Algebraic expressions

Unlike terms have different variables or different powers of the same variable. In 5x and 5y, the variables are x and y, so they are unlike terms. In 7a and 2a, and in 9p and 3p, the variable and its power are the same; only the coefficients differ, so they are like terms. Exam tip: Compare variables and their powers first, not the coefficients.

In the expression \(2a+3b\), \(a\) and \(b\) are letters whose values can vary, so they are variables. Therefore, there are 2 variables. The numbers 2 and 3 are coefficients, not variables. Exam tip: Count the distinct letters whose values can change to find the number of variables.

A numerical expression contains only numbers and operations; it has no variables such as \(x\), \(a\), \(p\), or \(q\). Since \(7+5\) contains only numbers, it is a numerical expression. In contrast, \(3x+2\) contains the variable \(x\), so it is an algebraic expression. Exam tip: if an expression includes a letter representing a variable, it is not purely numerical.

\(2r-5\) is an algebraic expression because it contains the variable \(r\), the numbers 2 and 5, and subtraction. \(12+7\), \(8\div4\), and \(6\times3\) are numerical expressions because they contain no variable. Exam tip: Look for a letter representing a number to identify an algebraic expression.

There are three like terms x: \(x+x+x=(1+1+1)x=3x\). Therefore, the simplified form is \(3x\). \(x^3\) is obtained from multiplication, \(x\times x\times x\), not from addition. Exam tip: when adding like terms, add their coefficients.

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 concept is translating verbal operations into an algebraic expression while preserving their order. “Four times s” means 4s, because multiplication of 4 and s is written by juxtaposition. The phrase “subtracting 7 from 4 times s” means start with 4s and take away 7, giving 4s − 7. Thus option B is correct. Option A adds 7 instead of subtracting it. Option C reverses the order and means subtracting 4s from 7, which is a different expression. Option D incorrectly multiplies s by 28 and does not represent the stated operation.

\(2a\) 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.