Algebraic expressions

Substituting the given values: \(2x^2-xy+y^2=2(1)^2-(1)(-3)+(-3)^2\). Thus, \(2+3+9=14\). Therefore, the correct answer is 14. Option 8 may result from incorrectly treating \(-xy\) as negative despite both \(x\) and \(y\) being multiplied with a negative sign. Exam tip: the square of a negative number is positive, and \(- (1\times -3)=+3\).

A minus sign precedes the second bracket, so the signs of all its terms change: \(5x^2-(2x^2-3x)+(4x-7)=5x^2-2x^2+3x+4x-7=3x^2+7x-7\). Therefore, the coefficient of \(x\) is \(7\). The number \(-7\) is the constant term, not the coefficient of \(x\). Exam tip: When a bracket is preceded by a minus sign, change the sign of every term inside it.

Like terms have exactly the same variables and powers. In the first option, every variable part is (a^2b).

On expanding the brackets, \(3(2a-b)-2(a+3b)+5b=6a-3b-2a-6b+5b\). Combining like terms gives \(6a-2a=4a\) and \(-3b-6b+5b=-4b\). Therefore, the simplified expression is \(4a-4b\). The option \(4a+4b\) results from handling the negative signs of the \(b\)-terms incorrectly. Exam tip: when a minus sign precedes a bracket, change the sign of every term inside it.

The value of the expression is 0 when \(x=3\). So, \(2(3)^2+3k-15=0\). This gives \(18+3k-15=0\), hence \(3+3k=0\) and \(k=-1\). If \(k=1\), the expression equals 6, not 0. Exam tip: When a polynomial is stated to be zero at a given value of \(x\), substitute that value directly and solve the resulting equation.

Combining like terms gives \(4x^2y+2x^2y=6x^2y\) and \(-6xy+9xy=3xy\). Thus, the simplified expression is \(6x^2y+3xy-5\). It contains three different types of terms: an \(x^2y\) term, an \(xy\) term, and the constant term \(-5\). The terms \(6x^2y\) and \(3xy\) are not like terms because their variable parts are different. Exam tip: check both variables and their exponents before combining terms.

In (7x^2-3x), both terms contain variables and there is no constant term. It has two terms with different powers.

Use the distributive property: \(2p(p-3)=2p^2-6p\) and \(4(p+1)=4p+4\). Combining like terms gives \(2p^2-6p+4p+4=2p^2-2p+4\). Therefore, option A is correct. Option C incorrectly omits the \(4p\) term from the second bracket. Exam tip: after expanding brackets, combine only like terms with the same power of the variable.

Given \,\(x+y=5\), substitute 5 for the complete group \,\(x+y\) in \,\(2(x+y)-3\): \,\(2\times 5-3=10-3=7\). Therefore, the correct answer is 7. Option 10 results from finding \,\(2(x+y)\) but forgetting to subtract 3. Exam tip: When a grouped expression is given a value, substitute the value for the entire group.

The like terms containing \(r^2s\) are \(9r^2s\) and \(-5r^2s\). Thus, \(9r^2s-5r^2s=4r^2s\), so the coefficient of \(r^2s\) is 4. Although \(-4rs+rs=-3rs\), this is not a like term of \(r^2s\). Exam tip: Combine only terms with exactly the same variables raised to the same powers.

The phrase “adding \((x+y)\) to four times \((3x-2y)\)” means first multiply the complete expression \((3x-2y)\) by 4, and then add \((x+y)\). Hence, \(4(3x-2y)+(x+y)\) is correct. In option B, 4 multiplies the entire sum, so it represents a different expression. Exam tip: When a whole expression is multiplied, enclose it in brackets.

Add the like terms in the two polynomials: \(x^2+2x^2=3x^2\), \(2x-x=x\), and \(-3+4=1\). Therefore, the sum is \(3x^2+x+1\). The option \(3x^4+x+1\) is incorrect because powers are not added while adding polynomials; only coefficients of like terms are combined. Exam tip: Arrange terms in descending powers before combining like terms.

To subtract the second polynomial, change the sign of every term in it: \((5x^2-3x+8)-(2x^2+4x-1)=5x^2-3x+8-2x^2-4x+1\). Combining like terms gives \(3x^2-7x+9\). Hence, option B is correct. Option C results from an incorrect sign while subtracting \(4x\). Exam tip: a minus sign before brackets changes the sign of every term inside them.

Given a-b=4 and b=3, substitute b=3 to get a-3=4; hence a=7. Therefore, 2a-b=2(7)-3=14-3=11. Option 14 results from forgetting to subtract b. Exam tip: First find the value of the unknown variable by substitution, then evaluate the required expression.

Combining like terms gives \(3xy+4xy=7xy\) and \(-2x+7x=5x\). Thus, the simplified expression is \(7xy+5x-5y\), so \(9x\) does not remain. \(5x\) is a close distractor because it is correctly obtained by adding the two \(x\)-terms. Exam tip: combine only terms that have the same variables with the same powers.

First apply the distributive property: \(2x(x-4)=2x^2-8x\). Adding \(3x\) gives \(2x^2-8x+3x=2x^2-5x\). Therefore, option A is correct. In option B, \(3x\) has incorrectly been treated as the constant term \(3\). Exam tip: combine only like terms; \(-8x+3x=-5x\).

At \(x=2\), the expression is given to have value \(16\). So, \(2^2+2k+6=16\), which gives \(10+2k=16\). Hence, \(2k=6\) and \(k=3\). If \(k=2\), the value would be 14, not 16. Exam tip: Substitute the given value of \(x\) into every term before simplifying.

Combine like terms: \(2a^2+a^2=3a^2\), \(-3ab+5ab=2ab\), and \(4b^2-b^2=3b^2\). Hence, the simplified expression is \(3a^2+2ab+3b^2\). In option D, the coefficient of \(ab\) is incorrectly taken as \(8\); its correct sum is \(2\). Exam tip: add or subtract only terms with the same variables raised to the same powers.

In the given polynomial, the term containing \(x^2\) is \(-4x^2\). The number multiplying \(x^2\) is \(-4\), so its coefficient is \(-4\). The numbers \(6\), \(1\), and \(-9\) belong to the \(x^3\) term, the \(x\) term, and the constant term respectively. Exam tip: To find a coefficient, first identify the term with the required power.

For a monomial containing more than one variable, the total degree is found by adding the exponents of all variables in that term. In option A, x²y³ has exponent 2 on x and exponent 3 on y, so its total degree is 2 + 3 = 5. In option B, x³y³ has degree 3 + 3 = 6. In option C, x⁴y² has degree 4 + 2 = 6. In option D, xy² can be written as x¹y², giving degree 1 + 2 = 3. Therefore only option A satisfies the condition. The question asks for the combined power, not the exponent of one variable alone, so simply noticing a large individual exponent would be incorrect.