Algebraic form of product of 7y and 4z is
Answer: D
\[ 7y \times 4z = (7 \times 4)(y \times z) = 28yz \]
The coefficients multiply and the variables are written side by side. Adding them instead would give \( 11yz \) - the first option.
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On solving expression 8c + 4d - c + 3d, answer will be
Answer: D
Collect the like terms.
\[ 8c - c = 7c \]
\[ 4d + 3d = 7d \]
\[ \therefore\ 7c + 7d \]
The \( c \) and \( d \) terms are unlike and cannot be merged into a single term such as \( 14cd \).
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On simplifying algebraic expression 2{4a + [-b -5(c + a -b)]}, answer will be
Answer: D
Work outward from the innermost bracket.
\[ -5(c + a - b) = -5c - 5a + 5b \]
\[ -b + (-5c - 5a + 5b) = -5a + 4b - 5c \]
\[ 4a + (-5a + 4b - 5c) = -a + 4b - 5c \]
\[ 2(-a + 4b - 5c) = -2a + 8b - 10c \]
The \( -5 \) flips the sign of \( -b \) inside, which is what turns the \( -b \) outside into a net \( +4b \).
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Which of the following is obtained by subtracting x²-y² from y² - x² ?
Answer: A
Subtracting means taking the first expression away from the second.
\[ (y^{2} - x^{2}) - (x^{2} - y^{2}) = y^{2} - x^{2} - x^{2} + y^{2} \]
\[ = 2y^{2} - 2x^{2} = -2(x^{2} - y^{2}) \]
The bracket's signs both flip when it is subtracted - missing that gives 0 instead.
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By factorizing algebraic expression mp - np + mq - nq, answer will be
Answer: B
Group the terms in pairs and factor each.
\[ mp - np + mq - nq = p(m - n) + q(m - n) \]
\[ = (p + q)(m - n) \]
The bracket \( (m-n) \) appears in both parts, so it comes out as the common factor.
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On solving algebraic expression 2(x - 2y) +4(2y + 3z) -3(x + 4y - 5z), answer will be
Answer: B
Expand each bracket, then collect.
\[ 2(x - 2y) = 2x - 4y \]
\[ 4(2y + 3z) = 8y + 12z \]
\[ -3(x + 4y - 5z) = -3x - 12y + 15z \]
\[ x: \ 2 - 3 = -1; \quad y: \ -4 + 8 - 12 = -8; \quad z: \ 12 + 15 = 27 \]
\[ \therefore\ -x - 8y + 27z \]
The \( -3 \) turns \( -5z \) into \( +15z \), which is what makes the \( z \) coefficient large and positive.
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Simplify (x - 9)(x + 10) ⁄ (x² - 81)
Answer: C
Factorise the denominator as a difference of squares, then cancel.
\[ x^{2} - 81 = (x-9)(x+9) \]
\[ \frac{(x-9)(x+10)}{(x-9)(x+9)} = \frac{x+10}{x+9} \]
The common factor \( (x-9) \) cancels, provided \( x \neq 9 \). Cancelling the wrong pair would leave \( \tfrac{x+10}{x-9} \).
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On simplifying the algebraic expression \( -13a \times -2 \), the answer will be
Answer: B
Multiplying two negatives gives a positive.
\[ -13a \times -2 = (13 \times 2)a = 26a \]
The variable \( a \) is carried through unchanged, so the answer keeps it - dropping it would leave the bare number 26.
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Answer of factorization of expression 4z(3a + 2b - 4c) + (3a + 2b - 4c) is
Answer: A
Both terms share the bracket \( (3a + 2b - 4c) \), so take it out as a common factor.
\[ 4z(3a+2b-4c) + 1 \cdot (3a+2b-4c) = (4z + 1)(3a + 2b - 4c) \]
The second term carries an invisible coefficient of 1, which is what becomes the \( +1 \) in the first bracket - dropping it would leave \( 4z(3a+2b-4c) \) alone.
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Sum of 5x + 3y and 4x - 2y is
Answer: B
Add like terms.
\[ (5x + 3y) + (4x - 2y) = (5x + 4x) + (3y - 2y) \]
\[ = 9x + y \]
The \( y \) terms leave a coefficient of 1, which is written without the digit.
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