrrr} &3a &+& 4b &-& 5c \\ \times & &&&&3a \\ \hline &9a^2 &+& 12ab &-& 15ac \end{array}$$

Hence, $\bm{(3a + 4b - 5c) \times 3a = 9a^2 + 12ab - 15ac}$.

Solving,

$$\begin{array}{rrrr} &-xy^2 &-& 6x^2y \\ \times & &&-5xy \\ \hline &5x^2y^3 &+& 30x^3y^2 \end{array}$$

Hence, $\bm{-5xy \times (-xy^2 - 6x^2y) = 5x^2y^3 + 30x^3y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (x + 2)(x + 10) \\[1em] \Rightarrow x^2 + 10x + 2x + 20 \\[1em] \Rightarrow x^2 + 12x + 20.\end{aligned}$$

Hence, $\bm{(x + 2)(x + 10) = x^2 + 12x + 20}$.

Solving,

$$\begin{aligned}\Rightarrow (x + 5)(x - 3) \\[1em] \Rightarrow x^2 - 3x + 5x - 15 \\[1em] \Rightarrow x^2 + 2x - 15.\end{aligned}$$

Hence, $\bm{(x + 5)(x - 3) = x^2 + 2x - 15}$.

Solving,

$$\begin{aligned}\Rightarrow (x - 5)(x + 3) \\[1em] \Rightarrow x^2 + 3x - 5x - 15 \\[1em] \Rightarrow x^2 - 2x - 15.\end{aligned}$$

Hence, $\bm{(x - 5)(x + 3) = x^2 - 2x - 15}$.

Solving,

$$\begin{aligned}\Rightarrow (x - 5)(x - 3) \\[1em] \Rightarrow x^2 - 3x - 5x + 15 \\[1em] \Rightarrow x^2 - 8x + 15.\end{aligned}$$

Hence, $\bm{(x - 5)(x - 3) = x^2 - 8x + 15}$.

Solving,

$$\begin{aligned}\Rightarrow (2x + y)(x + 3y) \\[1em] \Rightarrow 2x^2 + 6xy + xy + 3y^2 \\[1em] \Rightarrow 2x^2 + 7xy + 3y^2.\end{aligned}$$

Hence, $\bm{(2x + y)(x + 3y) = 2x^2 + 7xy + 3y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (3x - 5y)(x + 6y) \\[1em] \Rightarrow 3x^2 + 18xy - 5xy - 30y^2 \\[1em] \Rightarrow 3x^2 + 13xy - 30y^2.\end{aligned}$$

Hence, $\bm{(3x - 5y)(x + 6y) = 3x^2 + 13xy - 30y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (x + 9y)(x - 5y) \\[1em] \Rightarrow x^2 - 5xy + 9xy - 45y^2 \\[1em] \Rightarrow x^2 + 4xy - 45y^2.\end{aligned}$$

Hence, $\bm{(x + 9y)(x - 5y) = x^2 + 4xy - 45y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2x + 5y)(2x + 5y) \\[1em] \Rightarrow 4x^2 + 10xy + 10xy + 25y^2 \\[1em] \Rightarrow 4x^2 + 20xy + 25y^2.\end{aligned}$$

Hence, $\bm{(2x + 5y)(2x + 5y) = 4x^2 + 20xy + 25y^2}$.

Solving,

$$\begin{array}{rr} &3abc \\ \times &-5a^2b^2c \\ \hline &-15a^3b^3c^2 \end{array}$$

Hence, $\bm{3abc \times (-5a^2b^2c) = -15a^3b^3c^2}$.

Solving,

$$\begin{array}{rrrrrr} &x &-& y &+& z \\ \times & &&&&-2x \\ \hline &-2x^2 &+& 2xy &-& 2xz \end{array}$$

Hence, $\bm{(x - y + z) \times (-2x) = -2x^2 + 2xy - 2xz}$.

Solving,

$$\begin{array}{rrrrrr} &2x &-& 3y &-& 5z \\ \times & &&&&-2y \\ \hline &-4xy &+& 6y^2 &+& 10yz \end{array}$$

Hence, $\bm{(2x - 3y - 5z) \times (-2y) = -4xy + 6y^2 + 10yz}$.

Solving,

$$\begin{array}{rrrr} &-8xyz &+& 10x^2yz^3 \\ \times & &&xyz \\ \hline &-8x^2y^2z^2 &+& 10x^3y^2z^4 \end{array}$$

Hence, $\bm{(-8xyz + 10x^2yz^3) \times xyz = -8x^2y^2z^2 + 10x^3y^2z^4}$.

Solving,

$$\begin{array}{rrrrrr} &-13xy^2z &+& 15x^2yz &-& 6xyz^2 \\ \times & &&&&xyz \\ \hline &-13x^2y^3z^2 &+& 15x^3y^2z^2 &-& 6x^2y^2z^3 \end{array}$$

Hence, $\bm{xyz \times (-13xy^2z + 15x^2yz - 6xyz^2) = -13x^2y^3z^2 + 15x^3y^2z^2 - 6x^2y^2z^3}$.

Solving,

$$\begin{array}{rrrrrr} &4abc &-& 5a^2bc &-& 6ab^2c \\ \times & &&&&-2abc^2 \\ \hline &-8a^2b^2c^3 &+& 10a^3b^2c^3 &+& 12a^2b^3c^3 \end{array}$$

Hence, $\bm{(4abc - 5a^2bc - 6ab^2c) \times (-2abc^2) = -8a^2b^2c^3 + 10a^3b^2c^3 + 12a^2b^3c^3}$.

Solving,

$$\begin{aligned}\Rightarrow (xy - ab)(xy + ab) \\[1em] \Rightarrow xy \times xy + xy \times ab - ab \times xy - ab \times ab \\[1em] \Rightarrow x^2y^2 + abxy - abxy - a^2b^2 \\[1em] \Rightarrow x^2y^2 - a^2b^2.\end{aligned}$$

Hence, $\bm{(xy - ab)(xy + ab) = x^2y^2 - a^2b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2abc - 3xy)(2abc + 3xy) \\[1em] \Rightarrow 2abc \times 2abc + 2abc \times 3xy - 3xy \times 2abc - 3xy \times 3xy \\[1em] \Rightarrow 4a^2b^2c^2 + 6abcxy - 6abcxy - 9x^2y^2 \\[1em] \Rightarrow 4a^2b^2c^2 - 9x^2y^2.\end{aligned}$$

Hence, $\bm{(2abc - 3xy)(2abc + 3xy) = 4a^2b^2c^2 - 9x^2y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (a + b - c)(2a - 3b) \\[1em] \Rightarrow a(2a - 3b) + b(2a - 3b) - c(2a - 3b) \\[1em] \Rightarrow 2a^2 - 3ab + 2ab - 3b^2 - 2ac + 3bc \\[1em] \Rightarrow 2a^2 - ab - 2ac - 3b^2 + 3bc.\end{aligned}$$

Hence, $\bm{(a + b - c)(2a - 3b) = 2a^2 - ab - 2ac - 3b^2 + 3bc}$.

Solving,

$$\begin{aligned}\Rightarrow (5x - 6y - 7z)(2x + 3y) \\[1em] \Rightarrow 5x(2x + 3y) - 6y(2x + 3y) - 7z(2x + 3y) \\[1em] \Rightarrow 10x^2 + 15xy - 12xy - 18y^2 - 14xz - 21yz \\[1em] \Rightarrow 10x^2 + 3xy - 14xz - 18y^2 - 21yz.\end{aligned}$$

Hence, $\bm{(5x - 6y - 7z)(2x + 3y) = 10x^2 + 3xy - 14xz - 18y^2 - 21yz}$.

Solving,

$$\begin{aligned}\Rightarrow (5x - 6y - 7z)(2x + 3y + z) \\[1em] \Rightarrow 5x(2x + 3y + z) - 6y(2x + 3y + z) - 7z(2x + 3y + z) \\[1em] \Rightarrow 10x^2 + 15xy + 5zx - 12xy - 18y^2 - 6yz - 14zx - 21yz - 7z^2 \\[1em] \Rightarrow 10x^2 + 3xy - 9zx - 18y^2 - 27yz - 7z^2.\end{aligned}$$

Hence, $\bm{(5x - 6y - 7z)(2x + 3y + z) = 10x^2 + 3xy - 9zx - 18y^2 - 27yz - 7z^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2a + 3b - 4c)(a - b - c) \\[1em] \Rightarrow 2a(a - b - c) + 3b(a - b - c) - 4c(a - b - c) \\[1em] \Rightarrow 2a^2 - 2ab - 2ac + 3ab - 3b^2 - 3bc - 4ac + 4bc + 4c^2 \\[1em] \Rightarrow 2a^2 + ab - 6ac - 3b^2 + bc + 4c^2.\end{aligned}$$

Hence, $\bm{(2a + 3b - 4c)(a - b - c) = 2a^2 + ab - 6ac - 3b^2 + bc + 4c^2}$.

Solving,

$\Rightarrow \frac{3a}{a}$

$\Rightarrow 3$

Hence, $\bm{\frac{3a}{a} = 3}$.

Solving,

$\Rightarrow \frac{15x}{3x}$

$\Rightarrow 5$

Hence, $\bm{\frac{15x}{3x} = 5}$.

Solving,

$\Rightarrow \frac{16m}{4}$

$\Rightarrow 4m$

Hence, $\bm{\frac{16m}{4} = 4m}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{20x^2}{5x} \\[1em] \Rightarrow 4x^{2-1} \\[1em] \Rightarrow 4x\end{aligned}$$

Hence, $\bm{\frac{20x^2}{5x} = 4x}$.

Solving,

$\Rightarrow \frac{30p^2}{10p^2}$

$\Rightarrow 3$

Hence, $\bm{\frac{30p^2}{10p^2} = 3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{14a^3b^3}{2a^2} \\[1em] \Rightarrow 7a^{3-2}b^3 \\[1em] \Rightarrow 7ab^3\end{aligned}$$

Hence, $\bm{\frac{14a^3b^3}{2a^2} = 7ab^3}$.

Solving,

$\Rightarrow \frac{18pqr^2}{3pq}$

$\Rightarrow 6r^2$

Hence, $\bm{\frac{18pqr^2}{3pq} = 6r^2}$.

Solving,

$\Rightarrow \frac{100}{50b} = \frac{2}{b}$

Hence, $\bm{\frac{100}{50b} = \frac{2}{b}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x^5}{x^2} \\[1em] \Rightarrow 2x^{5-2} \\[1em] \Rightarrow 2x^3\end{aligned}$$

Hence, $\bm{2x^5 \div x^2 = 2x^3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{6a^8}{3a^3} \\[1em] \Rightarrow 2a^{8-3} \\[1em] \Rightarrow 2a^5\end{aligned}$$

Hence, $\bm{6a^8 \div 3a^3 = 2a^5}$.

Solving,

$\Rightarrow \frac{20xy}{-5xy}$

$\Rightarrow -4$

Hence, $\bm{20xy \div -5xy = -4}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-24a^2b^2c^2}{6ab} \\[1em] \Rightarrow -4a^{2-1}b^{2-1}c^2 \\[1em] \Rightarrow -4abc^2\end{aligned}$$

Hence, $\bm{-24a^2b^2c^2 \div 6ab = -4abc^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-5x^2y}{xy^2} \\[1em] \Rightarrow \frac{-5x^{2-1}}{y^{2-1}}\\[1em] \Rightarrow \frac{-5x}{y}\end{aligned}$$

Hence, $\bm{-5x^2y \div xy^2 = \frac{-5x}{y}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{40p^3q^4r^5}{10p^3q} \\[1em] \Rightarrow 4q^{4-1}r^5 \\[1em] \Rightarrow 4q^3r^5\end{aligned}$$

Hence, $\bm{40p^3q^4r^5 \div 10p^3q = 4q^3r^5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-64x^4y^3z}{4x^3y^2z} \\[1em] \Rightarrow -16x^{4-3}y^{3-2} \\[1em] \Rightarrow -16xy\end{aligned}$$

Hence, $\bm{-64x^4y^3z \div 4x^3y^2z = -16xy}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{35xy^5}{7x^2y^4} \\[1em] \Rightarrow \frac{5y^{5-4}}{x^{2-1}} \\[1em] \Rightarrow \frac{5y}{x}\end{aligned}$$

Hence, $\bm{35xy^5 \div 7x^2y^4 = \frac{5y}{x}}$.

Solving,

$$\begin{aligned}\Rightarrow -\frac{3m}{4} \div 2m \\[1em] \Rightarrow -\frac{3m}{4} \times \frac{1}{2m} \\[1em] \Rightarrow \frac{-3m}{8m} \\[1em] \Rightarrow \frac{-3}{8}\end{aligned}$$

Hence, $\bm{-\frac{3m}{4} \div 2m = \frac{-3}{8}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-15p^6q^8}{-5p^6q^7} \\[1em] \Rightarrow 3q^{8-7} \\[1em] \Rightarrow 3q\end{aligned}$$

Hence, $\bm{\frac{-15p^6q^8}{-5p^6q^7} = 3q}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-21m^5n^7}{14m^2n^2} \\[1em] \Rightarrow \frac{-3}{2}m^{5-2}n^{7-2} \\[1em] \Rightarrow \frac{-3}{2}m^3n^5\end{aligned}$$

Hence, $\bm{\frac{-21m^5n^7}{14m^2n^2} = \frac{-3}{2}m^3n^5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{36a^4x^5y^6}{4x^2a^3y^2} \\[1em] \Rightarrow 9a^{4-3}x^{5-2}y^{6-2} \\[1em] \Rightarrow 9ax^3y^4\end{aligned}$$

Hence, $\bm{\frac{36a^4x^5y^6}{4x^2a^3y^2} = 9ax^3y^4}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{20x^3a^6}{5xy} \\[1em] \Rightarrow \frac{4x^{3-1}a^6}{y} \\[1em] \Rightarrow \frac{4x^2a^6}{y}\end{aligned}$$

Hence, $\bm{\frac{20x^3a^6}{5xy} = \frac{4x^2a^6}{y}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{28a^2b^3}{c^2} \div 4abc \\[1em] \Rightarrow \frac{28a^2b^3}{c^2} \times \frac{1}{4abc} \\[1em] \Rightarrow \frac{28a^2b^3}{4abc^3} \\[1em] \Rightarrow \frac{7ab^2}{c^3}\end{aligned}$$

Hence, $\bm{\frac{28a^2b^3}{c^2} \div 4abc = \frac{7ab^2}{c^3}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2a^2}{9b^2} \div \frac{3b}{2a} \\[1em] \Rightarrow \frac{2a^2}{9b^2} \times \frac{2a}{3b} \\[1em] \Rightarrow \frac{4a^3}{27b^3}\end{aligned}$$

Hence, $\bm{\frac{2a^2}{9b^2} \div \frac{3b}{2a} = \frac{4a^3}{27b^3}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-5.5x^2}{y} \div \frac{11x}{y} \\[1em] \Rightarrow \frac{-5.5x^2}{y} \times \frac{y}{11x} \\[1em] \Rightarrow \frac{-5.5x^2y}{11xy} \\[1em] \Rightarrow \frac{-5.5x}{11} \\[1em] \Rightarrow -\frac{x}{2} \\[1em] \Rightarrow -0.5x\end{aligned}$$

Hence, $\bm{\frac{-5.5x^2}{y} \div \frac{11x}{y} = -\frac{x}{2} = -0.5x}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{64x^2y^2}{z^2} \div \frac{8xy}{z} \\[1em] \Rightarrow \frac{64x^2y^2}{z^2} \times \frac{z}{8xy} \\[1em] \Rightarrow \frac{64x^2y^2z}{8xyz^2} \\[1em] \Rightarrow \frac{8xy}{z}\end{aligned}$$

Hence, $\bm{\frac{64x^2y^2}{z^2} \div \frac{8xy}{z} = \frac{8xy}{z}}$.

Solving,

$\Rightarrow \frac{-15m^5n^2}{-3m^5}$

$\Rightarrow 5n^2$

Hence, $\bm{\frac{-15m^5n^2}{-3m^5} = 5n^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{35x^4y^2}{-15x^2y^2} \\[1em] \Rightarrow \frac{-7x^{4-2}}{3} \\[1em] \Rightarrow \frac{-7x^2}{3}\end{aligned}$$

Hence, $\bm{\frac{35x^4y^2}{-15x^2y^2} = \frac{-7x^2}{3}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-24x^6y^2}{6x^6y} \\[1em] \Rightarrow -4y^{2-1} \\[1em] \Rightarrow -4y\end{aligned}$$

Hence, $\bm{\frac{-24x^6y^2}{6x^6y} = -4y}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{9x^3 - 6x^2}{3x} \\[1em] \Rightarrow \frac{9x^3}{3x} - \frac{6x^2}{3x} \\[1em] \Rightarrow 3x^2 - 2x\end{aligned}$$

Hence, $\bm{\frac{9x^3 - 6x^2}{3x} = 3x^2 - 2x}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{6m^2 - 16m^3 + 10m^4}{-2m} \\[1em] \Rightarrow \frac{6m^2}{-2m} - \frac{16m^3}{-2m} + \frac{10m^4}{-2m} \\[1em] \Rightarrow -3m - (-8m^2) + (-5m^3) \\[1em] \Rightarrow -3m + 8m^2 - 5m^3\end{aligned}$$

Hence, $\bm{\frac{6m^2 - 16m^3 + 10m^4}{-2m} = -3m + 8m^2 - 5m^3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{15x^3y^2 + 25x^2y^3 - 36x^4y^4}{5x^2y^2} \\[1em] \Rightarrow \frac{15x^3y^2}{5x^2y^2} + \frac{25x^2y^3}{5x^2y^2} - \frac{36x^4y^4}{5x^2y^2} \\[1em] \Rightarrow 3x + 5y - \frac{36}{5}x^2y^2\end{aligned}$$

Hence, $\bm{\frac{15x^3y^2 + 25x^2y^3 - 36x^4y^4}{5x^2y^2} = 3x + 5y - \frac{36}{5}x^2y^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{36a^3x^5 - 24a^4x^4 + 18a^5x^3}{-6a^3x^3} \\[1em] \Rightarrow \frac{36a^3x^5}{-6a^3x^3} - \frac{24a^4x^4}{-6a^3x^3} + \frac{18a^5x^3}{-6a^3x^3} \\[1em] \Rightarrow -6x^2 - (-4ax) + (-3a^2) \\[1em] \Rightarrow -6x^2 + 4ax - 3a^2\end{aligned}$$

Hence, $\bm{\frac{36a^3x^5 - 24a^4x^4 + 18a^5x^3}{-6a^3x^3} = -6x^2 + 4ax - 3a^2}$.

Any factor or group of factors of a product is the coefficient of the remaining factors.

$-xyz = (-yz) \times x$, so the coefficient of $x$ in $-xyz$ is $-yz$.

Hence, option 2 is the correct option.

If a factor of a product is a numerical quantity, it is called the numeral coefficient of the remaining factors.

Solving,

$$\begin{aligned}\Rightarrow -8x \div 7z \\[1em] \Rightarrow \frac{-8x}{7z} \\[1em] \Rightarrow \left(\frac{-8}{7}\right) \times \frac{x}{z}\end{aligned}$$

So, the numeral coefficient is $-\frac{8}{7}$.

Hence, option 3 is the correct option.

The degree of a polynomial is the greatest of the exponents (powers) of its various terms.

Degrees of the terms are 1, (1 + 3) = 4 and (1 + 1) = 2. The greatest is 4.

Hence, option 4 is the correct option.

An expression with only one term is a monomial. Each of $x^3y, xy$ and $25yx^3$ has one term, whereas $5 + y^2x$ has two terms, so it is a binomial and not a monomial.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4(x^2 + 1) + 5(x + 1) + 5 \\[1em] \Rightarrow 4x^2 + 4 + 5x + 5 + 5 \\[1em] \Rightarrow 4x^2 + 5x + 14\end{aligned}$$

The term which does not contain any variable is the constant term, i.e. 14.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 5x + 5 \div x + x \\[1em] \Rightarrow 5x + \frac{5}{x} + x \\[1em] \Rightarrow (5 + 1)x + \frac{5}{x} \\[1em] \Rightarrow 6x + \frac{5}{x}\end{aligned}$$

On adding the like terms $5x$ and $x$, the expression has the two terms $6x$ and $\frac{5}{x}$.

Hence, option 1 is the correct option.

The given expression has the four terms $3x^2, -\frac{2}{x}, \frac{5}{x^2}$ and 7. An expression with two or more than two terms is a multinomial.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4x - y + 6z - (3x - 4y + 5z) \\[1em] \Rightarrow 4x - y + 6z - 3x + 4y - 5z \\[1em] \Rightarrow (4 - 3)x + (-1 + 4)y + (6 - 5)z \\[1em] \Rightarrow x + 3y + z\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (x + y) - (y + z) + (z + x) \\[1em] \Rightarrow x + y - y - z + z + x \\[1em] \Rightarrow (1 + 1)x + (1 - 1)y + (-1 + 1)z \\[1em] \Rightarrow 2x\end{aligned}$$

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (3x^2y) \times (50xy^2) \div (25x^2y^2) \\[1em] \Rightarrow \frac{(3 \times 50)(x^{2+1} \times y^{1+2})}{25x^2y^2} \\[1em] \Rightarrow \frac{150x^3y^3}{25x^2y^2} \\[1em] \Rightarrow 6x^{3-2}y^{3-2} \\[1em] \Rightarrow 6xy\end{aligned}$$

Hence, option 2 is the correct option.

Mother's age = 3 years more than thrice your age = $3 \times$ your age + 3

$\Rightarrow x = 3y + 3$

So, statement 1 is true.

The mother's age is 3 years more than 3 times your age, not 3 years less.

So, statement 2 is false.

Hence, option 3 is the correct option.

In $2x - 5y + 8z$, the terms $2x, -5y$ and $8z$ are unlike terms, so they cannot be combined into a single term. Thus, $2x - 5y + 8z \neq 5xyz$ and Assertion (A) is false.

As only like terms (terms having the same literal coefficients) can be added or subtracted to get a single term, and so $2x - 5y + 8z$ cannot be added to get a single term. So, Reason (R) is true.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (5x^2 - 7x) - (x^2 + x) + (-3x^2 + 5x) \\[1em] \Rightarrow 5x^2 - 7x - x^2 - x - 3x^2 + 5x \\[1em] \Rightarrow (5 - 1 - 3)x^2 + (-7 - 1 + 5)x \\[1em] \Rightarrow x^2 - 3x\end{aligned}$$

So, Assertion (A) is true.

The steps shown in Reason (R) are correct, so Reason (R) is true.

Hence, option 3 is the correct option.