Algebra Questions and Answers

Answer: C

The solution set is shown in (C): all x from a point just to the right of 0 onwards.

Multiply both sides by 3 and collect terms:

\[ \begin{aligned} 3(2 - 5x) &\le -(6x - 5) \\ 6 - 15x &\le -6x + 5 \\ 1 &\le 9x \\ x &\ge \frac{1}{9} \end{aligned} \]

So the bold part starts at a filled dot at the small positive value 1/9 and runs to the right, which is line (C).

Line (A) shows x at most a positive value, the inequality reversed. Line (B) starts the set at a negative value, which no correct rearrangement gives, since 1/9 is positive. Line (D) shows two separate rays, the shape of a quadratic inequality, not a linear one.

Answer: A

The correct option is f(2,2) = g(2,2).

f(p,q) is p raised to the power q, and g(p,q) is a tower of q copies of p. With two terms each:

\[ \begin{aligned} f(2,2) &= 2^{2} = 4 \\ g(2,2) &= 2^{2} = 4 \end{aligned} \]

The other options are false. f(g(2,2),2) = f(4,2) = 16 and f(2,g(2,2)) = f(2,4) = 16, so neither is less than the other. g(2,1) = 2 = f(2,1) by definition, so they are equal. f(3,2) = 9 while g(3,2) = \(3^{3}\) = 27, so f(3,2) is smaller, not greater.

Answer: D

Multiply numerators and denominators separately, then cancel.

\[ \frac{3x}{2y} \cdot \frac{x}{3y} = \frac{3x^{2}}{6y^{2}} = \frac{x^{2}}{2y^{2}} \]

The 3 in the numerator cancels one factor of the 6 - leaving it in gives \( \tfrac{x^{2}}{6y^{2}} \).

Answer: B

Only the \( y \) terms are alike.

\[ 13y + 3y = 16y \]

\[ \therefore\ 16y + 2z \]

The \( z \) term is unlike and stays as it is; it cannot be merged with the \( y \) terms.

Answer: B

\[ 4 \times (-3b) = -12b \]

A positive times a negative is negative, and the numbers multiply as 4 and 3 to give 12.

Answer: A

Both terms share a factor of 9.

\[ 54x - 45y = 9 \times 6x - 9 \times 5y = 9(6x - 5y) \]

9 is the highest common factor of 54 and 45; taking out 3 alone would leave \( 3(18x - 15y) \), which is not fully factorised.

Answer: B

Every term shares the factor \( 5a \).

\[ 5ab + 10ac + 15ad = 5a(b + 2c + 3d) \]

Checking: \( 5a \times b = 5ab \), \( 5a \times 2c = 10ac \), \( 5a \times 3d = 15ad \). Taking out 5 alone would leave the \( a \) behind in each term.

Answer: B

Expand each bracket, then collect.

\[ 4(a + 2b - 3c) = 4a + 8b - 12c \]

\[ -3(2b - 3c) = -6b + 9c \]

\[ (4a + 8b - 12c) + (-6b + 9c) + 2a = 6a + 2b - 3c \]

The \( -3 \) turns \( -3c \) into \( +9c \), which is what leaves only \( -3c \) at the end rather than \( -21c \).

Answer: A

Dividing by a fraction means multiplying by its reciprocal.

\[ \frac{1}{x^{2}-1} \div \frac{1}{x+1} = \frac{1}{x^{2}-1} \times (x+1) \]

Factorising the denominator:

\[ = \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1} \]

The \( (x+1) \) cancels, which is why the difference of squares has to be split before the simplification becomes visible.

Answer: B

\[ (x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3} \]

The two middle terms share a factor of \( 3xy \):

\[ 3x^{2}y + 3xy^{2} = 3xy(x + y) \]

\[ \therefore\ (x+y)^{3} = x^{3} + 3xy(x+y) + y^{3} \]

The sign inside the bracket must match the original: \( (x-y)^{3} = x^{3} - 3xy(x-y) - y^{3} \).