Algebraic expressions

Substitute the given values: \(2a+b=2\times2+3=4+3=7\). Therefore, the correct answer is 7. Getting 6 may result from an error in the order of multiplication and addition. Exam tip: substitute the values of variables first, then perform multiplication before addition.

The term 5x^2y can be written as 5 \(\times x^2y\). Thus, the numerical factor multiplying \(x^2y\) is 5, so its coefficient is 5. The number 2 is the exponent of x in \(x^2\), not the coefficient. Exam tip: The number multiplying the stated algebraic part of a term is its coefficient.

Half of \(x\) is \(\frac{x}{2}\). “3 more than” means adding 3 to that quantity, so the correct expression is \(\frac{x}{2}+3\). \(\frac{x+3}{2}\) represents half of \(x+3\), so it is different. Exam tip: For “more than,” write the stated quantity first, then add the given number.

“2 more than c” means adding 2 to c, so the correct expression is \(c+2\). In contrast, \(c-2\) means 2 less than c, and \(2c\) means twice c. Exam tip: the phrase “more than” indicates addition.

Given \(x=0\), substitute it into \(7x+2\): \(7(0)+2=0+2=2\). Therefore, the correct value is \(2\). The number \(7\) is only the coefficient, not the value of the expression. Exam tip: To evaluate an expression, substitute the given value of the variable everywhere it occurs.

In a polynomial, exponents of variables must be 0 or positive integers. Here \(a^{-1}=\frac{1}{a}\), so the negative exponent makes it non-polynomial. Exam tip: check every variable exponent.

Given x=-2, substitute it into the expression: x^2+3x+1=(-2)^2+3(-2)+1=4-6+1=-1. Therefore, the correct value is -1. The option 1 may result from incorrectly taking 3x as +6. Exam tip: always use brackets while squaring a negative number, for example, (-2)^2=4.

Terms in an expression are separated by addition or subtraction signs. Here the terms are \(7mn\), \(-3m\), and \(5\), so there are 3 terms. The term \(-3m\) must be counted separately, with its negative sign. Exam tip: Count parts separated by \(+\) or \(-\), not factors joined by multiplication.

Using the distributive property, \(2(x-5)=2x-10\). Adding \(3x\) to \(2x-10\), the like terms \(2x\) and \(3x\) combine to give \(5x\), so the simplified expression is \(5x-10\). \(5x+10\) is incorrect because \(2\times(-5)=-10\), not \(+10\). Exam tip: When opening brackets, multiply the outside number by every term inside the bracket.

The linear term containing x is -3x. Since -3x = (-3)×x, the coefficient of x is -3. Here, 4 is the coefficient of x², while 6 is the constant term. In exams, identify the number multiplying the variable and keep its sign.

Substituting the given values, \(3a-2b=3(2)-2(-1)=6+2=8\). Since \(b=-1\), \(2b=-2\), so subtracting \(2b\) means subtracting \(-2\), which gives \(+2\). Option 6 is only the value of \(3a\); it ignores the \(-2b\) term. Exam tip: When subtracting a negative number, use brackets first to avoid sign errors.

A constant term is a term that contains no variable. In \(5u^2+2u-8\), both \(5u^2\) and \(2u\) contain the variable \(u\), whereas \(-8\) does not. Therefore, the constant term is \(-8\). Also, \(8u\) is not a term in the given expression. Exam tip: always retain the sign while identifying a constant term.

The governing concept is translating words into algebraic notation while preserving the order of operations. “The sum of m and n” is represented by (m + n). “Three times” means multiply the entire sum by 3, giving 3(m + n). Therefore option C is correct. The brackets are important because they show that 3 multiplies both m and n; on expansion the expression becomes 3m + 3n. Option A multiplies only m, option B multiplies only n, and option D combines multiplication of m and n with addition of 3, which represents a different idea.

The governing concept is that the numerical coefficient is the number multiplying the variable part of a term. In 9x²y, the variable part is x²y and the number attached to it is 9, so the numerical coefficient is 9. Option A is the variable part, not a number. Option C is the exponent on x, not the coefficient of the whole term. Option D is only part of the original term and still contains a variable, so it cannot be the numerical coefficient. Therefore option B is correct. Separating the numerical factor from the literal factor avoids confusing an exponent with a coefficient.

Here, 4r^2 and -6r^2 are like terms, so their coefficients are combined: 4-6=-2. Thus, 4r^2-6r^2=-2r^2, while 5r remains unchanged because it has a different power of r. Therefore, the simplified expression is -2r^2+5r. In option D, the sign of 5r is incorrectly changed. Exam tip: Combine only terms with the same variable raised to the same power.

In 7ab and -2ab, the variable part ab is exactly the same: both a and b have exponent 1. Therefore, they are like terms even though their coefficients, 7 and -2, are different. In option C, the exponents of a and b are interchanged, so those terms are not like terms. Exam tip: compare only the variables and their exponents; coefficients need not be the same.

Substituting t=3 gives 2t^2-t=2(3^2)-3=2(9)-3=18-3=15. Therefore, the correct answer is 15. The nearby distractor 18 results from forgetting to subtract 3 in the final step. Exam tip: after substitution, evaluate powers first, then multiplication, and finally addition or subtraction.

On expanding the brackets, \(5(x+2)-2(x-1)=5x+10-2x+2\). Multiplying \(-2\) by both terms in \((x-1)\) gives \(-2x+2\). Combining like terms gives \(5x-2x+10+2=3x+12\). The option \(3x+8\) may result from using the wrong sign for the constant term in \(-2(x-1)\). Exam tip: when a negative coefficient is outside a bracket, distribute it carefully to every term inside.

“5 more than 2y” means add 5 to 2y. Therefore, the expression is 2y+5. The expression 2y-5 represents 5 less than 2y, so it is not correct. Exam tip: Use addition (+) for “more than.”

Given \(p=-3\), \(p^2+2p=(-3)^2+2(-3)=9-6=3\). Therefore, the correct value is \(3\). The option \(-3\) can result from not handling the square of the negative value correctly. Exam tip: Always use brackets when squaring a negative number, as in \((-3)^2\).