Core 1

elimination:

2x+3y= 8 you need to get the x's or y's to match

3x-y=23 x3 to get 3y in both equations              as they both +/- you add

2x+3y =8

6x-3y= 69

11x= 77  /11

x=7 now you put x=7 into original equation

(2x7) +3y=8

14+3y= 8

3y=8-14 (-6) /3

y=12

substitution:

x+2y= 3 make either x or y the subject of either equation + substitute

x2+3xy=10

x=3-2y        substitute into 2nd equation

(3-2y)2 + 3(3-2y)y = 10

9+4y2-12y +(9-6y)y = 10 

9+4y2-12y + 9y-6y2 = 10

-2y2-3y+9 = 10

2y2+3y+1 = 0 factorise

(2y+1)(y+1)

y=-1/2    or -1    now pick easier first equation and sub in ys to find x values

Bigger than >       Smaller than <

1. When you multiply or divide an inquality by a negative number, you need to change the inquality sign to its opposite        e.g. 

12-3x<27  minus 12

-3x<15  divide by -3  FLIP  

x>-5

2. To solve a quadratic inequality you:

-solve the corresponding

-sketch the graph of the quadratic function

-use your sketch to find the required set of values

Sometimes you need to find the set of values for x when two inequalities are true together.. USE A NUMBER LINE!                 

Find the set of values of x for which:  

3x-5<x+8    &    5x>x-8         Hollow dots on a number line= < or >

2x-5<8 4x>-8 Solid dots on a number line=<(or same as)>

2x<13 x>-2

x<6.5

-4 -------3-------2-------1-------0------1-------2-------3-------4------5------6------7-------8

o------------------------------------------------------------------------------->

<-----------------------------------------------------------------------------------------o

so the required set of values are -2<x<6.5

Straight lines equation: y=mx+c  m= gradient c= y intercept (where line crosses y)

If the gradient is positive the line will go up in a right diagonal line like /

If the gradient is negative the line will do down like \

e.g. what is the gradient and y intercept of 3y+2x=5?

3y=5-2x

y=5/3-2/3x

y=-2/3x+5/3

gradient= -2/3x    and     y intercept=5/3

parrallel lines have the same gradient.

 = length between two points      

equation:  y1-y2 / x1-x2  or  y2-y1/ x2-x1  y is always on top

e.g. (2, 6) & (5, 9)

9-6 / 5-2

3/3

=1

y=1x+c

Once you've found your gradient sub it into y=mx+c

e.g. (2, 6) & (5, 9)

9-6 / 5-2

3/3

=1

y=1x+c

pick either coordinates and sub in it y and x (2, 6)

6=1x2+c

c=4

y=1x+4

x co ordinate if midpoints= x1+x2 / 2

y co ordinate of midpoints= y1+y2 /2

e.g. (4, 7) & (8, 9)

(4+8)/2         (7+9)/2    

=6                 =8

Midpoint: (6,8)

A number times by it's reciprocal is always one

-5 = -1/5

0 does not have a reciprocal.

The gradient of a line x the gradient of it's perpendicular line always equals -1

e.g. y=3x+2

gradient of perpendicular = -1/3 (becomes negative also when it flips over)

we have the line y=2x+3

the line perpendicular goes through (0,5) Find the equation of the perpendic line

gradient= -1/2x+c

y=-1/2x+5 (as 5 is the y intercept)

Find where y=2x+1 and y=3x+5 cross

2x+1 = 3x+5 -2x

1=x+5  -5

x= -4

2x-4+1=-7

y=-7

crossing point = (-4,-7)

These have to go up or down by the same amount each time

equation: nth term= a+(n-1)d 

 a= starting number   d= common difference 

e.g.

4,7,10,13...

a=4    d= 3 as increasing by 3 each time

nth term= 4+(n-1)3

     =4+3n-3

     =3n+1

The 3rd term in a sequence is 20 and the 7th term is 12. 

Find a and d.

3rd 7th

a+2d=20 a+6d=12

put in a similtaneous equation and work out by elimination

a+2d=20 x3       3a+6d=60 as both are +s you -

a+6d=12    a+6d=12

 2a      =48  /2

   a      =24

24+2d=20 -24

2d=-4 /2               d      = -2

Formula:        Sn= 1/2n (a+L) or Sn= [2a+(n-1)d]

1) Find the sum of the first 100 odd numbers

1 + 3 + 5+...     a=1 d= 2

1/2x100[2x1+(100-1)2]

50[2+(99x2)]

50 x 200=   10,000

2) Salary= £15,000    Rise= £1,000peryear   Max Salary= £25,000

a) how much did she earn in her first eight years? b) 14 years?

15+16+17+18+19+20+21+22 =£148,000

b) (a) +23+24+25+25+25+25 =£295,000  repeat 25,000 because that is the max.

Tangents are straight lines that touch a curve at a single point. A normal is perpendicular to a tangent at the same point.

General Rule: dy/dx = nxn-1

e.g. y=x3   = 3x2

we use this to find the gradent of curves at different points

To find the gradient of different points:    e.g. y=x2 at x=5

1. Differentiate equation of the curve

dy/dx= 2x

2. Sub in x value of point you want to know gradient of

2x5=10

=10

Whenever you find a gradient of a curve using dy/dx you also get the gradient of a tangent.

eg. find the equation of a tangent to the curve y=2x2+4x at point (1,6)

1) find dy/dx   4x+4

2) find gradient of curve (sub in x=1) 4x1+4= 8

3) use y=mx+c to find equation of tangent (m=8)

    y=8x+c

    6=8x1+c

    c=-2

y=8x-2

d2y / dx2 means differentiate and then differentiate again

f(x) is a function

f '(x) means dy/dx

f ''(x) means d2y / dx2

e.g. Find d2y/dx of 3x3+ 2x2 -3/x

        3x3+2x2-3x-1

dy/dx=     9x2+4x+3x-2

d2y/dx2= 18x+4-6x-3

Intergration finds y when you have dy/dx

Add 1 to the power and then divide by the new power number

e.g.

dy/dx= 3x2

   3x3 /3

=x3 +C   remember the C!!

Given that f-1(x) = x2-2 / x1/2 and that curve C for f(x) passes through the point (4,5), find the equation for C

f-1(x) = x2/ x1/2   - 2/ x1/2

=x3/2-2x-1/2

intergrate as usual

x5/2 / 5/2 -2x1/2 / 1/2 + c

f(x) 2/5x5/2 -4x1/2 + c

sub in (4,5)

5=2/5x45/2 -4x41/2 + c

5=64/5-8+c  add 8 13=64/5 +c

c= 13-64/5 (13-12^4/5) c=1/5          y=2/5x5/2-41/2+1/5

Seen by exam question laid out as ax2 + bx + c

If there are real roots b2-4ac>0  will have two roots (touch x axis twice)

If the roots are equal b2-4ac=0  will have one root (touch the x axis once)

If there are no real roots b2-4ac<0 will have no real roots (won't touch x axis)