Algebraic expressions

(\frac{x}{5}=\frac{2x}{10}), so (\frac{7x}{10}-\frac{2x}{10}=\frac{x}{2}). Make a common denominator for fraction terms.

Given \(x+y=8\) and \(x-y=2\), substitute these values directly into the expression: \(4(x+y)-5(x-y)=4\times8-5\times2=32-10=22\). The value 24 may result from handling the subtraction term \(5(x-y)\) incorrectly. Exam tip: When values of grouped expressions such as \(x+y\) and \(x-y\) are given, substitute them directly as complete units.

Combining (x^2), (xy), and (y^2) terms separately gives (3x^2+5xy-5y^2). Do not mix different variable parts.

Substituting \(t=3\), we get \(2t^4-5t^3+t^2=2(3^4)-5(3^3)+3^2\). Since \(3^4=81\), \(3^3=27\), and \(3^2=9\), the value is \(2\times81-5\times27+9=162-135+9=36\). Therefore, \(36\) is correct. A value such as \(18\) can result from incorrectly evaluating powers or coefficients. Exam tip: calculate powers first, then perform multiplication and subtraction.

The sum is (a+b), and five times it is (5(a+b)). Subtracting the difference (a-b) gives (5(a+b)-(a-b)).

Like terms must have identical variables with identical exponents. Here, \(5a^2b=a^2b^1\), but \(8ab=a^1b^1\); the power of \(a\) differs. Coefficients may differ in like terms. Exam tip: compare every exponent.

Substitute \(p=-1\): \(p^4=1\), \(p^3=-1\), and \(p^2=1\). Therefore, \(2p^4-3p^3+4p^2-5=2(1)-3(-1)+4(1)-5=2+3+4-5=4\). Hence, option B is correct. \(2\) is only the value of the first term, \(2p^4\), not of the whole expression. Exam tip: an even power of \(-1\) is \(1\), while an odd power is \(-1\).

In \((6x-3y)-(2x+8y)\), the sign of every term in the second expression changes: \(6x-3y-2x-8y\). Combining like terms gives \(6x-2x=4x\) and \(-3y-8y=-11y\), so the result is \(4x-11y\). In option C, the sign of \(8y\) has not been changed correctly. Exam tip: When a minus sign appears before brackets, change the signs of all terms inside the brackets.

Combine like terms: \(4x^2y+5x^2y=9x^2y\) and \(-9xy^2+6xy^2=-3xy^2\). Hence, the simplified form is \(9x^2y-3xy^2\). The terms \(x^2y\) and \(xy^2\) are not like terms because the powers of \(x\) and \(y\) differ, so they cannot be combined. Exam tip: before adding terms, check that every variable and its exponent match.

On substituting \(q=4\), \(q^3=4^3=64\) and \(2q=2\times4=8\). Thus, \(\frac{q^3-2q}{4}=\frac{64-8}{4}=\frac{56}{4}=14\). Therefore, 14 is the correct option. A result such as 16 can arise from an error in finding or subtracting \(2q\). Exam tip: evaluate the exponent first, simplify the numerator, and then divide.

If \(n\) is odd, the four consecutive odd numbers are \(n\), \(n+2\), \(n+4\), and \(n+6\). Their sum is \(n+(n+2)+(n+4)+(n+6)=4n+12\). \(4n+6\) is not the sum of all four terms. Exam tip: consecutive odd numbers differ by 2.

Simplify the brackets first: \(3(4x-5)=12x-15\). In the square bracket, \(2x-3(1-x)=2x-3+3x=5x-3\). Therefore, the complete expression is \(12x-15-(5x-3)=12x-15-5x+3=7x-12\). The option \(7x-18\) can result from incorrectly treating \(-(-3)\) as \(-3\) in the final subtraction. Exam tip: when removing a bracket preceded by a minus sign, change the signs of every term inside it.

Given x=1 and y=3, we get x+y=4. Therefore, 3(x+y)^2=3×4^2=3×16=48, and 4xy=4×1×3=12. Hence, the value is 48−12=36. A value such as 38 can result from an error while squaring or multiplying. Exam tip: after substitution, evaluate brackets first, then powers, multiplication, and subtraction.

Combine like terms: the a-terms give 4a+9a-6a=7a, and the b-terms give -5b-2b=-7b. Hence, the simplified form is 7a-7b. In 19a-7b, the term -6a has not been subtracted while combining the a-terms. Exam tip: add or subtract only terms having the same variable and exponent.

The perimeter of a rectangle is \(2(\text{length}+\text{breadth})\). Thus, \(2[(4x-7)+(3x+2)] = 2(7x-5) = 14x-10\). Therefore, the correct expression is \(14x-10\). The expression \(7x-5\) is only the sum of the length and breadth, not the perimeter. Exam tip: A rectangle has two equal lengths and two equal breadths, so multiply \(l+b\) by 2.

Substituting x=1 gives 6(1)^3-9(1)^2+4(1)-8. Since every positive power of 1 is 1, this becomes 6-9+4-8=-7. Therefore, the correct answer is -7. The value -5 can result from an error while adding or subtracting the terms. Exam tip: after substitution, retain the sign of every term and simplify step by step.

First simplify the innermost bracket: \(2x-(x-6)=2x-x+6=x+6\). Then \(7x-3(x+6)=7x-3x-18=4x-18\). Hence, the correct answer is \(4x-18\). The option \(4x+18\) results from incorrectly multiplying \(-3\) by \(+6\). Exam tip: When removing brackets preceded by a minus sign, check the sign of every term carefully.

To subtract, change the sign of every term in the second expression: \((5x^2-3x+8)-(2x^2-7x-6)=5x^2-3x+8-2x^2+7x+6\). Combining like terms gives \(3x^2+4x+14\). Therefore, option A is correct. In option C, the sign of \(-7x\) has not been changed to \(+7x\). Exam tip: When a minus sign appears before brackets, reverse the signs of all terms inside the brackets.

Substituting x=2 gives 2(2+5)-3(2-1). Thus, 2×7-3×1=14-3=11, so the correct answer is 11. A value such as 13 may result from not subtracting the term 3(2-1) correctly. Exam tip: after substitution, evaluate brackets first, then multiplication and subtraction.

First simplify the innermost brackets: \(2x-(x-2)=2x-x+2=x+2\). The expression becomes \(2(5x-3)-4(x+2)\). Expanding gives \(10x-6-4x-8=6x-14\). Hence, \(6x-14\) is correct. A choice such as \(14x-14\) can result from not distributing \(-4\) across the complete term \((x+2)\). Exam tip: simplify nested brackets from the innermost bracket outward.

The governing concept is subtraction of algebraic expressions by distributing a negative sign through every term inside the bracket. Start with 2a² + 3ab − b² − (a² − 4ab + 2b²). The subtraction changes the bracket to −a² + 4ab − 2b², so the full expression becomes 2a² + 3ab − b² − a² + 4ab − 2b². Now combine like terms: 2a² − a² = a², 3ab + 4ab = 7ab, and −b² − 2b² = −3b². Thus the simplified result is a² + 7ab − 3b², making option B correct. The other choices arise from retaining an incorrect sign or combining unlike or incorrectly signed terms.

Because \(x\neq0\), \(\frac{x}{x}=1\). Therefore, \(x^2+\frac{x}{x}=x^2+1\). Option \(x^2+x\) is incorrect because \(\frac{x}{x}\) equals 1, not \(x\). Exam tip: before cancelling terms in a fraction, ensure that the denominator is non-zero.

As \(x\neq0\), \(\frac{2x^2}{x}=2x^{2-1}=2x\). Therefore, the expression becomes \(x^3+2x\). The option \(x^2+2x\) incorrectly changes \(x^3\) to \(x^2\). Exam tip: subtract exponents only within the term where powers of the same base are being divided.

Factorising gives \(x^2-1=(x-1)(x+1)\). Since \(x\neq-1\), \(\frac{(x-1)(x+1)}{x+1}=x-1\). Hence the expression is \(x^2+(x-1)=x^2+x-1\). The close distractor \(x^2+x+1\) has the wrong sign for the constant term. Exam tip: factorise before cancelling and retain values that make the original denominator zero.

Using the difference-of-squares identity, \(x^2-4=(x-2)(x+2)\). Since \(x\neq2\), the factor \(x-2\) can be cancelled, giving \(\frac{x^2-4}{x-2}=x+2\). Therefore, the expression is \(x^3+(x+2)=x^3+x+2\), so option A is correct. Option B has an incorrect sign on the constant term. Exam tip: even after cancellation, retain the original restriction \(x\neq2\).