### Archive

Archive for the ‘Algebra’ Category

## Simultaneous Equations: Substitution Method

This is exactly the same method we use to find where two lines intersect.

Process:

1. Rearrange one of the equations to make one of the variables a subject
2. Substitute for this variable into the other equation
3. Solve the resulting equation
4. Substitute these solutions into one of the original equations to find the values of the other variable

Example

Solve the system of equations

$y=4x-3\rightarrow 1$

$3y+x=7\rightarrow 2$

Substitute eqn 1 into eqn 2

$\implies 3(4x-3)+x=7$

$\displaystyle \implies 12x-9+x=7\implies 13x=16\implies \frac{16}{13}$

Substituting into eqn 1 gives

$\displaystyle y=4\left(\frac{16}{13}\right)-3=\frac{25}{13}$

Solve the system of equations

$y-3=x^2+4x\rightarrow 1$

$2x=y-6\rightarrow 2$

Rearrange eqn 2 to make y the subject

$\implies y=2x+6\rightarrow 3$

Substitute eqn 3 into eqn 1

$\implies (2x+6)-3=x^2+4x\implies x^2+2x-3=0$

Factorising gives

$(x+3)(x-1)=0\implies x=-3\mbox{ or }x=1$

Substituting these values into eqn 3 gives

$y=0\mbox{ and }y=8$

So the solutions are $(-3,0)\mbox{ and }(1,8)$

Solve the system of equations

$3x+2y=13\rightarrow 1$

$2x-y=-3\rightarrow 2$

Rearranging eqn 2 to make y the subject gives

$y = 2x+3\rightarrow 3$

Substituting eqn 3 into eqn 1 gives

$3x+2(2x+3)=13\implies 7x+6=13\implies x=1$

Substituting the value of x into eqn 3 gives

$y=5$

Categories: Equations

## Simultaneous equations: Elimination method

Simultaneous equations are sytems of equations where there are a number of variables.

Our task is to find the values of the variables that satisfy all the equations.
For this method we will solve equations of the type

$\displaystyle ax+by=c\rightarrow 1$

$\displaystyle dx+ey=f\rightarrow 2$

With these equations we can do one of three things

1. Multiply by a scaler.
3. Subtract one equation from another

Multiplying by a scaler

eqn 1 multiplied by 6 gives

$\displaystyle 6ax+6by=6c\rightarrow 3$

eqn 1 + eqn 2 gives

$\displaystyle (a+d)x+(b+e)y=c+f\rightarrow 4$

Subtract one equation from another

eqn 1 – eqn 2 gives

$\displaystyle (a-d)x+(b-e)y=c-f\rightarrow 5$

### Type 1 Simultaneous equations

With this type of problem the y terms are the same size but different signs.

To solve these we:

1. Add the equations together (this will cancel out the y terms)
2. Solve the resulting equation to find the value of x
3. Substitute the x value into one of the original equations
4. Solve the resulting equation to find the value of y
5. Check by substituting x and y into the other equation

Examples

Solve

$\displaystyle 3x+4y=18\rightarrow 1$

$\displaystyle 7x-4y=2\rightarrow 2$

This is a type 1 problem because we have two 4y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

$\displaystyle 10x=20$

$\displaystyle \implies x=2$

Substitute $x=2$ into eqn 1

gives $\displaystyle 6+4y=18$

$\implies 4y=12$

$\implies y=3$

Check in eqn 2

$14-12=2$ true

Solve

$\displaystyle 2x-5y=13\rightarrow 1$

$\displaystyle 3x+5y=7\rightarrow 2$

This is a type 1 problem because we have two 5y’s, but one is positive and the other is negative.

So eqn 1 + eqn 2 gives

$\displaystyle 5x=20$

$\displaystyle \implies x=4$

Substitute $x=4$ into eqn 2 (it is easier to deal with positives)

gives $\displaystyle 12+5y=7$

$\implies 5y=-5$

$\implies y=-1$

Check in eqn 1

$8--5=13$ true

### Type 2 Simultaneous equations

With this type of problem the y terms are the same size and sign.

To solve these we:

1. Subtract one equation from the other (this will cancel out the y terms)
2. Solve the resulting equation to find the value of x
3. Substitute the x value into one of the original equations
4. Solve the resulting equation to find the value of y
5. Check by substituting x and y into the other equation

Examples

Solve

$\displaystyle 4x+2y=20\rightarrow 1$

$\displaystyle 7x+2y=23\rightarrow 2$

This is a type 2 problem because we have two +2y’s.

So eqn 2 – eqn 1 gives

$\displaystyle 3x=3$

$\implies x=1$

Substitute $x=1$ into eqn 1

gives $\displaystyle 4+2y=20$

$\implies 2y=16$

$\implies y=8$

Check in eqn 2

$\displaystyle 7+16=23$ true

Solve

$\displaystyle 2x-4y=6\rightarrow 1$

$\displaystyle 5x-4y=3\rightarrow 2$

This is a type 2 problem because we have two -4y’s.

So eqn 2 – eqn 1 gives

$\displaystyle 3x=-3\rightarrow 1$

$\implies x=-1$

Substitute $x=-1$ into eqn 1

gives $\displaystyle -2-4y=6$

$\implies 4y=-8$

$\implies y=-2$

Check in eqn 2

$\displaystyle -5--8=3$ true

### Type 3 Simultaneous equations

With this type of problem the y terms are different.

To solve these we:

1. Multiply each equation by the other equations positive y coefficient
2. Solve the resulting type 1 or type 2 problem

Examples

Solve

$\displaystyle 3x+4y=-5\rightarrow 1$

$\displaystyle 5x+2y=1\rightarrow 2$

Multiply eqn 1 by 2 (that is the number of y’s in eqn 2)

$\displaystyle 6x+8y=-10\rightarrow 3$

Multiply eqn 2 by 4 (that is the number of y’s in eqn 1)

$\displaystyle 20x+8y=4\rightarrow 4$

Equations 3 and 4 now make a type 2 problem

Subtracting eqn 3 from eqn 4 gives

$\displaystyle 14x=14$

$\displaystyle \implies x=1$

Substitute $x=1$ into eqn 2

$\displaystyle \implies 5+2y=1$

$\displaystyle \implies 2y=-4$

$\displaystyle \implies y=-2$

Check in eqn 1

$\displaystyle 5-4=1$ true

Solve

$\displaystyle 4x-2y=7\rightarrow 1$

$\displaystyle x+3y=10\rightarrow 2$

Multiply eqn 1 by 3 (that is the number of y’s in eqn 2)

$\displaystyle 12x-6y=21\rightarrow 3$

Multiply eqn 2 by 2 (that is the number of y’s in eqn 1)

$\displaystyle 2x+6y=20\rightarrow 4$

Equations 3 and 4 now make a type 1 problem

Adding eqn 3 and eqn 4 gives

$\displaystyle 14x=41\rightarrow 4$

$\displaystyle x=\frac{41}{14}$

Substitute $x=\frac{41}{14}$ into eqn 2

$\displaystyle \implies \frac{41}{14}+3y=10$

$\displaystyle \implies 3y=\frac{99}{14}\implies y=\frac{33}{14}$

Check in eqn 1

$\displaystyle \frac{164}{14}-\frac{66}{14}=7$ true

Categories: Equations

## Multiplying out brackets: Questions

### For each set of questions expand and simplify

Set 1

1. $3(x+7)$
2. $5(x-6)$
3. $8(x-y)$
4. $-5(x^2+6)$
5. $x(4+x)$
6. $3x(x-8)$
7. $4+x(3+x)$
8. $6-x(x-2)$
9. $2+4(x-3)$
10. $2x+x(2x^2+x-5)$

Set 2

1. $(x+3)(x+5)$
2. $(x+6)(x-5)$
3. $(x-3)^2$
4. $(x-4)(x+4)$
5. $(x-9)(x+9)$
6. $(x+y)(x-y)$
7. $(2x+7)(3x-4)$
8. $(3x-5)(x-4)$
9. $(5-x)(4x+1)$
10. $(x-5)(4x+1)$

Set 3

1. $(x+1)(x-1)(x+6)$
2. $(x+1)(x+2)(x+3)$
3. $(x-4)^2(x+7)$
4. $(3-x)(x+5)(x-4)$
5. $(x+y)(y+z)(z+a)$
6. $(x+3)^2(x-3)^2$
7. $(x+a)^3$
8. $(x-a)^3$
9. $(1-x)(2-x)(3-x)$
10. $(a+b)^4$
Categories: Algebra

# Excel test

It uses macros to give 10 random questions.

Select the level you want from 1 to 4 and then click ‘new sheet,

Do the questions and then click ‘check’

If you get 100% the sheet will print for you

if not just try with a new sheet

Categories: Algebra

## Solving basic trigonometry equations

Let’s consider the sine function

As you can see, it gives positive values for $0^o and negative for values for $180^o

Now let’s consider the cosine function

Here you can see that is positive for values $0^o and negative for values $90^o

Finally, let’s consider the tangent function.

From the graph or by dividing the sine values by the cosine values you can see that it is positive for $0^o and negative for $90^o

### The Trigonometry Circle

The information above can be summarised with the diagram below

1. In the first quadrant ‘A’, all the ratios give a positive value
2. In the second quadrant ‘S’, only sine gives a positive value
3. In the third quadrant ‘T’, only tangent gives a positive value
4. In the forth quadrant ‘C’, only cosine gives a positive value

### Solving a simple problem

1. rearrange to ratio(angle) = value
2. solve for positive value to give the angle in each quadrant
3. Pick quadrants dependent on whether value was positive or negative
4. measure round to each angle in an anti-clockwise direction from the positive x axis

Example

Solve $sin x+1=0.3\mbox{ for } 0^o\leq x\leq 360^o$

Step 1 $\implies sin x=-0.7$

Step 2 $x=sin^{-1}(0.7)=44.4^o$ (Remember we use positive 0.7, because this will give us the angle in the first quadrant)

Step 3 Since -0.7 is negative, we want the third and forth quadrants (Sine is positive in the ‘ALL’ and ‘SINE’ quadrants)

Step 4 Solutions are $x=180^o+44.4^o=224.4^o\mbox{ and } x=360^o-44.4^o=315.6^o$

Solve $cos^3 x + 3=3.4\mbox{ for } 0^o\leq x\leq 360^o$

Step 1 $cos^3 x= 0.4$

$\implies cos x =(0.4)^{\frac{1}{3}}=0.737$

Step 2 $x=cos^{-1}(0.737)=42.5^o$

Step 3 Since 0.737 is positive, we need the first and forth quadrants (Cosine is positive in the ‘All’ and ‘COSINE’ quadrants)

Step 4 Solutions are $x=42.5^o\mbox{ and } x=360^o-42.5^o=317.5^o$

Solve Solve $tan x + 1=5.1\mbox{ for } 0\leq x\leq 2\pi$Step 1 $tan x=4.1$Step 2 $x=tan^{-1}(4.1)=1.33$

Step 3 Since 4.1 is positive we require the first and third quadrants (Tangent is positive in the ‘All’ and ‘TANGENT’ quadrants)

Step 4 Solutions are $x=1.33\mbox{ and } x=\pi + 1.33=4.47$

### Multiple Angles

When dealing with problems involving multiple angles we have to take into account the scaling that we will have to do at the end.  This is done by changing the range of values that we will consider.  For example if we go from 0 to 360 degrees for x we will go from 0 to 720 degrees  (twice round) for 2x.

Examples

Solve $sin(2x)=0.9\mbox{ for } 0^o\leq x\leq 360^o\implies 0^o\leq 2x\leq 720^o$

Step 2 $2x=sin^{-1}(0.9)=64.2^o$

Step 3 Since 0.9 is positive, we want the first and second quadrants (Sine is positive in the ‘ALL’ and ‘SINE’ quadrants). But we must go round twice.

Step 4 So $2x=64.2^o, 2x=180^o-64.2^o=115.8^o, 2x=360+64.2^o=424.2^o\mbox{ and }2x=540^o-64.2^o=475.8^o$

So dividing by 2 gives $x=32.1^o, x=57.9^o, x=212.1^o\mbox{ and }x=237.9^o$

Solve $tan\left(\frac{x}{2}\right)=-0.5\mbox{ for } 0^o\leq x\leq 360^o\implies 0^o\leq \frac{x}{2}\leq 180^o$

Step 2 $\frac{x}{2}=tan^{-1}(0.5)=26.6^o$

Step 3 Since -0.5 is negative we want quadrants ‘2’ and ‘4’.  However, we are only going up to 180 degrees so we only want the second quadrant.

Step 4 So $\frac{x}{2}=180^o-26.6^o=153.4^o\implies x=306.9^o$

Categories: Trigonometry

## Series Sigma Notation (basic)

A series can be described using sigma $\Sigma$ notation

So what does sigma notation look like? Let’s look at an example

$S=\displaystyle\sum_{r=3}^{7}r^2$

This tells us that we a dealing with series with terms that are of the form $r^2$

1. The first term is when $r=3$
2. The next is when $r=4$
3. We continue in this manner by adding one each time to r until r = 7
4. Each of these terms is added.

So we have $S=3^2+4^2+5^2+6^2+7^2=135$

### Summing a series given in sigma notation

The least technical method would be to use the one above.

Example

Find S, if $S=\displaystyle\sum_{r=5}^{8}(r+r^2)$

$\implies S=(5+5^2)+(6+6^2)+(7+7^2)+(8+8^2)$

$\implies S=30+42+56+72=200$

You have noticed that we could have written as a simplification:

$\implies S=(5+6+7+8)+(5^2+6^2+7^2+8^2)$

And it does follow that $\displaystyle\sum_{r=1}^{n}(f(r)+g(r))=\displaystyle\sum_{r=1}^{n}f(r)+\displaystyle\sum_{r=1}^{n}g(r)$

### Arithmetic representations in sigma notation

A sigma expression in the form $\displaystyle\sum_{r=1}^{n}(ar+b)$ where ‘a’ and ‘b’ are constants will produce an arithmetic series, with a common difference of ‘a’ and a starting value of ‘a+b’.  Using this imformation we could sum a series of this form using the formulae we learnt for arithmetic series.

Examples

Calculate $\displaystyle\sum_{r=1}^{10}(5r-2)$

This is an arithmetic series

$a=3$

$d=5$

$n=10$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$\implies S_{10}=\frac{10}{2}(2\times 3+5\times 9)=255$

Calculate $\displaystyle\sum_{r=11}^{19}(-3r+10)$

This is an arithmetic series

$a=7$

$d=-3$

$S_n=\frac{n}{2}(2a+(n-1)d)$

The required value is $S_{19}-S_{10}=\frac{19}{2}(2\times 7-3\times 18)-\frac{10}{2}(2\times 7-3\times 9)=-315$

### Geometric representations in sigma notation

A sigma expression in the form $\displaystyle\sum_{r=1}^{n}(ab^r)$, where the first term is $a\times b$ and the common ratio is $b$.

Example

Calculate $\displaystyle\sum_{r=1}^{20}(4\times 2^r)$

This is a geometric series

$a=8$

$r=2$

$n=20$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_{20}=\frac{8(1-2^{20})}{1-2}=8388600$

Categories: Sequences and Series

##### Set A
1. $5$
2. $3$
3. $6$
4. $10$
5. $4$
6. $6$
7. $-4$
8. $\frac{17}{6}$
##### Set B
1. $3$
2. $98$
3. $2$
4. $4$
5. $14$
6. $5$
7. $4$
8. $2$
##### Set C
1. $7$
2. $6$
3. $3$
4. $2$
5. $6$
6. $1$
7. $4$
8. $2$
##### Set D
1. $12$
2. $5$
3. $7$
4. $3$
5. $4$
6. $6$
7. $3$
8. $1$
Categories: Algebra