A Algebra — printable coverage pack

A1 · The basic processes of algebra, including use of the associative, commutative and distributive laws

The commutative laws allow the order to change in addition and multiplication: $a+b=b+a$ and $ab=ba$ .
The associative laws allow regrouping in addition and multiplication: $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$ .
The distributive law connects multiplication and addition: $a(b+c)=ab+ac$ .
Subtraction and division are not commutative or associative; treating them as though they are is a common source of sign and fraction errors.

Tier 1 · Easy

1. Use a basic algebraic law to work out $19\times37+19\times63$ .[2 marks]
Answer
- $1900$
Method: Use the distributive law to take out the common factor: $19\times37+19\times63=19(37+63)=19\times100=1900$ .

Tier 2 · Standard

1. Use algebraic laws to work out $25\times47+25\times55-25\times2$ without evaluating any product separately.[2 marks]
Answer
- $2500$
Method: All three terms share the factor $25$ , so distribute in reverse across both the addition and the subtraction: $25\times47+25\times55-25\times2=25(47+55-2)=25\times100=2500$ .

Tier 3 · Hard

1. Using algebraic laws, simplify $x(y-z)+z(x-y)$ to a single product.[3 marks]
Answer
- $y(x-z)$
Method: Distribute first: $x(y-z)+z(x-y)=xy-xz+zx-zy$ . Since multiplication is commutative, $xz=zx$ , so those terms cancel. The remainder is $xy-yz$ , which factorises by the distributive law to $y(x-z)$ .

A2 · Definition of a function; notation f(x)

A function assigns exactly one output to each permitted input.
In $f(x)$ , the letter $x$ is a placeholder for the input; $f(3)$ means substitute $3$ everywhere $x$ appears.
An equation such as $f(x)=k$ asks for input values whose output is $k$ ; it does not mean multiply $f$ by $x$ .
Use brackets when substituting a negative number or an algebraic expression to avoid sign errors.

Tier 1 · Easy

1. For $f(x)=4x-7$ , evaluate $f(5)$ .[1 mark]
Answer
- $f(5)=13$
Method: Substitute $x=5$ : $f(5)=4(5)-7=20-7=13$ .

Tier 2 · Standard

1. Given $f(x)=x^2-4x+1$ , work out $f(-2)$ and simplify $f(a+1)$ .[3 marks]
Answer
- $f(-2)=13$
- $f(a+1)=a^2-2a-2$
Method: $f(-2)=(-2)^2-4(-2)+1=4+8+1=13$ . Also, $f(a+1)=(a+1)^2-4(a+1)+1=a^2+2a+1-4a-4+1=a^2-2a-2$ .

Tier 3 · Hard

1. A function has the form $g(x)=ax+b$ . Given that $g(2)=7$ and $g(-1)=-2$ , work out $g(5)$ .[4 marks]
Answer
- $g(5)=16$
Method: The information gives $2a+b=7$ and $-a+b=-2$ . Subtracting the second equation from the first gives $3a=9$ , so $a=3$ . Then $-3+b=-2$ , so $b=1$ . Therefore $g(5)=3(5)+1=16$ .

A3 · Domain and range of a function

The domain is the set of allowed inputs; the range is the set of outputs actually produced.
Apply the function to every value in a finite domain, removing repeated outputs when writing the range as a set.
For a continuous restricted domain, check turning points and both endpoints before stating the range.
Values that make a denominator zero or an even root negative must be excluded from the domain; a common error is to list these as range restrictions instead.

Tier 1 · Easy

1. The function $f(x)=x+1$ has domain $\{-2,0,3\}$ . Write down its range.[1 mark]
Answer
- $\{-1,1,4\}$
Method: Apply $f(x)=x+1$ to each domain value: $f(-2)=-1$ , $f(0)=1$ and $f(3)=4$ . Hence the range is $\{-1,1,4\}$ .

Tier 2 · Standard

1. For $f(x)=x^2+2$ with domain $-2\leq x\leq3$ , work out the range of $f$ .[3 marks]
Answer
- $2\leq f(x)\leq11$
Method: The minimum of $x^2$ occurs at $x=0$ , which is in the domain, so the minimum output is $2$ . At the endpoints, $f(-2)=6$ and $f(3)=11$ , so the maximum is $11$ . Therefore $2\leq f(x)\leq11$ .

Tier 3 · Hard

1. The function $h(x)=\dfrac{6}{x-2}$ has domain $3\leq x\leq8$ . Determine its range.[3 marks]
Answer
- $1\leq h(x)\leq6$
Method: On this domain, $x-2$ is positive and increases from $1$ to $6$ , so $6/(x-2)$ decreases. The endpoint outputs are $h(3)=6$ and $h(8)=1$ . Every intermediate output occurs, so $1\leq h(x)\leq6$ .

A4 · Composite functions (the result of two or more functions acting in succession)

A composite function applies one function and then uses its output as the input to another.
AQA writes composites as $fg(x)$ , which means $f(g(x))$ : $g$ acts first and $f$ acts on the result.
To find a composite expression, substitute the entire inner function into every occurrence of the variable in the outer function.
In general $fg(x)$ and $gf(x)$ are different; reversing the order is the most common error.

Tier 1 · Easy

1. Let $f(x)=2x+1$ and $g(x)=x^2$ . Work out $fg(3)$ .[2 marks]
Answer
- $fg(3)=19$
Method: Apply $g$ first: $g(3)=3^2=9$ . Then apply $f$ : $f(9)=2(9)+1=19$ .

Tier 2 · Standard

1. Let $f(x)=x-4$ and $g(x)=3x^2$ . Find and simplify $gf(x)$ .[3 marks]
Answer
- $gf(x)=3x^2-24x+48$
Method: In $gf(x)$ , $f$ acts first: $gf(x)=g(x-4)=3(x-4)^2$ . Expanding gives $3(x^2-8x+16)=3x^2-24x+48$ .

Tier 3 · Hard

1. Let $f(x)=3x-2$ and $g(x)=x^2+1$ . Solve $fg(x)=13$ .[4 marks]
Answer
- $x=-2$ or $x=2$
Method: First form the composite: $fg(x)=3(x^2+1)-2=3x^2+1$ . Hence $3x^2+1=13$ , so $3x^2=12$ and $x^2=4$ . Therefore $x=-2$ or $x=2$ .

A5 · Inverse functions (domains chosen for f to make f one-one)

An inverse function reverses the original function, so $f^{-1}(f(x))=x$ on the chosen domain.
To find an inverse, write $y=f(x)$ , rearrange for $x$ , and then interchange the variable labels.
The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .
A function must be one-one to have an inverse function; a quadratic therefore needs a restricted domain such as one side of its turning point.

Tier 1 · Easy

1. Given $f(x)=5x-4$ , find $f^{-1}(x)$ .[2 marks]
Answer
- $f^{-1}(x)=\dfrac{x+4}{5}$
Method: Write $y=5x-4$ . Then $y+4=5x$ , so $x=(y+4)/5$ . Interchanging the labels gives $f^{-1}(x)=(x+4)/5$ .

Tier 2 · Standard

1. The function $f(x)=(x-3)^2+1$ has domain $x\geq3$ . Find $f^{-1}(x)$ and state its domain.[3 marks]
Answer
- $f^{-1}(x)=3+\sqrt{x-1}$
- Domain: $x\geq1$
Method: Write $y=(x-3)^2+1$ , so $(x-3)^2=y-1$ . The restriction $x\geq3$ requires the positive square root, giving $x=3+\sqrt{y-1}$ . Hence $f^{-1}(x)=3+\sqrt{x-1}$ . The range of $f$ is $y\geq1$ , so this is the inverse's domain.

Tier 3 · Hard

1. For $f(x)=\dfrac{2x+5}{x-1}$ , find $f^{-1}(x)$ and state the value excluded from its domain.[4 marks]
Answer
- $f^{-1}(x)=\dfrac{x+5}{x-2}$
- Excluded value: $x=2$
Method: Let $y=(2x+5)/(x-1)$ . Then $yx-y=2x+5$ , so $x(y-2)=y+5$ and $x=(y+5)/(y-2)$ . Interchange the labels to obtain $f^{-1}(x)=(x+5)/(x-2)$ . Its denominator is zero at $x=2$ , so that value is excluded.

A6 · Expanding brackets and collecting like terms

To expand brackets, multiply every term in one bracket by every term in the other.
Track signs explicitly, especially when subtracting a whole bracket: the subtraction changes every sign inside it.
Collect terms only when they have identical variable parts and powers; for example, $x^2$ and $x$ are not like terms.
For three brackets, expand two first, simplify, and then multiply by the remaining bracket.

Tier 1 · Easy

1. Expand and simplify $(x+4)(x-3)$ .[2 marks]
Answer
- $x^2+x-12$
Method: Multiply each pair of terms: $x^2-3x+4x-12$ . Collecting the linear terms gives $x^2+x-12$ .

Tier 2 · Standard

1. Expand and simplify $(2x-3)(x+5)-(x-1)^2$ .[3 marks]
Answer
- $x^2+9x-16$
Method: Expand $(2x-3)(x+5)=2x^2+7x-15$ and $(x-1)^2=x^2-2x+1$ . Subtract the entire second expression: $2x^2+7x-15-x^2+2x-1=x^2+9x-16$ .

Tier 3 · Hard

1. Expand and simplify $(x+2)(x-1)(2x-3)$ .[4 marks]
Answer
- $2x^3-x^2-7x+6$
Method: First, $(x+2)(x-1)=x^2+x-2$ . Then $(x^2+x-2)(2x-3)=2x^3-3x^2+2x^2-3x-4x+6=2x^3-x^2-7x+6$ .

A7 · Expand (a + b)^n for positive integer n; use of Pascal's triangle

The entries in row $n$ of Pascal's triangle are the coefficients in the expansion of $(a+b)^n$ .
In successive terms, the power of the first term decreases while the power of the second term increases; the two powers always add to $n$ .
Include numerical multipliers and signs inside the powers, for example terms from $(2x-1)^n$ use powers of both $2x$ and $-1$ .
A common error is to use only the Pascal coefficient and forget the powers of constants multiplying the variables.

Tier 1 · Easy

1. Expand $(x+2)^3$ .[2 marks]
Answer
- $x^3+6x^2+12x+8$
Method: Use Pascal coefficients $1,3,3,1$ : $x^3+3x^2(2)+3x(2^2)+2^3=x^3+6x^2+12x+8$ .

Tier 2 · Standard

1. Find the coefficient of $x^2$ in the expansion of $(2x-1)^4$ .[3 marks]
Answer
- Coefficient of $x^2$ : $24$
Method: The $x^2$ term uses two factors of $2x$ and two factors of $-1$ . Its Pascal coefficient is $\binom{4}{2}=6$ , so the term is $6(2x)^2(-1)^2=24x^2$ . The required coefficient is $24$ .

Tier 3 · Hard

1. The coefficient of $x^2$ in $(2x+k)^4$ is $216$ , where $k$ is an integer. Work out the possible values of $k$ .[4 marks]
Answer
- $k=-3$ or $k=3$
Method: The $x^2$ term uses two factors of $2x$ and two factors of $k$ . It is $\binom{4}{2}(2x)^2k^2=24k^2x^2$ . Hence $24k^2=216$ , so $k^2=9$ and $k=-3$ or $k=3$ .

A8 · Factorising

Factorising reverses expansion by writing an expression as a product.
Always remove the greatest common factor first, including a common power of the variable.
For a quadratic, choose factors whose product gives the constant term and whose cross-terms give the middle coefficient.
Recognise the difference of two squares: $a^2-b^2=(a-b)(a+b)$ , and continue until every factor is irreducible over the required number set.

Tier 1 · Easy

1. Factorise fully $6x^2-15x$ .[2 marks]
Answer
- $3x(2x-5)$
Method: The greatest common factor of $6x^2$ and $-15x$ is $3x$ . Taking it out gives $6x^2-15x=3x(2x-5)$ .

Tier 2 · Standard

1. Factorise $2x^2-7x-15$ .[3 marks]
Answer
- $(2x+3)(x-5)$
Method: The product $2\times(-15)=-30$ and the required sum is $-7$ , so split the middle term using $3$ and $-10$ : $2x^2+3x-10x-15=x(2x+3)-5(2x+3)=(2x+3)(x-5)$ .

Tier 3 · Hard

1. Factorise fully $x^4-5x^2+4$ .[4 marks]
Answer
- $(x-2)(x-1)(x+1)(x+2)$
Method: Treat the expression as a quadratic in $x^2$ : $x^4-5x^2+4=(x^2-1)(x^2-4)$ . Each factor is a difference of two squares, so $(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$ .

A9 · Manipulation of rational expressions: use of + - x / for algebraic fractions with numeric, linear or quadratic denominators

Factor numerators and denominators before cancelling; only common factors, not separate terms, may be cancelled.
To add or subtract algebraic fractions, use a common denominator and apply signs to the entire numerator.
Multiplication is performed across numerators and denominators; division is multiplication by the reciprocal of the divisor.
State excluded values from the original denominators, even if a factor later cancels.

Tier 1 · Easy

1. Simplify $\dfrac{x^2-9}{x+3}$ , stating the excluded value.[2 marks]
Answer
- $x-3$ , where $x\neq-3$
Method: Factor the numerator: $x^2-9=(x-3)(x+3)$ . Cancel the common factor $x+3$ to get $x-3$ . The original denominator is zero at $x=-3$ , so this value remains excluded.

Tier 2 · Standard

1. Simplify $\dfrac{2}{x-1}-\dfrac{3}{x+2}$ into one fraction.[3 marks]
Answer
- $\dfrac{7-x}{(x-1)(x+2)}$
Method: Use the common denominator $(x-1)(x+2)$ . The numerator is $2(x+2)-3(x-1)=2x+4-3x+3=7-x$ . Therefore the result is $(7-x)/((x-1)(x+2))$ .

Tier 3 · Hard

1. Simplify fully $\dfrac{x^2+5x+6}{x^2-4}\div\dfrac{x^2+6x+9}{2x-4}$ , stating the excluded values of $x$ .[4 marks]
Answer
- $\dfrac{2}{x+3}$ , where $x\neq2$ , $x\neq-2$ and $x\neq-3$
Method: Factorise everything: $\dfrac{(x+2)(x+3)}{(x-2)(x+2)}\div\dfrac{(x+3)^2}{2(x-2)}$ . Division is multiplication by the reciprocal: $\dfrac{x+3}{x-2}\times\dfrac{2(x-2)}{(x+3)^2}$ . Cancel the common factors $x-2$ and $x+3$ to get $\dfrac{2}{x+3}$ . The original denominators exclude $x=2$ and $x=-2$ , and the divisor must be non-zero, excluding $x=-3$ .

A10 · Use and manipulation of formulae and expressions

Substitute values with units consistently and keep enough exact working to avoid premature rounding.
To change the subject, undo operations in reverse order while performing the same operation on both sides.
If the new subject occurs in more than one term, collect those terms and factorise before dividing.
When removing a square or square root, consider whether the context or stated domain determines a sign.

Tier 1 · Easy

1. Rearrange $y=3x-7$ to make $x$ the subject.[2 marks]
Answer
- $x=\dfrac{y+7}{3}$
Method: Add $7$ to both sides to obtain $y+7=3x$ . Divide both sides by $3$ , giving $x=(y+7)/3$ .

Tier 2 · Standard

1. The formula $P=2\pi\sqrt{\dfrac{k}{m}}$ has positive variables. Make $k$ the subject.[3 marks]
Answer
- $k=\dfrac{mP^2}{4\pi^2}$
Method: Divide by $2\pi$ : $P/(2\pi)=\sqrt{k/m}$ . Square both sides to get $P^2/(4\pi^2)=k/m$ . Multiplying by $m$ gives $k=mP^2/(4\pi^2)$ .

Tier 3 · Hard

1. Given $y=\dfrac{3x-4}{x+2}$ , make $x$ the subject and hence find $x$ when $y=-5$ .[4 marks]
Answer
- $x=\dfrac{2y+4}{3-y}$
- When $y=-5$ , $x=-\dfrac{3}{4}$
Method: Multiply by $x+2$ : $yx+2y=3x-4$ . Collect the $x$ -terms: $x(y-3)=-4-2y$ . Thus $x=(-4-2y)/(y-3)=(2y+4)/(3-y)$ . For $y=-5$ , $x=(-10+4)/(3+5)=-6/8=-3/4$ .

A11 · Use of the factor theorem for rational values of the variable for polynomials

The factor theorem states that $x-a$ is a factor of $f(x)$ exactly when $f(a)=0$ .
For a factor $px-q$ , substitute $x=q/p$ ; the relevant root may therefore be rational rather than an integer.
Use substitution to find an unknown coefficient or confirm a factor before dividing the polynomial by that factor.
A zero remainder verifies the factor, but further factorisation is needed if the question asks for every factor or every root.

Tier 1 · Easy

1. Use the factor theorem to show that $x-2$ is a factor of $x^3-3x^2-4x+12$ .[2 marks]
Answer
- $f(2)=0$ , so $x-2$ is a factor
Method: Let $f(x)=x^3-3x^2-4x+12$ . Then $f(2)=2^3-3(2^2)-4(2)+12=8-12-8+12=0$ . By the factor theorem, $x-2$ is a factor.

Tier 2 · Standard

1. The polynomial $p(x)=x^3+2x^2+kx-6$ has factor $x+1$ . Work out $k$ .[3 marks]
Answer
- $k=-5$
Method: Since $x+1=x-(-1)$ is a factor, the factor theorem gives $p(-1)=0$ . Thus $-1+2-k-6=0$ , so $-5-k=0$ and $k=-5$ .

Tier 3 · Hard

1. Use the factor theorem, with a rational value of $x$ , to verify that $2x-1$ divides $f(x)=2x^3+x^2-13x+6$ , then factorise $f(x)$ fully.[4 marks]
Answer
- $f(\tfrac12)=0$ , so $2x-1$ is a factor; $f(x)=(2x-1)(x+3)(x-2)$
Method: For the factor $2x-1$ , substitute the rational root $x=\tfrac12$ : $f(\tfrac12)=2(\tfrac18)+\tfrac14-\tfrac{13}{2}+6=\tfrac14+\tfrac14-\tfrac{13}{2}+6=0$ , so $2x-1$ is a factor. Dividing $f(x)$ by $2x-1$ gives $x^2+x-6$ with zero remainder, and $x^2+x-6=(x+3)(x-2)$ . Therefore $f(x)=(2x-1)(x+3)(x-2)$ .

A12 · Completing the square

Completing the square rewrites a quadratic as $a(x-h)^2+k$ , making its turning point visible.
For $x^2+bx+c$ , halve $b$ inside the bracket and compensate outside: $(x+\frac b2)^2+c-\frac{b^2}{4}$ .
If the coefficient of $x^2$ is not $1$ , factor it from the quadratic and linear terms before completing the square.
A common error is to forget the compensating constant, especially when a factor remains outside the bracket.

Tier 1 · Easy

1. Write $x^2+8x+3$ in the form $(x+a)^2+b$ .[2 marks]
Answer
- $(x+4)^2-13$
Method: $x^2+8x+3=(x+4)^2-16+3=(x+4)^2-13$ .

Tier 2 · Standard

1. Write $x^2-6x+11$ in completed-square form and state the minimum point of $y=x^2-6x+11$ .[3 marks]
Answer
- $x^2-6x+11=(x-3)^2+2$
- Minimum point: $(3,2)$
Method: $x^2-6x+11=(x-3)^2-9+11=(x-3)^2+2$ . Since $(x-3)^2\ge 0$ , the minimum occurs when $x=3$ , giving $y=2$ .

Tier 3 · Hard

1. Write $2x^2+12x-5$ in the form $a(x+b)^2+c$ . Hence solve $2x^2+12x-5=13$ , giving exact answers.[4 marks]
Answer
- $2x^2+12x-5=2(x+3)^2-23$
- $x=-3\pm3\sqrt2$
Method: $2x^2+12x-5=2(x^2+6x)-5=2[(x+3)^2-9]-5=2(x+3)^2-23$ . Setting this equal to $13$ gives $2(x+3)^2=36$ , so $(x+3)^2=18$ and $x=-3\pm3\sqrt2$ .

A13 · Drawing and sketching of functions; interpretation of graphs (linear, quadratic, exponential y = ab^x and y = ab^-x, functions restricted to no more than 3 domains)

A sketch must show the function's defining shape and important features such as intercepts, turning points and asymptotes.
For $y=ab^x$ with $a>0$ and $b>1$ , the graph passes through $(0,a)$ , stays positive and approaches $y=0$ as $x$ decreases.
For $y=ab^{-x}$ with $a>0$ and $b>1$ , the graph is decreasing and still has horizontal asymptote $y=0$ .
For a function defined on separate domains, check each endpoint and use a filled point when it is included and an open point when it is excluded.

Tier 1 · Easy

1. For $y=3\times2^x$ , state the $y$ -intercept and the equation of the horizontal asymptote.[2 marks]
Answer
- $y$ -intercept: $(0,3)$
- Horizontal asymptote: $y=0$
Method: At the $y$ -intercept, $x=0$ , so $y=3\times2^0=3$ . As $x$ becomes increasingly negative, $2^x$ approaches $0$ , so the horizontal asymptote is $y=0$ .

Tier 2 · Standard

1. Sketch $y=2^{-x}+1$ . Label its $y$ -intercept and its horizontal asymptote, and state whether it is increasing or decreasing.[3 marks]
Answer
- $y$ -intercept: $(0,2)$
- Horizontal asymptote: $y=1$
- The function is decreasing
Method: When $x=0$ , $y=2^0+1=2$ . The term $2^{-x}$ approaches $0$ as $x$ increases, so the graph approaches $y=1$ from above. Since $2^{-x}$ gets smaller as $x$ increases, the curve is decreasing.

Tier 3 · Hard

1. On one set of axes, sketch the three-domain rule $f(x)=x+4$ for $x<-1$ , $f(x)=x^2$ for $-1\le x\le2$ , and $f(x)=7-x$ for $x>2$ . Mark endpoint types and solve $f(x)=3$ .[4 marks]
Answer
- Open points at $(-1,3)$ and $(2,5)$ ; closed points at $(-1,1)$ and $(2,4)$
- $x=\sqrt3$ or $x=4$
Method: Draw $y=x+4$ only for $x<-1$ , ending with an open point at $(-1,3)$ . Draw the section of $y=x^2$ from the closed point $(-1,1)$ through $(0,0)$ to the closed point $(2,4)$ . Draw $y=7-x$ for $x>2$ , beginning open at $(2,5)$ . The first domain would give $x=-1$ , which is excluded. In the middle domain, $x^2=3$ gives $x=\sqrt3$ because $-\sqrt3<-1$ . In the last domain, $7-x=3$ gives $x=4$ .

A14 · Solution of linear and quadratic equations (by factorisation, graph, completing the square or formula)

A linear equation has the unknown to the first power; keep the equation balanced while isolating it.
A quadratic equation can be solved by factorisation, a graph, completing the square or $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Before using a quadratic method, rearrange the equation into $ax^2+bx+c=0$ .
A common error with the formula is to put only $\sqrt{b^2-4ac}$ over $2a$ instead of the whole numerator.

Tier 1 · Easy

1. Solve $x^2-x-12=0$ .[2 marks]
Answer
- $x=-3$ or $x=4$
Method: $x^2-x-12=(x-4)(x+3)$ . Therefore $(x-4)(x+3)=0$ , so $x=4$ or $x=-3$ .

Tier 2 · Standard

1. Solve $2x^2+5x-4=0$ , giving exact answers.[3 marks]
Answer
- $x=\frac{-5\pm\sqrt{57}}{4}$
Method: Here $a=2$ , $b=5$ and $c=-4$ . The quadratic formula gives $x=\frac{-5\pm\sqrt{5^2-4(2)(-4)}}{2(2)}=\frac{-5\pm\sqrt{57}}{4}$ .

Tier 3 · Hard

1. The roots of $3x^2-7x-2=0$ are $p$ and $q$ , where $p>q$ . Work out the exact value of $p-q$ .[4 marks]
Answer
- $p-q=\frac{\sqrt{73}}{3}$
Method: The quadratic formula gives $x=\frac{7\pm\sqrt{(-7)^2-4(3)(-2)}}{6}=\frac{7\pm\sqrt{73}}{6}$ . Thus $p=\frac{7+\sqrt{73}}6$ and $q=\frac{7-\sqrt{73}}6$ , so $p-q=\frac{2\sqrt{73}}6=\frac{\sqrt{73}}3$ .

A15 · Algebraic and graphical solution of simultaneous equations in two unknowns, where the equations could both be linear or one linear and one second order

A simultaneous solution is a pair of values satisfying both equations; graphically it is an intersection point.
For two linear equations, eliminate one unknown or substitute an expression from one equation into the other.
When one equation is second order, substitution usually produces a quadratic, so there may be two intersection points.
Always substitute each candidate back to find its paired coordinate; do not report unpaired $x$ -values alone.

Tier 1 · Easy

1. Solve simultaneously $2x+y=7$ and $x-y=2$ .[2 marks]
Answer
- $x=3$ , $y=1$
Method: Add the equations to eliminate $y$ : $3x=9$ , so $x=3$ . Substituting into $x-y=2$ gives $3-y=2$ , so $y=1$ .

Tier 2 · Standard

1. Solve simultaneously $y=x+2$ and $y=x^2-4$ .[3 marks]
Answer
- $(x,y)=(-2,0)$ or $(3,5)$
Method: Equate the two expressions for $y$ : $x^2-4=x+2$ , so $x^2-x-6=0$ . Factorising gives $(x-3)(x+2)=0$ , hence $x=3$ or $x=-2$ . Using $y=x+2$ gives the pairs $(3,5)$ and $(-2,0)$ .

Tier 3 · Hard

1. Solve simultaneously $x+y=7$ and $xy=10$ .[4 marks]
Answer
- $(x,y)=(2,5)$ or $(5,2)$
Method: From $x+y=7$ , write $y=7-x$ . Substitute into $xy=10$ : $x(7-x)=10$ , so $x^2-7x+10=0$ . Since $(x-2)(x-5)=0$ , $x=2$ or $x=5$ . Then $y=7-x$ , giving $(2,5)$ or $(5,2)$ .

A16 · Algebraic solution of linear equations in three unknowns

A solution in three unknowns must satisfy all three linear equations at the same time.
Eliminate the same unknown from two different pairs of equations to obtain two equations in two unknowns.
Solve the resulting pair, then substitute back into an original equation to find the third unknown.
Sign errors are common during elimination, so check the final triple in all three original equations.

Tier 1 · Easy

1. Solve $x+y+z=9$ , $x-y=2$ and $z=3$ .[2 marks]
Answer
- $x=4$ , $y=2$ , $z=3$
Method: Using $z=3$ in the first equation gives $x+y=6$ . Together with $x-y=2$ , addition gives $2x=8$ , so $x=4$ . Then $y=2$ and $z=3$ .

Tier 2 · Standard

1. Solve $x+y+z=6$ , $2x-y+z=3$ and $x+2y-z=2$ .[3 marks]
Answer
- $x=1$ , $y=2$ , $z=3$
Method: Subtract the first equation from the second: $x-2y=-3$ , so $x=2y-3$ . From the first equation, $z=6-x-y=9-3y$ . Substitute both into the third: $(2y-3)+2y-(9-3y)=2$ , so $7y=14$ and $y=2$ . Hence $x=1$ and $z=3$ .

Tier 3 · Hard

1. Solve $2x+3y-z=-2$ , $x-y+2z=9$ and $3x+2y+2z=10$ .[4 marks]
Answer
- $x=2$ , $y=-1$ , $z=3$
Method: Add twice the first equation to the second to eliminate $z$ : $5x+5y=5$ , so $x+y=1$ . Subtract the second equation from the third to get $2x+3y=1$ . Subtracting $2(x+y)=2$ gives $y=-1$ , so $x=2$ . Substitution into $x-y+2z=9$ gives $3+2z=9$ , hence $z=3$ .

A17 · Solution of linear and quadratic inequalities

Solve a linear inequality like an equation, but reverse the inequality sign when multiplying or dividing by a negative number.
For a quadratic inequality, find the roots and use the sign of the quadratic in the intervals they create.
A positive-leading quadratic is negative between two distinct roots and positive outside them.
Use open endpoints for $<$ or $>$ and included endpoints for $\le$ or $\ge$ .

Tier 1 · Easy

1. Solve $5-2x<11$ .[2 marks]
Answer
- $x>-3$
Method: Subtract $5$ to get $-2x<6$ . Dividing by $-2$ reverses the inequality, giving $x>-3$ .

Tier 2 · Standard

1. Solve $(x-4)(x+1)\le0$ .[3 marks]
Answer
- $-1\le x\le4$
Method: The roots are $x=-1$ and $x=4$ . The quadratic has a positive $x^2$ coefficient, so it is non-positive between the roots. Equality is allowed, hence $-1\le x\le4$ .

Tier 3 · Hard

1. Find the values of $x$ satisfying both $x^2-5x-6>0$ and $2x+1\le9$ .[4 marks]
Answer
- $x<-1$
Method: $x^2-5x-6=(x-6)(x+1)$ , so the first inequality gives $x<-1$ or $x>6$ . The second inequality gives $2x\le8$ , so $x\le4$ . Intersecting these sets leaves $x<-1$ .

A18 · Index laws, including fractional and negative indices and the solution of equations

For the same base, $a^m a^n=a^{m+n}$ , $a^m/a^n=a^{m-n}$ and $(a^m)^n=a^{mn}$ .
A negative index means a reciprocal: $a^{-n}=1/a^n$ for $a\ne0$ .
A fractional index represents a root: $a^{1/n}=\sqrt[n]{a}$ and $a^{m/n}=\sqrt[n]{a^m}$ when defined.
To solve an index equation, rewrite both sides with the same base before equating exponents.

Tier 1 · Easy

1. Work out $16^{-1/2}$ .[1 mark]
Answer
- $\frac14$
Method: $16^{1/2}=4$ , and the negative index takes the reciprocal, so $16^{-1/2}=1/4$ .

Tier 2 · Standard

1. Solve $9^{x+1}=27^{2x-1}$ .[3 marks]
Answer
- $x=\frac54$
Method: Write both sides with base $3$ : $3^{2x+2}=3^{6x-3}$ . Therefore $2x+2=6x-3$ , so $5=4x$ and $x=5/4$ .

Tier 3 · Hard

1. Given $a>0$ and $b>0$ , simplify fully $\frac{(81a^{-8}b^6)^{1/4}}{3a^{-1}b^{1/2}}$ .[4 marks]
Answer
- $\frac ba$
Method: The numerator is $81^{1/4}a^{-8/4}b^{6/4}=3a^{-2}b^{3/2}$ . Dividing by $3a^{-1}b^{1/2}$ gives $a^{-1}b=b/a$ .

A19 · Algebraic proof

An algebraic proof starts from a general representation, not from a selection of examples.
Write an even integer as $2n$ and an odd integer as $2n+1$ , where $n$ is an integer.
To prove divisibility by $k$ , rearrange the expression into $k$ multiplied by an integer.
State the conclusion explicitly and keep the integer assumptions visible; unexplained numerical examples do not prove a general claim.

Tier 1 · Easy

1. Prove algebraically that the sum of two odd integers is even.[2 marks]
Answer
- $(2m+1)+(2n+1)=2(m+n+1)$ , so the sum is even
Method: Let the two odd integers be $2m+1$ and $2n+1$ , where $m$ and $n$ are integers. Their sum is $2m+2n+2=2(m+n+1)$ . Since $m+n+1$ is an integer, the sum is even.

Tier 2 · Standard

1. Prove that the product of three consecutive integers is divisible by $6$ .[3 marks]
Answer
- $n(n+1)(n+2)$ contains a factor of $2$ and a factor of $3$ , so it is divisible by $6$
Method: Represent the integers as $n$ , $n+1$ and $n+2$ . Among any three consecutive integers, one is a multiple of $3$ . At least one is even, so the product also has a factor of $2$ . Since $2$ and $3$ are coprime, $n(n+1)(n+2)$ is divisible by $6$ .

Tier 3 · Hard

1. Given $x\ne0$ , prove algebraically that $x^2+\frac1{x^2}\ge2$ .[4 marks]
Answer
- $x^2+\frac1{x^2}-2=(x-\frac1x)^2\ge0$ , so $x^2+\frac1{x^2}\ge2$
Method: For real $x\ne0$ , the square $(x-1/x)^2$ is non-negative. Expanding gives $x^2-2+1/x^2\ge0$ . Adding $2$ to both sides gives $x^2+1/x^2\ge2$ , as required.

A20 · Using nth terms of sequences; limiting value of a sequence as n approaches infinity

An nth-term formula gives a sequence value directly by substituting a positive integer for $n$ .
For a rational expression in $n$ , divide numerator and denominator by the highest power of $n$ to find its limiting value.
Terms containing $1/n$ , $1/n^2$ and higher negative powers approach $0$ as $n$ approaches infinity.
The limiting value describes what the terms approach; it need not itself be a term of the sequence.

Tier 1 · Easy

1. State the limiting value of $u_n=\frac{5n-2}{n}$ as $n$ approaches infinity.[1 mark]
Answer
- $5$
Method: $u_n=5-2/n$ . As $n$ approaches infinity, $2/n$ approaches $0$ , so $u_n$ approaches $5$ .

Tier 2 · Standard

1. For $v_n=\frac{3n^2+2}{n^2+4n}$ , work out $v_2$ and the limiting value as $n$ approaches infinity.[2 marks]
Answer
- $v_2=\frac76$
- Limiting value: $3$
Method: Substitution gives $v_2=\frac{3(2)^2+2}{(2)^2+4(2)}=\frac{14}{12}=\frac76$ . For the limit, divide top and bottom by $n^2$ : $v_n=\frac{3+2/n^2}{1+4/n}$ . Both $2/n^2$ and $4/n$ approach $0$ , so the limiting value is $3$ .

Tier 3 · Hard

1. The sequence $w_n=\frac{2n+1}{n+3}$ has limiting value $L$ . Find $L$ and the least positive integer $n$ for which $w_n$ differs from $L$ by less than $0.01$ .[4 marks]
Answer
- $L=2$
- Least $n=498$
Method: Dividing by $n$ shows $L=2$ . Also $2-w_n=\frac{2n+6-(2n+1)}{n+3}=\frac5{n+3}$ , which is positive. Differing by less than $0.01$ therefore means $5/(n+3)<1/100$ . Hence $n+3>500$ , so $n>497$ and the least integer is $498$ .

A21 · nth terms of linear sequences

A linear sequence has a constant first difference and nth term $an+b$ .
The coefficient $a$ is the common difference; use any known term to find $b$ .
Equivalently, if the first term is $u_1$ and the common difference is $d$ , then $u_n=u_1+(n-1)d$ .
Check a formula against at least two supplied terms, since an indexing error often changes the constant term.

Tier 1 · Easy

1. Find the nth term of $7,11,15,19,\ldots$ .[2 marks]
Answer
- $4n+3$
Method: The common difference is $4$ , so start with $4n$ . This gives $4$ when $n=1$ , which is $3$ below the first term $7$ . Therefore the nth term is $4n+3$ .

Tier 2 · Standard

1. A linear sequence has 12th term $-17$ and 30th term $-71$ . Find its nth term.[3 marks]
Answer
- $19-3n$
Method: There are $18$ steps from the 12th to the 30th term, and the total change is $-71-(-17)=-54$ . Thus the common difference is $-54/18=-3$ . Writing the term as $-3n+b$ and using $n=12$ gives $-36+b=-17$ , so $b=19$ .

Tier 3 · Hard

1. A linear sequence has 5th term $18$ . The sum of its 8th and 12th terms is $76$ . Find the nth term and determine which term equals $202$ .[4 marks]
Answer
- $u_n=4n-2$
- $202$ is the 51st term
Method: Let $u_n=a+(n-1)d$ . Then $a+4d=18$ . Also $(a+7d)+(a+11d)=76$ , so $2a+18d=76$ . Doubling the first equation gives $2a+8d=36$ ; subtraction gives $10d=40$ , hence $d=4$ and $a=2$ . Thus $u_n=2+4(n-1)=4n-2$ . Solving $4n-2=202$ gives $n=51$ .

A22 · nth terms of quadratic sequences

A quadratic sequence has constant second differences and nth term $an^2+bn+c$ .
If the constant second difference is $D$ , then the coefficient of $n^2$ is $a=D/2$ .
Subtract the values of $an^2$ from the sequence; the remainder is linear and can be written as $bn+c$ .
A common error is to use the second difference itself as $a$ instead of halving it.

Tier 1 · Easy

1. Find the nth term of $4,13,28,49,\ldots$ .[2 marks]
Answer
- $3n^2+1$
Method: The first differences are $9,15,21$ , so the second difference is $6$ and the $n^2$ coefficient is $3$ . Subtracting $3n^2$ from the terms leaves $1$ each time, so the nth term is $3n^2+1$ .

Tier 2 · Standard

1. Find the nth term of $2,9,20,35,54,\ldots$ .[3 marks]
Answer
- $2n^2+n-1$
Method: The first differences are $7,11,15,19$ , so the second difference is $4$ and the leading term is $2n^2$ . Subtracting $2n^2$ from the sequence gives $0,1,2,3,4$ , whose nth term is $n-1$ . Therefore the original nth term is $2n^2+n-1$ .

Tier 3 · Hard

1. A quadratic sequence has nth term $an^2+bn+c$ . Its 1st, 2nd and 4th terms are $6$ , $15$ and $45$ . Work out its 10th term.[4 marks]
Answer
- $u_n=2n^2+3n+1$
- $u_{10}=231$
Method: The data give $a+b+c=6$ , $4a+2b+c=15$ and $16a+4b+c=45$ . Subtracting consecutive equations gives $3a+b=9$ and $12a+2b=30$ , so $6a+b=15$ . Hence $3a=6$ and $a=2$ ; then $b=3$ and $c=1$ . Therefore $u_{10}=2(10)^2+3(10)+1=231$ .