2.c1.3 algebra and functions 3

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Contents
Linear simultaneous equations
Simultaneous equations involving one linear and
one quadratic equation
Linear inequalities
Quadratic inequalities
Polynomials
Examination-style questions

An equation with two unknowns has an infinite number of
solution pairs. For example:
x + y = 3
is true when x = 1 and y = 2
x = –4 and y = 7
x = 6.4 and y = –3.4 and so on.
We can represent the set of
solutions graphically.
0
3
3 x
y
The coordinates of every point
on the line satisfy the equation.
x + y = 3

Similarly, an infinite number of solution pairs exist for the
equation
y – x = 1
Again, we can represent the set of
solutions graphically.
There is one pair of values that
satisfies both these equations
simultaneously.
This pair of values corresponds to
the point where the lines x + y = 3
and y – x = 1 intersect:
This is the point (1, 2). At this point x = 1 and y = 2.
0 x
y
-1
1
y – x = 1
x + y = 3

Two linear equations with two unknowns, such as x and y, can
be solved simultaneously to give a single pair of solutions.
Linear simultaneous equations can be solved algebraically
using:
The substitution method.
The elimination method, or
The solution to the equations can be illustrated graphically by
finding the points where the two lines representing the
equations intersect.
When will a pair of linear simultaneous
equations have no solutions?
In the case where the lines corresponding to the equations are
parallel, they will never intersect and so there are no solutions.

The elimination method
If two equations are true for the same values, we can add or
subtract them to give a third equation that is also true for the
same values. For example:
Subtracting gives: 3x + 7y = 22
3x + 4y = 10–
3y = 12
The terms in x have
been eliminated.
y = 4
Substituting y = 4 into the first equation gives:
3x + 28 = 22
x = –2
3x = –6
Solve the simultaneous equations
3x + 7y = 22 and 3x + 4y = 10.

We can check whether x = –2 and y = 4 solves both
3x + 7y = 22
3x + 4y = 10
by substituting them into the second equation.
LHS = 3 × –2 + 4 × 4
= –6 + 16
= 10
= RHS
So the solution is x = –2, y = 4.

Solve: 5x – 2y = 31
4x + 3y = 11
Sometimes we need to multiply one or both of the equations
before we can eliminate one of the variables. For example:
We need to have the same number in front of either the x or
the y terms before adding or subtracting the equations.
1
2
15x – 6y = 93
Call these equations 1 and 2 .
3 × 1
+ 8x + 6y = 222 × 2
23x =
3
4
3 + 4
x = 5
115

Substitute x = 5 in 1 :
5 × 5 – 2y = 31
25 – 2y = 31
–2y = 6
x = –3
Check by substituting x = 5 and y = –3 into 2 :
LHS = 4 × 5 + 3 × –3
= 20 – 9
= 11
= RHS
So the solution is x = 5, y = –3.

y = 2x – 3
2x + 3y = 23
Solve:
The substitution method
Two simultaneous equations can also be solved by
substituting one equation into the other. For example:
Substitute 1 into 2 :
2x + 3(2x – 3) = 23
2x + 6x – 9 = 23
8x – 9 = 23
8x = 32
x = 4
1
2
y = 2x – 3

y = 2 × 4 – 3
y = 5
Substituting x = 4 into 1 gives
Check by substituting x = 4 and y = 5 into 2 :
So the solution is x = 4, y = 5.
LHS = 2 × 4 + 3 × 5
= 8 + 15
= 23
= RHS

One of the equations needs to be arranged in the form x = …
or y = … before it can be substituted into the other equation.
Rearrange equation 1 :
3x – y = 9
– y = 9 – 3x
y = 3x – 9
3x – y = 9
8x + 5y = 1
1
2
Solve:

Now substitute y = 3x – 9 into equation 2 :
8x + 5(3x – 9) = 1
8x + 15x – 45 = 1
23x – 45 = 1
23x = 46
x = 2
Substitute x = 2 into equation 1 to find the value of y:
6 – y = 9
–y = 3
y = –3
So the solution is x = 2, y = –3.

Contents
Linear inequalities
Polynomials
One linear and one quadratic equation

Suppose one of the equations in a pair of simultaneous
equations is linear and the other is a quadratic of the form
y = ax2
+ bx + c.
By considering the points
where the graphs of the two
equations might intersect
we can see that there could
be two, one or no pairs of
solutions.

If the second equation contains terms in xy or y2
the shape
of the corresponding graph will not be a parabola but a
circle, a hyperbola or an ellipse:
A line and a
circle
A line and a
hyperbola
A line and an
ellipse
Again we can have two, one or no pairs of solutions.

y = x2
+ 1
y = x + 3
Solve:
When a pair of simultaneous equations contains one linear
and one quadratic equation, we usually solve them by
substitution. For example:
x2
+ 1 = x + 3
Rearranging to give a quadratic equation of the form
ax2
+ bx + c = 0 gives
x2
– x – 2 = 0
(x + 1)(x – 2) = 0
x = –1 or x = 2
Substituting equation 1 into equation 2 gives
1
2

We can substitute these values of x into one of the equations
y = x + 3
y = x2
+ 1 1
2
When x = –1:
It is easiest to substitute into equation 2 because it is linear.
to find the corresponding values of y.
y = –1 + 3
y = 2
When x = 2:
y = 2 + 3
y = 5
The solutions are x = –1, y = 2 and x = 2, y = 5.

We can demonstrate
the solutions to
using a graph.
y = x2
+ 1
y = x + 3
0
y = x2
+ 1
y = x + 3
(–1, 2)
(2, 5)
It is difficult to sketch a parabola accurately. For this reason, it
is difficult to solve simultaneous equations with quadratic terms
using graphs, particularly when the solutions are not integers.
x
y

Look at this pair of simultaneous equations:
x2
+ y2
= 13
y – x = 1 1
2
What shape is the graph given by x2
+ y2
= 13?
We can solve this pair of simultaneous equations algebraically
using substitution.
We can then sketch the graphs of the equations to demonstrate
where they intersect.
The graph of x2
+ y2
= 13 is a circular graph with its centre at
the origin and a radius of .13

x2
+ y2
= 13
y – x = 1 1
2
x2
+ (x + 1)2
= 13
(x + 3)(x – 2) = 0
x = –3 or x = 2
Substituting into 2 gives
x2
+ x2
+ 2x + 1 = 13
2x2
+ 2x – 12 = 0
x2
+ x – 6 = 0
Rearranging 1 gives y = x + 1

We can substitute these values of x into one of the equations
x2
+ y2
= 13
y = x + 1 1
2
When x = –3:
It is easiest to substitute into equation 1 because it is linear.
to find the corresponding values of y.
y = –3 + 1
y = –2
When x = 2:
y = 2 + 1
y = 3
The solutions are x = –3, y = –2 and x = 2, y = 3.

Demonstrating these solutions graphically gives:
The graphs intersect at the points (–3, –2) and (2, 3).
x
y
0
x2
+ y2
= 13 y = x + 1
(–3, –2)
(2, 3)

Using the discriminant
In summary, to solve a pair of simultaneous equations where
one equation is linear and the other is quadratic:
Substitute the linear equation into the quadratic equation
to give a single equation of the form ax2
+ bx + c = 0.
Rearrange the linear equation so that one of the
variables is written in terms of the other.
We can find the determinant of this equation to find how
many times the line and the curve will intersect. When
b2
– 4ac > 0, there are two distinct points of intersection.
b2
– 4ac = 0, there is one point of intersection (or two
coincident points). The line is a tangent to the curve.
b2
– 4ac < 0, there are no points of intersection.

Using the discriminant
Show that the line y – 4x + 7 = 0 is a
tangent to the curve y = x2
– 2x + 2.
y – 4x + 7 = 0 1
y = x2
– 2x + 2 2
4x – 7 = x2
– 2x + 2Substituting into 2 gives
Rearranging 1 gives y = 4x – 7
x2
– 6x + 9 = 0
The discriminant = b2
– 4ac = (–6)2
– 4(9)
= 36 – 36
= 0
b2
– 4ac = 0 and so the line is a tangent to the curve.

Contents
Linear inequalities
Polynomials
Linear inequalities

Linear inequalities
An inequality links two or more expressions with
the symbols: <, >, ≤ or ≥.
Inequalities are linear if the expressions they contain can be
written in the form ax + b where a and b are constants.
For example:
3x + 2 > 5
Solving this inequality involves finding the values of x that
make the inequality true.
In this example, the inequality is true when x > 1.
The solution can be illustrated using a number line as follows:
–3 –2 –1 0 1 2 3 4 5

Solving linear inequalities
Like an equation, we can solve an inequality by adding or
subtracting the same value to both sides of the inequality sign.
We can also multiply or divide both sides of the inequality by
a positive value. For example:
Solve 4x – 7 > 11 – 2x.
4x > 18 – 2x
6x > 18
x > 3
How could we check this solution?
Add 7 to both sides:
Add 2x to both sides:
Divide both sides by 6:

Checking solutions
To verify that
substitute a value just above 3 into the inequality and then
substitute a value just below 3.
x > 3
is the solution to 4x – 7 > 11 – 2x
Substituting x = 4 into the inequality gives
4 × 4 – 7 > 11 – 2 × 4
16 – 7 > 11 – 8
9 > 3 This is true.
Substituting x = 2 into the inequality gives
4 × 2 – 7 > 11 – 2 × 2
8 – 7 > 11 – 4
1 > 7 This is not true.

Multiplying or dividing by negatives
Although most inequalities can be solved like equations we
have to take great care when multiplying or dividing both sides
of an inequality by a negative value.
The following simple inequality is true:
–3 < 5
Look what happens if we multiply both sides by –1:
–3 × –1 < 5 × –1
3 < –5
To keep the inequality true we have to reverse the inequality
sign:
3 > –5
This is not true.

Multiplying or dividing by negatives
Remember: when both sides of an inequality are multiplied
or divided by a negative number the inequality is reversed.
Remember: when both sides of an inequality are multiplied
or divided by a negative number the inequality is reversed.
For example: 4 – 3x ≤ 10
–3x ≤ 6
We could also solve this type of inequality by collecting x terms
on the right and reversing the inequality sign at the end.
4 – 3x ≤ 10
4 ≤ 10 + 3x
–6 ≤ 3x
–2 ≤ x
x ≥ –2 The inequality sign is reversed.
x ≥ –2

Solving combined linear inequalities
The two inequalities 4x + 3 ≥ 5 and 4x + 3 < 15 can be written
as a single combined inequality:
We can solve this inequality as follows:
5 ≤ 4x + 3 < 15
Subtract 3 from each part: 2 ≤ 4x < 12
Divide each part by 4: 0.5 ≤ x < 3
We can illustrate this solution on a number line as follows:
–1 –0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

Solving combined linear inequalities
Some combined inequalities contain variables in more than
one part. For example:
Treat this as two separate inequalities:
x – 2 ≤ 3x + 2 ≤ 2x + 7
x – 2 ≤ 3x + 2
– 2 ≤ 2x + 2
– 4 ≤ 2x
– 2 ≤ x
and 3x + 2 ≤ 2x + 7
x + 2 ≤ 7
x ≤ 5
We can write the complete solution as –2 ≤ x ≤ 5 and
illustrate it on a number line as follows:
–3 –2 –1 0 1 2 3 4 5 6 7

Overlapping solutions
Solve the following inequality and illustrate the solution on a
number line:
Treating as two separate inequalities gives
2x – 1 ≤ x + 2 < 7
2x – 1 ≤ x + 2
x – 1 ≤ 2
x ≤ 3
and x + 2 < 7
x < 5
If x < 5 then it is also ≤ 3. The whole solution set is therefore
given by x < 5. This can be seen on the number line:
–3 –2 –1 0 1 2 3 4 5 6 7

Solutions in two parts
Solve the following inequality and illustrate the solution on a
number line:
Treating as two separate inequalities gives
4x + 5 < 3x + 5 ≤ 4x + 3
4x + 5 < 3x + 5
4x < 3x
x < 0
and 3x + 5 ≤ 4x + 3
5 ≤ x + 3
2 ≤ x
We cannot write these solutions as a single combined
inequality. The solution has two parts.
–3 –2 –1 0 1 2 3 4 5 6 7
x ≥ 2

Contents
Linear inequalities
Polynomials

Quadratic inequalities contain terms in both x2
and x. For
example:
x2
+ x – 6 ≥ 0
Factorizing gives (x + 3)(x – 2) ≥ 0
x2
+ x – 6 is equal to 0 when:
x + 3 = 0 and x – 2 = 0
These values give the end points of the solution set:
x = –3 x = 2
–5 –4 –3 –2 –1 0 1 2 3 4 5

To find the solution set we can substitute a value from each of
the following three regions:
into the original inequality x2
+ x – 6 ≥ 0.
When x = –4:
–5 –4 –3 –2 –1 0 1 2 3 4 5
16 – 4 – 6 ≥ 0
region 1 region 2 region 3
–42
+ –4 – 6 ≥ 0
6 ≥ 0
This is true and so values in region 1 satisfy the inequality.
region 1

region 2 region 3
When x = 0
–6 ≥ 0
02
+ 0 – 6 ≥ 0
This is not true and so values in region 2 do not satisfy the
inequality.
–5 –4 –3 –2 –1 0 1 2 3 4 5
region 1
When x = 3
9 + 3 – 6 ≥ 0
32
+ 3 – 6 ≥ 0
6 ≥ 0
This is true and so values in region 3 satisfy the inequality.
region 3region 2

We have shown that values in region 1 and region 3 satisfy the
inequality x2
+ x – 6 ≥ 0.
We can show the complete solution set as follows:
–5 –4 –3 –2 –1 0 1 2 3 4 5
region 1 region 2 region 3
–5 –4 –3 –2 –1 0 1 2 3 4 5
So the solution to x2
+ x – 6 ≥ 0 is:
x ≤ –3 or x ≥ 2

An alternative method for solving inequalities involves using
graphs. For example:
Solve x2
+ x – 3 > 4x + 1.
The first step is to rearrange the inequality so that all the terms
are on one side and 0 is on the other.
x2
– 3x – 4 > 0
Sketching the graph of y = x2
– 3x – 4 will help us to solve this
inequality.
The coefficient of x2
> 0 and so the graph will be ∪-shaped.

Next, we find the roots by solving x2
– 3x – 4 = 0.
Factorizing gives (x + 1)(x – 4) = 0
x = –1 or x = 4
We can now sketch the graph.
The inequality
x2
– 3x – 4 > 0
is true for the parts of the
curve that lie above the
x-axis.
0
y
x
So, the solution to x2
+ x – 3 > 4x + 1 is
x < –1 or x > 4
(–1, 0) (4, 0)
0
y
x
(–1, 0) (4, 0)

Solving quadratic inequalities using graphs

Contents
Linear inequalities
Polynomials
Polynomials

Polynomials
A polynomial in x is an expression of the form
where a, b, c, … are constant coefficients and
n is a positive integer.
1 2 2
+ + +...+ + +n n n
ax bx cx px qx r− −
Examples of polynomials include:
Polynomials are usually written in descending powers of x.
3x7
+ 4x3
– x + 8 x11
– 2x8
+ 9x 5 + 3x2
– 2x3
.and
The value of a is called the leading coefficient.
They can also be written in ascending powers of x, especially
when the leading coefficient is negative, as in the last example.

Polynomials
A polynomial of degree 1 is called linear and has the
general form ax + b.
A polynomial of degree 2 is called quadratic and has the
general form ax2
+ bx + c.
A polynomial of degree 3 is called cubic and has the
general form ax3
+ bx2
+ cx + d.
A polynomial of degree 4 is called quartic and has the
general form ax4
+ bx3
+ cx2
+ dx + e.
The degree, or order, of a polynomial is given by the highest
power of the variable.

Using function notation
Polynomials are often expressed using function notation.
For example, consider the polynomial function:
f(x) = 2x2
– 7
We can use this notation to substitute given values of x.
For example:
Find f(x) when a) x = –2 b) x = t + 1
a) f(–2) = 2(–2)2
– 7
= 8 – 7
= 1
b) f(t + 1) = 2(t + 1)2
– 7
= 2(t2
+ 2t + 1) – 7
= 2t2
+ 4t + 2 – 7
= 2t2
+ 4t – 5

Adding and subtracting polynomials
When two or more polynomials are added, subtracted or
multiplied, the result is another polynomial.
Find a) f(x) + g(x) b) f(x) – g(x)
a) f(x) + g(x)
= 2x3
– 5x + 4 + 2x – 4
Polynomials are added and subtracted by collecting like
terms.
= 2x3
– 3x
For example: f(x) = 2x3
– 5x + 4 and g(x) = 2x – 4
b) f(x) – g(x)
= 2x3
– 5x + 4 – (2x – 4)
= 2x3
– 5x + 4 – 2x + 4
= 2x3
– 7x + 8

Multiplying polynomials
When two polynomials are multiplied together every term in the
first polynomial must by multiplied by every term in the second
polynomial. For example:
f(x) = 3x3
– 2 and g(x) = x3
+ 5x – 1
f(x)g(x) = (3x3
– 2)(x3
+ 5x – 1)
= 3x6
+ 15x4
– 3x3
– 2x3
– 10x + 2
= 3x6
+ 15x4
– 5x3
– 10x + 2

Multiplying polynomials
Sometimes we only need to find the coefficient of a single term.
For example:
Find the coefficient of x2
when x3
– 4x2
+ 2x
is multiplied by 2x3
+ 5x2
– x – 6.
We don’t need to multiply this out in full. We only need to
decide which terms will multiply together to give terms in x2
.
(x3
– 4x2
+ 2x)(2x3
+ 5x2
– x – 6)
We have: 24x2
– 2x2
= 22x2
So, the coefficient of x2
is 22.

Contents
Linear inequalities
Polynomials

Examination-style question 1
a) Solve the simultaneous equations
x – 2y = 2
x2
+ 4y2
= 100
b) Interpret your solution geometrically.
a) Label the equations
1x – 2y = 2
2x2
+ 4y2
= 100
Rearranging equation 1
x = 2 + 2y
Substituting into equation 2
(2 + 2y)2
+ 4y2
= 100

4 + 8y + 4y2
+ 4y2
= 100
8y2
+ 8y – 96 = 0
y2
+ y – 12 = 0
(y + 4)(y – 3) = 0
y = –4 or y = 3
Substituting into equation 1
When y = –4, x = –6
When y = 3, x = 8
b) The line x – 2y = 2 crosses the curve x2
+ 4y2
= 100 at the
points (–4, –6) and (3, 8).

a) Write an expression for the area A of the following
rectangle:
b) If the area satisfies the inequality
5 < A < 12
find the range of possible values for
x.
x – 2
x + 2
a) A = (x + 2)(x – 2)
= x2
– 4
b) The range of possible values for x is given by
5 < x2
– 4 < 12

We have to solve 5 < x2
– 4 and then solve x2
– 4 < 12
5 < x2
– 4
x2
– 9 > 0
(x + 3)(x – 3) > 0
Sketching y = x2
– 9
x2
– 9 > 0 when x < –3 or x > 3
x2
– 4 < 12
x2
– 16 < 0
(x + 4)(x – 4) < 0
Sketching y = x2
– 16
0
y
x
(3, 0)(–3, 0)
0
y
x
(4, 0)(–4, 0)
x2
– 16 < 0 when –4 < x < 4
So the range of possible value for x is
3 < x < 4 (ignoring negative solutions)

57
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2.c1.3 algebra and functions 3

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