Algebraic expressions

Use the distributive property: \(7(2r-3)=14r-21\) and \(4(r+1)=4r+4\). Thus, \(14r-21+4r+4=18r-17\). Therefore, the correct expression is \(18r-17\). The option \(18r+25\) may result from incorrectly adding the constant terms with their signs. Exam tip: while expanding brackets, multiply the outside number by every term inside the bracket.

\(5a^2b\) and \(3a^2b\) are like terms, so they add to \(8a^2b\). Similarly, \(2ab-7ab=-5ab\). Therefore, the simplified expression is \(8a^2b-5ab\). \(10a^3b\) is incorrect because exponents are not added when terms are added or subtracted. Exam tip: combine only terms with identical variable parts and powers.

Given x=2 and y=-1, x^2+xy+y^2=2^2+(2)(-1)+(-1)^2=4-2+1=3. Therefore, the correct answer is 3. The value 1 is only y^2; the terms x^2 and xy must also be included. Exam tip: the square of a negative number is positive, so (-1)^2=1.

A linear term has a variable with exponent 1. In the given expression, 5x^2 has exponent 2, 2 is a constant term, and -3x has x raised to the power 1. Therefore, -3x is the linear term. In exams, identify the type of each term by checking the exponent of its variable.

On expanding the bracket, \(2(a+b)-a=2a+2b-a\). Combining like terms gives \(2a-a=a\), so the expression equals \(a+2b\). The option \(2a+b\) does not simplify the \(a\)-terms correctly. Exam tip: while expanding brackets, multiply the outside coefficient by every term inside the bracket.

The powers of the variable s in the expression are 3, 2, and 1, while the constant term −8 has power 0. Therefore, the highest power is 3, found in the term 4s^3. The number 8 is only part of the constant term, not a power. Exam tip: To find the degree of a polynomial, identify the greatest exponent of its variable.

In an algebraic term, the number that multiplies the variable part is called the numerical coefficient. In 6x^2y, 6 multiplies x^2y, so the numerical coefficient is 6. Here, x^2y is the variable part, while 2 is only the exponent of x. Exam tip: To identify a coefficient, separate the variables and their exponents from the term; the remaining number is the coefficient.

Using the distributive property, \(3(4x-5)=12x-15\) and \(-2(2x+1)=-4x-2\). Therefore, \(12x-15-4x-2=8x-17\). Hence, \(8x-17\) is correct. In \(8x-13\), the constant terms have been combined incorrectly. Exam tip: when a negative coefficient is outside a bracket, apply it to every term inside the bracket.

The difference between m and n is \(m-n\). Taking five times this entire difference gives \(5(m-n)\), so option C is correct. In \(5m-n\), only m is multiplied by 5, not the whole difference. Exam tip: When a phrase says “times the difference,” put the difference in brackets before multiplying.

Combine like terms: \(10z^2-4z^2=6z^2\) and \(z-6z=-5z\). There is no other constant term to combine with \(3\), so it remains \(3\). Therefore, the simplified expression is \(6z^2-5z+3\). Option B incorrectly adds the \(z^2\) terms. Exam tip: add or subtract only terms having the same variable and exponent.

Given \(c=-2\), \(2c^3+c^2=2(-2)^3+(-2)^2=2(-8)+4=-16+4=-12\). Therefore, \(-12\) is correct. Since \(c^3\) has an odd exponent, it remains negative, whereas \(c^2\) has an even exponent and is positive. Exam tip: When substituting a negative value, check whether each exponent is odd or even.

Like terms are algebraic terms that have exactly the same variables with exactly the same exponents. Their coefficients may be different, but the variable pattern must match. In \(a^2b\), the exponent of a is 2 and the exponent of b is 1. In \(ab^2\), the exponent of a is 1 and the exponent of b is 2. The exponents are therefore interchanged, so the terms are not like terms. Option C states this idea.

The terms may look similar because both contain a and b, but merely having the same letters is not enough. For example, \(3a^2b\) and \(-5a^2b\) are like terms, while \(a^2b\) and \(ab^2\) are not. Options A, B and D do not describe the actual difference: the coefficients need not be the issue, variables are present, and these are not constants. Hence option C is correct.

On opening the bracket, \(-3(4-x)=-12+3x\), because \(-3\) multiplies both 4 and \(-x\). Therefore, \(2x-3(4-x)=2x-12+3x=5x-12\). In \(5x+12\), the sign of the constant term is incorrect. Exam tip: when multiplying a bracket by a negative number, check the sign of every term.

In \(8rs-5r\), the term \(8rs\) can be written as \(8\times rs\). Hence, the coefficient of \(rs\) is 8. In \(rs+8\), the coefficient of \(rs\) is 1 because \(rs=1\times rs\). Exam tip: To find a coefficient, identify the numerical factor multiplying the term.

Here, \(2x^2\) and \(-4x^2\) are like terms, so \(2x^2-4x^2=-2x^2\). Similarly, \(3xy\) and \(5xy\) are like terms, giving \(3xy+5xy=8xy\). Therefore, the simplified expression is \(-2x^2+8xy\). The terms \(x^2\) and \(xy\) are not like terms, so they cannot be combined. Exam tip: add or subtract only terms with identical variables and powers.

Substituting the given values, (2x+y^2)=2(1)+5^2=2+25=27. Hence, the correct answer is 27. A result such as 12 may come from incorrectly treating y² as 2y. In exams, evaluate powers first, then multiplication and addition.

Use the distributive property: \(6(x+1)-x=6x+6-x\). Combining the like terms \(6x\) and \(-x\) gives \(5x\), so the expression equals \(5x+6\). The option \(7x+6\) results from incorrectly treating \(6x-x\) as \(7x\). Exam tip: after opening brackets, add or subtract coefficients only of like terms.

Simplifying a polynomial means combining only like terms. Like terms have the same variables with the same powers, so their coefficients can be added. In the expression \(x^2+x+x^2+x+1\), the two \(x^2\) terms combine to give \(2x^2\). The two x terms combine to give \(2x\). The number 1 is a constant and has no matching constant, so it remains 1. Therefore the simplified expression is \(2x^2+2x+1\), which is option A.

A useful check is to group the terms: \((x^2+x^2)+(x+x)+1\). Adding each group gives \(2x^2+2x+1\). We must not multiply the terms or change their powers; addition combines coefficients of identical variable parts. Thus option B incorrectly changes the coefficient and loses the x term, while C and D use incorrect powers. The grouping confirms option A.

Substituting \(n=-4\), we get \(n^2=(-4)^2=16\). Hence, \(\frac{n^2+n}{2}=\frac{16+(-4)}{2}=\frac{12}{2}=6\). Therefore, 6 is the correct answer. The distractor 8 results from wrongly ignoring the \(n=-4\) term in \(n^2+n\). Exam tip: always put a negative number in brackets before squaring it.

On expanding the brackets, \(3(a-b)+2(2a+b)=3a-3b+4a+2b\). Combining like terms gives \(3a+4a=7a\) and \(-3b+2b=-b\). Therefore, the simplified expression is \(7a-b\). The option \(7a+b\) results from combining the terms containing \(b\) incorrectly. Exam tip: multiply the number outside a bracket by every term inside it.

\(4x^2\) and \(3x^2\) are like terms, so they add to \(7x^2\). Similarly, \(-6xy\) and \(+xy\) are like terms, giving \(-6xy+xy=-5xy\). Hence, the simplified expression is \(7x^2-5xy\). \(12x^2-5xy\) is incorrect because \(4+3=7\), not 12. Exam tip: combine only terms with identical variable parts and exponents.

A constant term is a term that contains no variable. In \(2x^2+5x-1\), both \(2x^2\) and \(5x\) contain \(x\), whereas \(-1\) has no variable. Therefore, the constant term is \(-1\). The number \(5\) is the coefficient of \(5x\), not the constant term. Exam tip: identify the term with no letter or variable to find the constant term.

Answer: option A, −4a − 30. Simplify from the innermost brackets outward. First, 2(a−5)=2a−10. Therefore 4a−2(a−5) = 4a−(2a−10) = 4a−2a+10 = 2a+10. Substitute this into the original expression: 2a−3(2a+10). Now distribute −3 to both terms: −3(2a+10)=−6a−30. Combining like terms gives 2a−6a−30 = −4a−30. Thus A is correct. Option B has the wrong sign for the constant term. Options C and D have a positive coefficient of a because the negative multiplier was not distributed correctly. The common warning is that subtracting a bracket changes the signs of every term inside it.

The governing concepts are the distributive law and the rule that a minus sign outside brackets changes the signs of every term inside. Begin with the innermost product: 2(4 − x) = 8 − 2x. The square-bracket expression is therefore 3x − (8 − 2x) = 3x − 8 + 2x = 5x − 8. Substituting into the complete expression gives 10 − (5x − 8). Apply the outer subtraction to both terms: 10 − 5x + 8 = 18 − 5x. Therefore option B is correct. Option A does not combine the x-terms correctly. Option C reverses the effect of the outer subtraction, and option D ignores the coefficient 5 formed from 3x + 2x.

The governing concepts are the distributive law, nested grouping symbols, and sign reversal when subtracting a bracketed expression. First simplify the braces: 4x − {3x + 1} = 4x − 3x − 1 = x − 1. Also expand the first product: 6(x − 2) = 6x − 12. The whole expression becomes (6x − 12) − (x − 1). Because the second bracket is subtracted, both of its signs change: 6x − 12 − x + 1 = 5x − 11. Thus option A is correct. Option B keeps the constant as −13 incorrectly. Options C and D use 7x, which results from adding x instead of subtracting it after removing the outer minus sign.