GSEB Class 7 Maths Solutions Chapter 12 બીજગણિતીય પદાવલિ Exercise 12.2

Chapter 12 બીજગણિતીય પદાવલિ Ex 12.2

1. સજાતીય પદ સાથે ગોઠવી સાદું રૂપ આપોઃ

Question (i) 21b - 32 + 7b - 20b
Answer:
\( = 21b + 7b - 20b - 32 \)
\( = (21 + 7 - 20)b - 32 \)
\( = (8)b - 32 \)
\( = 8b - 32 \)
In simple words: First, we group all the similar terms that have 'b'. Then, we do the math with the numbers in front of 'b'. Finally, we write down the simplified expression.

Exam Tip: To simplify expressions, always combine like terms by adding or subtracting their coefficients while keeping the variable part unchanged. Constants are grouped separately.

Question (ii) -z² + 13z² - 5z + 7z³ – 15z
Answer:
\( = 7z^3 - z^2 + 13z^2 - 5z - 15z \)
\( = 7z^3 + (- 1 + 13)z^2 + (-5 - 15) z \)
\( = 7z^3 + (12)z^2 + (-20)z \)
\( = 7z^3 + 12z^2 - 20z \)
In simple words: We first put the terms in order, starting with the highest power of 'z'. Then, we gather all the terms that have the same power of 'z' and combine their numbers. This gives us a shorter, easier expression.

Exam Tip: When simplifying, arrange terms in descending order of their variable's power (e.g., \( z^3 \), then \( z^2 \), then \( z \)). This helps in clear organization and reduces calculation errors.

Question (iii) p - (p - q) - q -(q – p)
Answer:
\( = p - p + q - q - q + p \)
\( = p - p + p + q - q - q \)
\( = (1 - 1 + 1) p + (1 - 1 - 1) q \)
\( = (1)p + (-1)q \)
\( = p - q \)
In simple words: First, we remove all the brackets, being careful with the minus signs. Then, we gather all the 'p' terms together and all the 'q' terms together. After adding and subtracting their coefficients, we get the final, simpler expression.

Exam Tip: Always pay close attention to negative signs in front of parentheses; they reverse the sign of every term inside when the parentheses are removed.

Question (iv) 3a - 2b - ab – (a – b + ab) + 3ab + b - a
Answer:
\( = 3a - 2b - ab - a + b - ab + 3ab + b - a \)
\( = (3a - a - a) + (- 2b + b + b) + (- ab - ab + 3ab) \)
\( = (3 - 1 - 1) a + (- 2 + 1 + 1) b + (- 1 - 1 + 3)ab \)
\( = (1)a + (0)b + (1)ab \)
\( = a + ab \)
In simple words: First, we get rid of the parentheses, remembering to change signs if a minus is outside. Then, we gather and combine all the terms with 'a', all the terms with 'b', and all the terms with 'ab' separately. This process simplifies the long expression.

Exam Tip: Be methodical: first distribute signs into parentheses, then identify and group like terms, and finally perform addition/subtraction on the coefficients of each group.

Question (v) 5x²y - 5x² + 3yx² - 3y² + x² − y² + 8xy² - 3x²y²
Answer:
\( = (5x^2y + 3yx^2) + (8xy^2) + (- 5x^2 + x^2) + (-3y^2 – y^2) - 3x^2y^2 \)
\( = (5 + 3)x^2y + (8)xy^2 + (- 5 + 1)x^2 + (- 3 - 1)y^2 - 3x^2y^2 \)
\( = (8)x^2y + (8)xy^2 + (-4)x^2 + (-7)y^2 - 3x^2y^2 \)
\( = 8x^2y + 8xy^2 - 4x^2 - 7y^2 - 3x^2y^2 \)
In simple words: We start by finding all the terms that are exactly alike, such as \( x^2y \) or \( y^2 \). We then group these similar terms together and add or subtract their numerical parts. This helps to reduce the expression to its simplest form.

Exam Tip: Like terms must have the exact same variables raised to the exact same powers. For example, \( x^2y \) is different from \( xy^2 \) and \( x^2 \).

Question (vi) (3y² + 5y – 4) – (8y – y² – 4)
Answer:
Here, we will first bring like terms together in each case. Then we will perform addition/subtraction operations.
\( = 3y^2 + 5y – 4 – 8y + y^2 + 4 \)
\( = (3y^2 + y^2) + (5y – 8y) + (- 4 + 4) \)
\( = (3 + 1)y^2 + (5 – 8)y + (- 4 + 4) \)
\( = (4)y^2 + (-3)y + 0 \)
\( = 4y^2 - 3y \)
In simple words: First, we remove the parentheses, making sure to flip the signs of all terms in the second bracket because of the minus sign. Next, we group terms that look alike, such as all the \( y^2 \) terms and all the 'y' terms. Finally, we combine these groups to get the simplified answer.

Exam Tip: When subtracting an entire expression in parentheses, distribute the negative sign to every term inside the parentheses before combining like terms.

2. સરવાળા કરો:

Question (i) 3mn, - 5mn, 8mn, – 4mn
Answer:
\( = 3mn + (-5mn) + 8mn + (- 4mn) \)
\( = [3 + (-5) + 8 + (-4)]mn \)
\( = [11 + (-9)]mn \)
\( = [2]mn \)
\( = 2mn \)
In simple words: To add these terms, since all of them have 'mn', we just add their numbers together. We sum up all the positive numbers and all the negative numbers, then combine those results.

Exam Tip: When adding terms with identical variable parts, simply sum or subtract their numerical coefficients. The variable part remains unchanged.

Question (ii) t - 8tz, 3tz - z, z-t
Answer:
\( = (t - 8tz) + (3tz - z) + (z - t) \)
\( = t - 8tz + 3tz - z + z - t \)
\( = t - t + 3tz - 8tz + z - z \)
\( = (1 - 1)t + (3 - 8) tz + (1 - 1) z \)
\( = (0)t + (-5) tz + (0)z \)
\( = -5tz \)
In simple words: First, we remove any brackets. Then we group all the terms that have 't' together, all the terms with 'tz' together, and all the terms with 'z' together. Finally, we add or subtract their numbers to get the simplest answer.

Exam Tip: Organize the terms by variable before combining. For example, group all 't' terms, then 'tz' terms, and then 'z' terms to avoid mistakes.

Question (iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
Answer:
\( = (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3) \)
\( = -7mn + 5 + 12mn + 2 + 9mn - 8 – 2mn - 3 \)
\( = (-7mn + 12mn + 9mn – 2mn) + (5 + 2 - 8 - 3) \)
\( = (-7 + 12 + 9 – 2)mn + (5 + 2 - 8 - 3) \)
\( = (21 – 9)mn + (7 – 11) \)
\( = 12 mn + (-4) \)
\( = 12 mn - 4 \)
In simple words: We first remove all parentheses, since we are adding. Then, we gather all the 'mn' terms and all the constant numbers. We add and subtract the coefficients for 'mn' and the constants separately to get the final simple expression.

Exam Tip: When combining multiple expressions, always group similar terms (like \( mn \) or constants) together before performing any arithmetic operations.

Question (iv) a + b - 3, b-a + 3, a − b + 3
Answer:
\( = (a + b - 3) + (b - a + 3) + (a - b + 3) \)
\( = a + b - 3 + b - a + 3 + a - b + 3 \)
\( = (a - a + a) + (b + b - b) + (- 3 + 3 + 3) \)
\( = (1 - 1 + 1) a + (1 + 1 - 1) b + (- 3 + 6) \)
\( = (2 - 1) a + (2 - 1) b +(-3 + 6) \)
\( = (1) a + (1) b + (3) \)
\( = a + b + 3 \)
In simple words: We open all the brackets, then group all the 'a' terms, all the 'b' terms, and all the plain numbers separately. We add or subtract the numbers in each group to simplify the entire expression.

Exam Tip: Be mindful of coefficients for terms. If a variable has no explicit number in front, its coefficient is 1 (e.g., 'a' is \( 1a \)).

Question (v) 14x + 10y - 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
Answer:
\( = (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy \)
\( = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy \)
\( = (14x – 7x) + (10y – 10y) +(- 12xy + 8xy + 4xy) + (- 13 + 18) \)
\( = (14 – 7)x + (10 – 10)y + (- 12 + 8 + 4) xy + (5) \)
\( = (7)x + (0)y + (- 12 + 12)xy + 5 \)
\( = 7x + 0y + (0)xy + 5 \)
\( = 7x + 5 \)
In simple words: First, we write all expressions together for addition. Then, we find and group all terms that are exactly alike, such as 'x' terms, 'y' terms, 'xy' terms, and constant numbers. After adding their numerical parts, we write the simplified answer.

Exam Tip: Organize your work by grouping all like terms vertically or in separate lines to reduce the chance of missing a term or making an arithmetic error.

Question (vi) 5m - 7n, 3n - 4m + 2, 2m - 3mn – 5
Answer:
\( = (5m – 7n) + (3n – 4m + 2) + (2m - 3mn – 5) \)
\( = 5m - 7n + 3n - 4m + 2 + 2m - 3mn - 5 \)
\( = (5m - 4m + 2m) + (-7n + 3n) – 3mn + (2 – 5) \)
\( = (5 – 4 + 2) m + (- 7 + 3) n – 3mn + (-3) \)
\( = (7–4) m + (-4)n – 3mn - 3 \)
\( = 3m - 4n - 3mn - 3 \)
In simple words: We begin by writing all the given expressions for addition. Next, we group terms that are similar, like 'm' terms, 'n' terms, 'mn' terms, and plain numbers. Finally, we combine these groups by adding or subtracting their coefficients to get the simplest form.

Exam Tip: When adding multiple polynomials, make sure each term is combined only with other terms that have the exact same variables and exponents.

Question (vii) 4x²y, - 3xy², -5xy², 5x²y
Answer:
\( = 4x^2y + (-3xy^2) + (- 5xy^2) + 5x^2y \)
\( = 4x^2y – 3xy^2 – 5xy^2 + 5x^2y \)
\( = (4x^2y + 5x^2y) + [(- 3xy^2) + (-5xy^2)] \)
\( = (4 + 5) x^2y + [(-3) + (-5)]xy^2 \)
\( = (9)x^2y + (-8)xy^2 \)
\( = 9x^2y – 8xy^2 \)
In simple words: We group the terms with \( x^2y \) together and the terms with \( xy^2 \) together. Then, we add the numbers in front of the \( x^2y \) terms and the numbers in front of the \( xy^2 \) terms separately. This gives us the final simplified answer.

Exam Tip: Remember that \( x^2y \) and \( xy^2 \) are different types of terms and cannot be combined directly. Only terms with identical variable parts can be added or subtracted.

Question (viii) 3p²q² – 4pq + 5, – 10p²q², 15 + 9pq + 7p²q²
Answer:
\( = (3p^2q^2 - 4pq + 5) + (-10p^2q^2) + (15 + 9pq + 7p^2q^2) \)
\( = 3p^2q^2 – 4pq +5 – 10p^2q^2 + 15 + 9pq + 7p^2q^2 \)
\( = (3p^2q^2 – 10p^2q^2 + 7p^2q^2) + (-4pq + 9pq) + (5 + 15) \)
\( = (3 – 10 + 7)p^2q^2 + (- 4 + 9) pq + 20 \)
\( = (0)p^2q^2 + 5pq + 20 \)
\( = 5pq + 20 \)
In simple words: First, we remove all the brackets. Then, we group terms that are alike, such as those with \( p^2q^2 \), those with \( pq \), and the plain numbers. After combining the numbers in each group, we write the simplified expression.

Exam Tip: When combining several polynomial expressions, ensure each type of term (e.g., \( p^2q^2 \), \( pq \), constants) is correctly identified and grouped before summing their coefficients.

Question (ix) ab - 4a, 4b - ab, 4a – 4b
Answer:
\( = (ab - 4a) + (4b – ab) + (4a – 4b) \)
\( = ab - 4a + 4b – ab + 4a - 4b \)
\( = (ab – ab) +(-4a + 4a) + (4b – 4b) \)
\( = (1 - 1) ab + (- 4 + 4) a + (4 – 4) b \)
\( = (0) ab + (0) a + (0) \)
\( = 0 + 0 + 0 \)
\( = 0 \)
In simple words: We first put all the expressions together for addition. Then, we gather all the 'ab' terms, all the 'a' terms, and all the 'b' terms separately. When we add the numbers for each group, all of them cancel out, giving us a final answer of zero.

Exam Tip: Be careful with terms that cancel each other out. For instance, \( +4a \) and \( -4a \) sum to zero and thus disappear from the expression.

Question (x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
Answer:
\( = (x^2 – y^2 – 1) + (y^2 – 1 – x^2) + (1 – x^2 - y^2) \)
\( = x^2 – y^2 – 1 + y^2 – 1 – x^2 + 1 – x^2 – y^2 \)
\( = (x^2 - x^2 - x^2) + (- y^2 + y^2 – y^2) + (- 1 – 1 + 1) \)
\( = (1 - 1 - 1)x^2 + (- 1 + 1 - 1)y^2 + (-2 + 1) \)
\( = (1 – 2)x^2 + (-2 + 1)y^2 + (-1) \)
\( = (-1)x^2 + (-1)y^2 + (-1) \)
\( = -x^2 – y^2 – 1 \)
In simple words: First, we write all three expressions in one line for addition. Then, we group all the \( x^2 \) terms, all the \( y^2 \) terms, and all the constant numbers. We add or subtract the numbers for each group separately to get the simplified expression.

Exam Tip: When combining multiple terms, especially those with negative signs, it helps to write the coefficients in parentheses (e.g., \( (1 - 1 - 1)x^2 \)) to clearly track the arithmetic.

3. બાદબાકી કરોઃ

Question (i) y² માંથી – 5y²
Answer:
\( \therefore y^2 – (-5y^2) \)
\( = y^2 + 5y^2 \)
\( = (1 + 5)y^2 \)
\( = 6y^2 \)
In simple words: To subtract a negative term, it's like adding the positive version of that term. So, \( y^2 \) minus negative \( 5y^2 \) becomes \( y^2 \) plus \( 5y^2 \), which equals \( 6y^2 \).

Exam Tip: Subtracting a negative number is equivalent to adding its positive counterpart. Remember that "minus a minus makes a plus."

Question (ii) -12xy માંથી 6xy
Answer:
\( \therefore - 12xy - 6xy \)
\( = (- 12 - 6)xy \)
\( = - 18xy \)
In simple words: We are subtracting \( 6xy \) from \( -12xy \). Since both terms are similar, we combine their numerical parts. When you have negative twelve and you subtract six more, you get negative eighteen.

Exam Tip: When subtracting terms with the same variable part, simply subtract their coefficients. Keep track of negative signs carefully.

Question (iii) (a + b) માંથી (a – b)
Answer:
\( \therefore (a + b) – (a – b) \)
\( = a + b - a + b \)
\( = (a – a) + (b + b) \)
\( = (1 - 1) a + (1 + 1) b \)
\( = (0) a + (2) b \)
\( = 2b \)
In simple words: We take off the brackets. The minus sign before the second bracket changes the signs of the terms inside. Then, we gather the 'a' terms and the 'b' terms. The 'a' terms cancel each other out, leaving only the 'b' terms.

Exam Tip: When subtracting one expression from another, remember to distribute the negative sign to every term in the expression being subtracted. \( -(a - b) \) becomes \( -a + b \).

Question (iv) b(5 – a) માંથી a (b – 5).
Answer:
\( \therefore b (5 – a) – a (b – 5) \)
\( = 5b - ab – ab + 5a \)
\( = 5b + 5a + (- ab – ab) \)
\( = (5b + 5a) + (- 1 – 1)ab \)
\( = 5a + 5b + (-2) ab \)
\( = 5a + 5b – 2ab \)
In simple words: First, we multiply 'b' into the first bracket and 'a' into the second bracket. Then, we gather any terms that are exactly alike, in this case, the 'ab' terms. Finally, we combine these similar terms to simplify the expression.

Exam Tip: Always distribute the multiplication (like \( b(5-a) \)) correctly before attempting to combine like terms. Pay attention to signs during distribution.

Question (v) 4m² – 3mn + 8 માંથી – m² + 5mn
Answer:
\( \therefore (4m^2 – 3mn) + 8 – (-m^2 + 5mn) \)
\( = 4m^2 - 3mn + 8 + m^2 – 5mn \)
\( = (4m^2 + m^2) + (- 3mn – 5mn) + 8 \)
\( = (4 + 1)m^2 + (- 3 – 5) mn + 8 \)
\( = (5)m^2 + (-8)mn + 8 \)
\( = 5m^2 - 8mn + 8 \)
In simple words: First, we open the brackets, changing signs in the second bracket because of the subtraction. Then, we group together all the \( m^2 \) terms, all the 'mn' terms, and the constant numbers. We perform the addition and subtraction for each group to simplify the expression.

Exam Tip: When subtracting an expression, treat it as adding the negative of that expression. This helps correctly distribute the minus sign to all terms within the parentheses.

Question (vi) 5x – 10 માંથી – x² + 10x − 5
Answer:
\( \therefore (5x – 10) – [-x^2 + 10x – 5] \)
\( = 5x – 10 + x^2 – 10x + 5 \)
\( = x^2 + (5x – 10x) + (- 10 + 5) \)
\( = x^2 + (5 – 10) x + (- 5) \)
\( = x^2 – 5x – 5 \)
In simple words: We begin by removing the square brackets, which means we change the sign of every term inside because of the minus sign outside. Then, we group similar terms like \( x^2 \), 'x' terms, and constant numbers. Finally, we combine their coefficients to get the simplified answer.

Exam Tip: Be extra careful when distributing a negative sign to an expression that itself starts with a negative term; for example, \( -[-x^2] \) becomes \( +x^2 \).

Question (vii) 3ab – 2a² – 2b² માંથી 5a² – 7ab + 5b²
Answer:
\( \therefore (3ab – 2a^2 – 2b^2) – (5a^2 – 7ab + 5b^2) \)
\( = 3ab – 2a^2 – 2b^2 – 5a^2 + 7ab – 5b^2 \)
\( = (3ab + 7ab) + (-2a^2 – 5a^2) + (-2b^2 – 5b^2) \)
\( = (3 + 7)ab + (- 2 – 5)a^2 + (- 2 – 5)b^2 \)
\( = (10) ab + (-7) a^2 + (-7) b^2 \)
\( = 10ab - 7a^2 – 7b^2 \)
In simple words: First, we remove the parentheses, changing the signs of the terms in the second expression due to subtraction. Then, we gather all the 'ab' terms, all the \( a^2 \) terms, and all the \( b^2 \) terms. We add or subtract the coefficients for each group to simplify.

Exam Tip: When subtracting polynomials, ensure that the negative sign is applied to *every* term within the second set of parentheses. Double-check each sign change.

Question (viii) 5p² + 3q² – pq માંથી 4pq – 5q² – 3p²
Answer:
\( \therefore (5p^2 + 3q^2 – pq) – (4pq – 5q^2 - 3p^2) \)
\( = 5p^2 + 3q^2 – pq – 4pq + 5q^2 + 3p^2 \)
\( = (5p^2 + 3p^2) + (3q^2 + 5q^2) + (- pq – 4pq) \)
\( = (5 + 3)p^2 + (3 + 5) q^2 + (- 1 – 4)pq \)
\( = (8)p^2 + (8)q^2 + (-5) pq \)
\( = 8p^2 + 8q^2 – 5pq \)
In simple words: First, we get rid of the brackets, making sure to change the signs of the terms being subtracted. Then, we group similar terms together: \( p^2 \) terms, \( q^2 \) terms, and 'pq' terms. Finally, we add or subtract their numbers to simplify the expression.

Exam Tip: For expressions with multiple variables and powers, such as \( p^2q^2 \), \( p^2 \), and \( q^2 \), remember that each specific combination of variables and powers constitutes a unique like term.

4.

Question (a) x² + xy + y² માં શું ઉમેરવાથી 2x² + 3xy મેળવી શકાય?
Answer:
To find what should be added to \( x^2 + xy + y^2 \) to get \( 2x^2 + 3xy \), we need to subtract \( x^2 + xy + y^2 \) from \( 2x^2 + 3xy \).
\( \therefore (2x^2 + 3xy) – (x^2 + xy + y^2) \)
\( = 2x^2 + 3xy – x^2 – xy – y^2 \)
\( = (2x^2 – x^2) – y^2 + (3xy – xy) \)
\( = (2 – 1)x^2 – y^2 + (3 – 1)xy \)
\( = (1)x^2 – y^2 + (2)xy \)
\( = x^2 - y^2 + 2xy \)
Thus, adding \( x^2 - y^2 + 2xy \) to \( x^2 + xy + y^2 \) will give \( 2x^2 + 3xy \).
In simple words: To find out what to add, we subtract the first expression from the second one. We take off the brackets, being careful with the minus sign, then group and combine the similar terms. The result is the expression you need to add.

Exam Tip: "What should be added to A to get B?" implies the operation \( B - A \). Similarly, "What should be subtracted from A to get B?" implies \( A - B \).

Question (b) 2a + 8b + 10 માંથી શું બાદ કરવાથી – 3a + 7b + 16 મેળવી શકાય?
Answer:
To find what should be subtracted from \( 2a + 8b + 10 \) to get \( -3a + 7b + 16 \), we need to subtract \( -3a + 7b + 16 \) from \( 2a + 8b + 10 \).
\( \therefore (2a + 8b + 10) – (-3a + 7b + 16) \)
\( = 2a + 8b + 10 + 3a – 7b - 16 \)
\( = (2a + 3a) + (8b – 7b) + (10 – 16) \)
\( = (2 + 3) a + (8 – 7) b + (- 6) \)
\( = (5) a + (1) b + (- 6) \)
\( = 5a + b - 6 \)
Thus, subtracting \( 5a + b – 6 \) from \( 2a + 8b + 10 \) will give \( -3a + 7b + 16 \).
In simple words: To figure out what to subtract, we perform subtraction: \( (2a + 8b + 10) - (-3a + 7b + 16) \). We remove the brackets, changing signs due to the subtraction. Then, we gather and combine similar terms (like 'a' terms, 'b' terms, and numbers) to get the answer.

Exam Tip: When the problem asks "What should be subtracted from A to get B?", the required expression is \( A - B \). Always be precise with which expression comes first in subtraction.

Question 5. 3x² – 4y² + 5xy + 20 માંથી શું લઈ લેવાથી – x² – y² + 6xy + 20 મેળવી શકાય?
Answer:
To find what should be removed from \( 3x^2 – 4y^2 + 5xy + 20 \) to get \( -x^2 – y^2 + 6xy + 20 \), we need to subtract \( -x^2 – y^2 + 6xy + 20 \) from \( 3x^2 – 4y^2 + 5xy + 20 \).
\( \therefore (3x^2 – 4y^2 + 5xy + 20) – (- x^2 – y^2 + 6xy + 20) \)
\( = 3x^2 - 4y^2 + 5xy + 20 + x^2 + y^2 - 6xy – 20 \)
\( = (3x^2 + x^2) + (-4y^2 + y^2) + (5xy – 6xy) + (20 – 20) \)
\( = (3 + 1)x^2 + (-4 + 1)y^2 + (5 – 6)xy + (0) \)
\( = 4x^2 + (-3)y^2 + (-1) xy + 0 \)
\( = 4x^2 – 3y^2 – xy \)
In simple words: To discover what needs to be taken away, we subtract the target expression from the initial one. We remove the brackets, remembering to switch the signs for the terms being subtracted. Then, we group similar terms (like \( x^2 \), \( y^2 \), 'xy', and numbers) and combine them.

Exam Tip: Problems like "What should be taken away from A to get B?" are solved by calculating \( A - B \). Ensure all signs are correctly managed during the subtraction of polynomials.

6.

Question (a) 3x − y + 11 અને − y – 11ના સરવાળામાંથી 3x − y -11 બાદ કરો.
Answer:
First, find the sum of \( 3x − y + 11 \) and \( − y – 11 \):
\( = (3x − y + 11) + (- y – 11) \)
\( = 3x - y + 11 - y - 11 \)
\( = 3x − y − y + 11 – 11 \)
\( = 3x – 2y + 0 \)
\( = 3x - 2y \)
Now, subtract \( 3x − y – 11 \) from \( 3x – 2y \):
\( = (3x – 2y) – (3x – y – 11) \)
\( = 3x – 2y – 3x + y + 11 \)
\( = (3x – 3x) + (-2y + y) + 11 \)
\( = 0 + (-y) + 11 \)
\( = - y + 11 \)
In simple words: We first add the two expressions together. Then, from that total, we subtract the third expression. It's really important to correctly handle the minus signs when opening the brackets during subtraction.

Exam Tip: Break down complex problems into smaller, manageable steps. First, calculate the sum, then perform the subtraction, ensuring accurate handling of negative signs at each stage.

Question (b) 4 + 3x અને 5 – 4x + 2x² ના સરવાળામાંથી 3x² – 5x અને -x² + 2x + 5નો સરવાળો બાદ કરો.
Answer:
First, find the sum of \( 4 + 3x \) and \( 5 – 4x + 2x^2 \):
\( = (4 + 3x) + (5 – 4x + 2x^2) \)
\( = 4 + 3x + 5 - 4x + 2x^2 \)
\( = (4 + 5) + (3x – 4x) + 2x^2 \)
\( = 9 + (-x) + 2x^2 \)
\( = 9 − x + 2x^2 \)
Now, find the sum of \( 3x^2 – 5x \) and \( -x^2 + 2x + 5 \):
\( = (3x^2 – 5x) + (- x^2 + 2x + 5) \)
\( = 3x^2 – 5x – x^2 + 2x + 5 \)
\( = (3x^2 − x^2) + (- 5x + 2x) + 5 \)
\( = (3 – 1)x^2 + (-5 + 2) x + 5 \)
\( = 2x^2 + (-3) x + 5 \)
\( = 2x^2 – 3x + 5 \)
Finally, subtract the second sum (\( 2x^2 – 3x + 5 \)) from the first sum (\( 9 − x + 2x^2 \)):
\( = (9 − x + 2x^2) – (2x^2 – 3x + 5) \)
\( = 9 − x + 2x^2 - 2x^2 + 3x – 5 \)
\( = (9 – 5) + (- x + 3x) + (2x^2 – 2x^2) \)
\( = (4) + (-1 + 3) x + (2 – 2)x^2 \)
\( = 4 + (2) x + (0)x^2 \)
\( = 4 + 2x \)
In simple words: This problem involves two steps of addition and one step of subtraction. First, we add the first two expressions. Then, we add the third and fourth expressions. Finally, we subtract the second sum from the first sum. Be careful with signs when opening brackets, especially during subtraction.

Exam Tip: For problems with multiple operations, calculate each sum/difference in separate, distinct steps. Clearly label each intermediate result to maintain clarity and accuracy.