Algebraic expressions

To subtract the second expression, change the sign of each of its terms: \((3x^2-2x+7)-(x^2+4x-5)=3x^2-2x+7-x^2-4x+5\). Combining like terms gives \(2x^2-6x+12\), so option A is correct. In option B, the \(4x\) term has effectively been added instead of subtracted. Exam tip: When a minus sign precedes brackets, change the signs of all terms inside the brackets.

First expand the brackets: \(7(2x-3)=14x-21\) and \(-4(3x+5)=-12x-20\). Combining like terms gives \(14x-12x-21-20=2x-41\). Therefore, the correct answer is \(2x-41\). In \(2x-1\), the constant terms have been combined incorrectly. Exam tip: When a negative multiplier is before a bracket, apply it to every term inside the bracket.

Substituting x=-4 gives 3x^2+2x-7=3(-4)^2+2(-4)-7. Since (-4)^2=16, we get 3×16-8-7=48-15=33. Therefore, 33 is correct. A value such as 49 can result from a sign error while handling 2x and 7. Exam tip: always write a negative value in brackets before squaring it.

Combining like terms gives (9a^2-4a^2=5a^2), (-6a+13a=7a), and (11-8=3). Group terms by powers.

We subtract \((4x-15)\) from \((11x-9)\): \((11x-9)-(4x-15)\). Since there is a minus sign before the second bracket, the signs of both terms inside it change: \(11x-9-4x+15\). Combining like terms gives \(11x-4x=7x\) and \(-9+15=6\), so the result is \(7x+6\). In \(7x-24\), the sign of \(-15\) has been handled incorrectly. Exam tip: while removing a bracket preceded by a minus sign, change the sign of every term inside it.

On substitution, \(p^2q=3^2\times(-2)=-18\), \(-2pq^2=-2\times3\times(-2)^2=-24\), and \(q=-2\). Therefore, \(-18-24-2=-44\). The value \(-40\) may result from incorrectly omitting the final \(q=-2\) term. Exam tip: the square of a negative number is positive, so \((-2)^2=4\).

Using the distributive property, \(6(3m-2n)=18m-12n\) and \(-5(m-4n)=-5m+20n\). Thus, \(18m-12n-5m+20n=13m+8n\). Therefore, \(13m+8n\) is correct. \(23m+8n\) results from incorrectly treating \(-5m\) as positive. Exam tip: when a minus sign precedes brackets, apply it to every term inside the brackets.

The difference of \(x\) and \(5\) is \(x-5\). Three times this difference is \(3(x-5)\). Adding \(2x\) gives \(3(x-5)+2x\), so option C is correct. In option A, only \(x\) is multiplied by 3, not the complete difference \((x-5)\). Exam tip: When a multiple of a difference is required, keep the whole difference in brackets.

Substitute \(r=-3\): \(2r^3+5r^2-4r=2(-3)^3+5(-3)^2-4(-3)\). Thus, \(2(-27)+5(9)+12=-54+45+12=3\). Therefore, the correct answer is 3. Getting \(9\) is a common error caused by mishandling powers or coefficients. Exam tip: an odd power of a negative number is negative, while an even power is positive.

First simplify the innermost grouping: \(3x-\{2x+7\}=3x-2x-7=x-7\). Then, \(5x-[x-7]=5x-x+7=4x+7\). Hence, the correct answer is \(4x+7\). The distractor \(4x-7\) results from forgetting that the minus sign before \([x-7]\) changes both signs inside it. Exam tip: When a bracket is preceded by a minus sign, change the sign of every term inside the bracket.

Combining like terms gives 8ab + 6ab = 14ab and -5a - 9a = -14a. Therefore, the simplified expression is 14ab - 14a. The terms 14ab and -14a cannot be combined because their variable parts are different. Exam tip: Add or subtract only terms with the same variables raised to the same powers.

Given \(a+b=12\) and \(ab=20\), substitute these values directly: \(4(a+b)-3ab=4\times12-3\times20=48-60=-12\). Hence, the correct answer is \(-12\). Option 8 may result from an error while subtracting 60 from 48. Exam tip: When values of compound terms such as \(a+b\) and \(ab\) are given, substitute each complete term directly into the expression.

Using the distributive property, \(3x(4x-5)=12x^2-15x\) and \(2x(5x+1)=10x^2+2x\). Therefore, the full expression is \(12x^2-15x-(10x^2+2x)=2x^2-17x\). The option \(22x^2-17x\) results from adding \(12x^2\) and \(10x^2\) instead of subtracting them. Exam tip: when a minus sign occurs before brackets, change the sign of every term inside the brackets.

Both terms are like terms because each has the variable part x^2. So, add only their coefficients: 4 + (-11) = -7. Hence, the sum is -7x^2. The option -7x^4 is incorrect because adding like terms does not change the exponent of x. Exam tip: Before adding terms, check that both the variable and its exponent are the same, then add the coefficients.

On substituting \(x=6\), \(x^2=36\) and \(3x=18\). Thus, \(\frac{x^2-3x+6}{3}=\frac{36-18+6}{3}=\frac{24}{3}=8\). Therefore, 8 is the correct option. The value 6 can result from mistakenly leaving out the \(+6\) in the numerator. Exam tip: evaluate the entire numerator before dividing by the denominator.

In the expression, \(-(4x-7)=-4x+7\) and \(3(x-5)=3x-15\). Therefore, \(12-4x+7+3x-15=-x+4\). Hence, \(-x+4\) is correct. \(x+4\) results from incorrectly handling the minus sign before the first bracket. Exam tip: When a bracket is preceded by a minus sign, reverse the signs of all terms inside it.

The perimeter of a triangle is the sum of its three sides: \((3x+2)+(2x-7)+(x+9)=3x+2x+x+2-7+9=6x+4\). Therefore, the correct expression is \(6x+4\). The option \(6x+18\) incorrectly adds the constant terms; \(2-7+9=4\). Exam tip: Combine like terms separately—first the terms containing \(x\), then the constants.

First simplify the innermost bracket: \(4p-(p+6)=4p-p-6=3p-6\). Then, \(3p-(3p-6)=3p-3p+6=6\). Therefore, the whole expression becomes \(6p-6\). The result \(6p+6\) occurs if the minus sign before a bracket is not distributed correctly. Exam tip: when a bracket is preceded by a minus sign, change the sign of every term inside it.

Substituting the given values: (u^2v+uv^2-3u) = 2^2(-5)+2(-5)^2-3(2) = 4(-5)+2(25)-6 = -20+50-6 = 24. Therefore, the correct answer is 24. Note that (-5)^2 = 25 because the square of a negative number is positive. Exam tip: Evaluate powers first, then perform multiplication and addition or subtraction.

The number 6 more than \(y\) is \(y+6\). Since the question asks for the square of this entire number, the expression is \((y+6)^2\). In \(y^2+6\), only \(y\) is squared, not the complete sum. Exam tip: When a phrase says “square of,” put the entire quantity in parentheses before squaring it.