Algebraic expressions

Multiplying any variable or number by 0 gives 0. Hence, \(0x=0\), so \(0\) is the correct option. Although \(x^0=1\) for \(x\ne 0\), it is not equal to \(0x\). Exam tip: a term with coefficient 0 contributes nothing to the value of an expression.

Reena is incorrect. Like terms must have the same variables with the same powers. In \(p^2q\) and \(pq^2\), the powers of \(p\) and \(q\) differ. Exam tip: compare exponents, not just the variables.

In \(2x^2y^3\), the factor containing \(y\) is \(y^3\), so the power of \(y\) is 3. The power of \(x\) is 2, and their sum, 5, gives the total degree of the term, not the power of \(y\). Exam tip: To find the power of a variable, look at the exponent written directly on that variable.

Half of x is \(\frac{x}{2}\). Adding 7 to this quantity gives \(\frac{x}{2}+7\), so option C is correct. In \(\frac{x+7}{2}\), both x and 7 are divided by 2, while \(7-\frac{x}{2}\) represents a difference rather than a sum. Exam tip: translate “half of x” first, then add 7 for “sum.”

Given \(x=5\), substitute it into the expression: \(\frac{x+3}{2}=\frac{5+3}{2}=\frac{8}{2}=4\). Therefore, the correct answer is 4. The value 8 is only the numerator \(x+3\); it must still be divided by 2. Exam tip: After substitution, simplify step by step and pay attention to the order of operations.

Combine like terms: \(3x-x=2x\) and \(2y+5y=7y\). Therefore, the simplified expression is \(2x+7y\). \(4x+7y\) is incorrect because the coefficient in \(3x-x\) is \(3-1=2\), not 4. Exam tip: Add or subtract only terms with the same variable and the same power.

Using the distributive property, \(2(x+4)=2\cdot x+2\cdot4=2x+8\). Therefore, \(2x+8\) and \(2\cdot x+2\cdot4\) are equal to the given expression. Also, by the commutative property of addition, \(8+2x=2x+8\). However, in \(2x+4\), the constant 4 has not been multiplied by 2, so it is not equal to the given expression. Exam tip: When expanding brackets, multiply the outside factor by every term inside the bracket.

For like terms, variables and their powers must be exactly the same. Here swapped powers make the terms unlike.

Substituting n=-1 gives 4n^2+3n-2=4(-1)^2+3(-1)-2. Since (-1)^2=1, the value is 4(1)-3-2=4-3-2=-1. Getting 1 usually results from using the wrong sign for 3(-1). Exam tip: Always use brackets when squaring a negative number.

Like terms must have exactly the same variables raised to the same powers; only their numerical coefficients may differ. In 2ab, 3ab, and 5ab, both a and b have power 1, so they are like ab. However, -4a has no b, so it is not a like term of ab. Exam tip: Compare the variable part and exponents, not the coefficients.

Using the distributive property, \(3(2x-3)=3\times2x-3\times3=6x-9\). Therefore, option A is correct. In option D, the constant term becomes \(+9\), whereas the required expression has \(-9\). Exam tip: When expanding brackets, multiply the outside number by every term inside the bracket.

Combining like terms gives \(x^2-x^2=0\) and \(-2x+5x=3x\). The constant term \(4\) remains unchanged, so the simplified expression is \(3x+4\). Option \(7x+4\) is incorrect because \(-2x+5x=3x\), not \(7x\). Exam tip: add or subtract only terms with the same variable and exponent.

Given \(x=2\), substitute it in \(4x+3\): \(4(2)+3=8+3=11\). Therefore, the correct answer is 11. The value 8 is only \(4\times2\); the additional 3 must still be added. Exam tip: Substitute the value of the variable first, then follow the order of operations.

\(6a\) and \(-2a\) are like terms, so their coefficients are subtracted: \(6a-2a=4a\). The constant term \(+5\) remains unchanged. Therefore, the simplified form is \(4a+5\). \(8a+5\) would result from adding \(6a\) and \(2a\), not subtracting them. Exam tip: Add or subtract only terms with the same variable and exponent.

The product of \(m\) and \(n\) is \(mn\). Subtracting 7 from this product gives \(mn-7\), so option D is correct. In \(7-mn\), the product \(mn\) is subtracted from 7, which is a different expression. Exam tip: In phrases such as “subtract from,” write the quantity from which subtraction is made first.

In \(9-5x\), the term containing \(x\) is \(-5x\). The number multiplying \(x\) is \(-5\), so the coefficient of \(x\) is \(-5\). Choosing \(5\) would ignore the negative sign. Exam tip: always include the sign of the term while identifying a coefficient.

Terms in an expression are parts separated by addition or subtraction signs. Here, the terms are \(2p^2\), \(-3q\), and \(8\). Therefore, the expression has 3 terms. Counting only the variables \(p\) and \(q\) would be incorrect because the constant \(8\) is also a term. Exam tip: Count the parts separated by + and − signs to find the number of terms.

The term containing x² is 4x². The number multiplying x² is 4, so the coefficient of x² is 4. Here, -3 is the coefficient of x, while 7 is the constant term. Exam tip: To identify a coefficient, look at the number and sign multiplying the required variable part.

A constant term is a term that contains no variable. In this expression, 6a contains a and -5b contains b, whereas 9 has no variable. Therefore, 9 is the constant term. Exam tip: identify the term with no letter or variable to find the constant term.

In an expression, terms are separated by plus (+) or minus (−) signs. Here the terms are 3p^2, 2p, and −8, so there are 3 terms. The coefficient 2 in 2p is not counted as a separate term. Exam tip: Count each complete part separated by + or − as one term.