2.c1.3 algebra and functions 3

© Boardworks Ltd 20051 of 56

1


Co

nte

nts


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 = 223x + 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 = –23x = –6

Solve the simultaneous equations3x + 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 = 5115



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) = 18x + 15x – 45 = 1

23x – 45 = 123x = 46

x = 2

Substitute x = 2 into equation 1 to find the value of y: 6 – y = 9

–y = 3y = –3

So the solution is x = 2, y = –3.


Co

nte

nts




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.

Call these equations 1 and 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.


Co

nte

nts




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 gives2x – 1 ≤ x + 2 < 7

2x – 1 ≤ x + 2x – 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 + 54x < 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


Co

nte

nts




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 thex-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


Co

nte

nts




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 nax 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.


Co

nte

nts




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 equations1x – 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

Do you find this slides were useful?

One second of your life , can bring a smile in a girl life

If Yes ,Join Dreams School “Campaign for Female Education”

Help us in bringing a change in a girl life, because “When someone takes away your pens you realize quite how important education is”.

Just Click on any advertisement on the page, your one click can make her smile. Eliminate Inequality “Not Women”

One second of your life , can bring a smile in her life!!

Do you find these slides were useful?

If Yes ,Join Dreams School “Campaign for Female Education”

Help us in bringing a change in a girl life, because “When someone takes away your pens you realize quite how important education is”.

Just Click on any advertisement on the page, your one click can make her smile. We our doing our part & u ?

Eliminate Inequality “Not Women”