Algebraic expressions

Here, 4c and c are like terms because both contain c to the first power. The coefficient of c is 1, so 4c+c = 4c+1c = 5c. The option 4c^2 is incorrect because adding like terms does not increase the exponent. Exam tip: When adding like terms, add only their coefficients.

All the terms contain the same variable \(q\), so they are like terms. Adding their coefficients gives \(10-3+1=8\). Therefore, \(10q-3q+q=8q\). The result \(7q\) would occur if the final \(+q\) were omitted. Exam tip: When simplifying like terms, operate on the coefficients and keep the variable unchanged.

The variable parts of 2x and 3y are x and y respectively. Like terms must have exactly the same variables with the same powers; only their coefficients may differ. Since the variables are different, these are unlike terms. A constant term has no variable. Exam tip: Compare the variable parts before deciding whether terms are like terms.

The expression 15 contains no variable such as x or y, and its value always remains 15. Therefore, it is a constant expression. A binomial and a trinomial have two and three terms respectively, whereas 15 has only one term. Exam tip: An expression with no variable is identified as a constant expression.

The numerical coefficient is the number that multiplies the variable part. Here, 6x^2 = 6 \(\times\) x^2, so the numerical coefficient is 6. The number 2 is the exponent of x, not the coefficient. Exam tip: In a term, the number left after identifying the variable part is its numerical coefficient.

In \(5p^2-3p+7\), the powers of \(p\) are 2, 1, and 0, all non-negative integers, so it is a polynomial. In \(4p^{-1}+2\), the power is \(-1\), so it is not a polynomial. Exam tip: check for negative powers first.

To find half of a quantity, divide it by 2. Therefore, half of k is \(\frac{k}{2}\). The expression \(2k\) represents twice k, not half of it. Exam tip: When you see “half,” divide the quantity by 2.

Twice \(r\) is \(2r\). Adding 9 to this quantity gives \(2r+9\). In \(2(r+9)\), 9 is added to \(r\) first and then the entire sum is doubled, so it represents a different expression. Exam tip: when a number is added “to twice” a variable, add it outside the doubled term.

The governing skill is translating a verbal statement into an algebraic expression while preserving the order indicated by the language. “Four times s” means 4s. The phrase “subtracting 7 from four times s” means that 4s is the starting quantity and 7 is removed from it, giving 4s − 7. Therefore option B is correct. Option A changes subtraction into addition. Option C reverses the order and represents 7 − 4s, which means subtracting 4s from 7. Option D incorrectly changes the multiplication into a different subtraction and does not contain the required term 4s. Careful attention to the word “from” prevents the common reversal error.

\(2a\) and \(5a\) are like terms because both have the variable \(a\) to the first power. Adding their coefficients gives \(2+5=7\), so \(2a+5a=7a\). The constant term \(-3\) remains unchanged; therefore, the simplified form is \(7a-3\). \(10a-3\) would result from incorrectly multiplying the coefficients. Exam tip: add or subtract coefficients only of like terms.

Like terms must have exactly the same variables raised to the same powers. Here, the power of b is 2 in b^2, whereas it is 1 in 2b. Therefore, they are unlike terms. Having the same variable b alone does not make them like terms. Exam tip: Compare variables and their exponents, not just the coefficients.

Like terms have the same variables raised to the same powers. Both \(3x\) and \(2x\) contain \(x\) to the power 1, so they are like terms. The term \(5\) is a constant, so it is not like either term containing \(x\). Exam tip: coefficients may differ; compare the variables and their powers to identify like terms.

Substituting
(x=1) gives
(x^2+2=1^2+2=1+2=3). Therefore, the correct answer is 3. Option 2 could result from incorrectly treating the squared term
(x^2) as 0. Exam tip: evaluate powers first, then perform addition or subtraction.

Given \(y=2\), \(y^2+y=2^2+2=4+2=6\). Therefore, the correct answer is 6. Option 4 is only the value of \(y^2\); the \(+y\) term must also be added. Exam tip: after substituting a variable's value, evaluate powers before addition or subtraction.

\(5u\) and \(u\) are like terms, so their coefficients are added: \(5u+u=6u\). The constant term \(-2\) remains unchanged. Therefore, the simplified expression is \(6u-2\). In \(6u+2\), the sign of the constant term is incorrect. Exam tip: Combine only terms with the same variable and the same power.

The sum of (x) and (y) is (x+y). Twice this entire sum is 2(x+y). In 2x+y, only x is multiplied by 2, not y. Exam tip: Use brackets when words such as “sum of” or “entire sum” occur.

In the expression \(7-3v\), \(v\) is the variable because its value can change. Here, \(7\) is the constant term and \(-3\) is the coefficient of \(v\), so \(-3\) is not a variable. Exam tip: Identify the letter or symbol whose value can change.

The governing concept is recognising algebraic operations from notation. The plus sign in a + b explicitly indicates addition, so a + b is a sum. When two variables are written together as ab without an operation sign, multiplication is implied; thus ab is the product of a and b. Option A is therefore correct. The expressions are not always equal: if a = 2 and b = 3, then a + b = 5 while ab = 6. Option C reverses the meanings of addition and multiplication. Option D is also incorrect because a and b are generally variables, not necessarily fixed constants.

Direct answer: Option A, \(x^2+3\). The phrase “the square of \(x\)” means multiplying \(x\) by itself, written \(x^2\). The phrase “adding 3 to” means place 3 after the square and use plus: \(x^2+3\). Step by step: identify the number being squared, which is \(x\); write its square as \(x^2\); add 3; obtain \(x^2+3\). Option A is correct. Option B, \(2x+3\), doubles \(x\) and then adds 3; it does not contain the square of \(x\). Option C, \(x+3^2\), squares 3 instead of squaring \(x\), and would equal \(x+9\). Option D, \(3x^2\), means multiplying the square by 3, not adding 3. A plus sign and a multiplication sign express different operations. Memory cue: “square of x, then add 3” means square first and add later: \(x^2+3\).

A constant term is a term with no variable. In 4x^2+3x+2, both 4x^2 and 3x contain x, whereas 2 contains no variable. Therefore, 2 is the constant term. Although x^0 equals 1, it is not a term in the given expression. Exam tip: identify the term that has no variable before selecting the constant term.

Substituting p=3 gives 4p-1=4×3-1=12-1=11. Therefore, the correct answer is 11. Option 12 is a close distractor if the subtraction of 1 is missed. Exam tip: After substitution, perform multiplication before addition or subtraction.

Here, 2x, 3x, and 4x are like terms because each term contains x to the power 1. So, add their coefficients: 2+3+4=9. Hence, the simplified expression is 9x. The option 24x comes from multiplying the coefficients, which is not required here. Exam tip: Only terms with the same variable and the same exponent can be added.

Zero is the additive identity: adding 0 to any number or variable does not change its value. Therefore, \(x+0=x\), so \(x\) is correct. \(1\) and \(x^0\) generally represent 1, so they are not the simplified form of \(x+0\). Exam tip: In addition, a term added with 0 can be removed.

1 is the multiplicative identity. Hence, multiplying any variable by 1 leaves it unchanged: \(1\cdot x=x\). The close distractor \(x+1\) is wrong because it adds 1 rather than multiplying by 1. Exam tip: multiplying an expression by 1 does not change it.

The product of 0 and any number or variable is 0. Therefore, \(0\cdot y=0\), and \(0\cdot y+6=0+6=6\). The option \(y+6\) is incorrect because the term containing \(y\) becomes 0 when multiplied by 0. Exam tip: simplify multiplication first, then perform addition or subtraction.