Algebra Questions and Answers

Answer: C

The expression equals \(-4x^{2}\).

The numerator is a difference of two squares, \(a^{2} - b^{2} = (a + b)(a - b)\), with \(a = x^{2} + y^{2} - z^{2}\) and \(b = x^{2} - y^{2} + z^{2}\):

\[ \begin{aligned} a + b &= 2x^{2} \\ a - b &= 2y^{2} - 2z^{2} = 2(y + z)(y - z) \end{aligned} \]

\[ \begin{aligned} \frac{4x^{2}(y + z)(y - z)}{(y + z)(z - y)} &= \frac{4x^{2}(y - z)}{z - y} \\ &= -4x^{2} \end{aligned} \]

The minus sign comes from \(z - y = -(y - z)\) in the denominator; missing it gives \(4x^{2}\). The options with \(2x^{2}\) drop a factor of 2 from \(a + b\) or \(a - b\).

Answer: B

One value of \((a + b)\) is \(2\frac{4}{5}\).

Use the identity linking the sum and difference of two numbers with their product.

\[ \begin{aligned} (a + b)^{2} &= (a - b)^{2} + 4ab \\ &= \frac{121}{25} + 4 \times \frac{3}{4} \\ &= \frac{121}{25} + \frac{75}{25} = \frac{196}{25} \end{aligned} \]

So \(a + b = \pm \frac{14}{5}\), and one of the values is \(\frac{14}{5} = 2\frac{4}{5}\). The question says 'one of the values' because \(-\frac{14}{5}\) also works. Forgetting the factor 4 on \(ab\) gives \(\sqrt{121/25 + 3/4}\), which is none of the options.

Answer: C

The value is 31.

Simplify the bracket first, then clear the fractions by multiplying by 12.

\[ \begin{aligned} 2x - \frac{1 - x}{2} &= \frac{5x - 1}{2} \\ \frac{7x - 1}{4} - \frac{5x - 1}{6} &= \frac{10}{3} \\ 3(7x - 1) - 2(5x - 1) &= 40 \\ 11x - 1 &= 40 \\ 11x &= 41 \end{aligned} \]

The expression asked for needs only \(11x\), so there is no need to find \(x\) itself:

\[ \frac{11x + 21}{2} = \frac{41 + 21}{2} = 31 \]

The usual slip is in the sign of the last term: \(-2(5x - 1) = -10x + 2\), not \(-10x - 2\). Getting it wrong gives \(11x = 45\) and a value of 33, which is not offered.

Answer: A

One factor is \(2x + 3y - 7\).

The first three terms form a perfect square: \(4x^2 + 12xy + 9y^2 = (2x + 3y)^2\). Put \(t = 2x + 3y\).

\[ \begin{aligned} t^2 - 5t - 14 &= (t - 7)(t + 2) \\ &= (2x + 3y - 7)(2x + 3y + 2) \end{aligned} \]

So the factors are \(2x + 3y - 7\) and \(2x + 3y + 2\). Check: \(-7 \times 2 = -14\) and \(-7 + 2 = -5\). The options with + 7, + 14 and − 2 have the wrong signs or numbers.

Answer: A

The value of (3x + 4) is 2.

Expand and solve:

\[ \begin{aligned} 15x + 20 - 48x - 56 &= 9x - 8 \\ -33x - 36 &= 9x - 8 \\ -42x &= 28 \\ x &= -\tfrac{2}{3} \end{aligned} \]

\[ 3x + 4 = 3\left(-\tfrac{2}{3}\right) + 4 = 2 \]

Check: 5(2) − 8(6x + 7) = 10 − 8(3) = −14, and 9x − 8 = −6 − 8 = −14. The other values of 3x + 4 (3, 0 and 1) do not satisfy the equation.

Answer: A

The value of (a − b) is \(\frac{1}{6}\).

Use the identity \((a - b)^2 = (a + b)^2 - 4ab\).

\[ \begin{aligned} (a - b)^2 &= \left(\frac{17}{6}\right)^2 - 4(2) \\ &= \frac{289}{36} - \frac{288}{36} = \frac{1}{36} \end{aligned} \]

Since a > b, a − b is positive, so it equals \(\frac{1}{6}\). Check: a = 3/2 and b = 4/3 add to 17/6, multiply to 2 and differ by 1/6. Stopping at \((a - b)^2 = \frac{1}{36}\) without taking the square root is the trap; 1/12, 1/3 and 2/3 do not square to 1/36.

Answer: A

The value of \((8x + 5)\) is 6.

Clear the fractions by multiplying both sides by 12, the LCM of 3 and 4.

\[ \begin{aligned} 4(5x - 7) + 24 &= 3(4x - 3) + 48x \\ 20x - 4 &= 60x - 9 \\ 5 &= 40x, \quad x = \tfrac{1}{8} \end{aligned} \]

\[ 8x + 5 = 1 + 5 = 6 \]

Check: with \(x = \frac{1}{8}\) both sides equal \(-\frac{1}{8}\). The usual error is forgetting to multiply the lone terms (2 and \(4x\)) by 12, which gives a wrong x and hence a wrong value of \(8x + 5\).

Answer: D

\((P + Q) - R\) equals \(20xy - 3x^{2} - 3y^{2}\).

Collect like terms in \(P + Q\), then subtract each term of R.

\[ \begin{aligned} P + Q &= 21xy + 2y^{2} - 4x^{2} \\ -R &= -5y^{2} + x^{2} - xy \\ \text{total} &= 20xy - 3x^{2} - 3y^{2} \end{aligned} \]

The xy terms give \(12 + 9 - 1 = 20\), the \(x^{2}\) terms give \(-18 + 14 + 1 = -3\), and the \(y^{2}\) terms give \(-10 + 12 - 5 = -3\).

Options starting with \(22xy\) add the xy term of R instead of subtracting it, and \(-7x^{2}\) does not come from any correct grouping of the \(x^{2}\) terms.

Answer: B

One value of \(\left(x - \frac{1}{x}\right)\) is 4.

Use the identity \(\left(t - \frac{1}{t}\right)^{2} = t^{2} + \frac{1}{t^{2}} - 2\) and its plus form twice.

\[ \begin{aligned} \left(x^{2} + \tfrac{1}{x^{2}}\right)^{2} &= 322 + 2 = 324 \\ x^{2} + \tfrac{1}{x^{2}} &= 18 \\ \left(x - \tfrac{1}{x}\right)^{2} &= 18 - 2 = 16 \end{aligned} \]

So \(x - \frac{1}{x} = \pm 4\), and 4 is one of its values. (\(x^{2} + \frac{1}{x^{2}}\) is positive, so it is 18, not \(-18\).)

Check: if \(x - \frac{1}{x} = 4\), then \(x^{2} + \frac{1}{x^{2}} = 16 + 2 = 18\) and \(x^{4} + \frac{1}{x^{4}} = 324 - 2 = 322\). The values 2, 6 and 8 give squares of 4, 36 and 64, none of which equals 16.

Answer: B

The value of \((2A + B - C - 2D)\) is 1.

Split the middle term: the product \(15 \times 8 = 120\) has the factor pair \(-6\) and \(-20\), which add to \(-26\).

\[ \begin{aligned} 15x^{2} - 26x + 8 &= 15x^{2} - 6x - 20x + 8 \\ &= 3x(5x - 2) - 4(5x - 2) \\ &= (3x - 4)(5x - 2) \end{aligned} \]

Taking \(A = 3, B = -4, C = 5, D = -2\):

\[ 2(3) + (-4) - 5 - 2(-2) = 6 - 4 - 5 + 4 = 1 \]

Writing the factors the other way round (\(A = 5, B = -2, C = 3, D = -4\)) gives 13, which is not offered, so the intended factorisation is the one above. 0, 2 and 3 come from sign slips with the negative B and D.