8 Simultaneous Equations and Intersections

advertisement
“Teach A Level Maths”
Vol. 1: AS Core Modules
8: Simultaneous Equations
and Intersections
© Christine Crisp
Simultaneous Equations and Intersections
Module C1
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with
permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"
Simultaneous Equations and Intersections
Suppose we want to find where 2 lines meet.
e.g. 1
y   x  3 and y  2 x  5
Sketching the
lines gives
The point of intersection has an x-value between
-1 and 0 and a y-value between 3 and 4.
The exact values can be found by solving the
equations simultaneously
Simultaneous Equations and Intersections
At the point of intersection, we notice that the
x-values on both lines are the same and the yvalues are also the same.
yy   x  3
yy  2 x  5
 23 , 113 
As the y-values are the
same, the right-hand
sides of the equations
must also be the same.
  x  3  2x  5

 2  3x
2

x

3
Substituting into one of the original equations, we can
y  x  3
find y:

y   23   3  y  11
3
The point of intersection is
 23 , 113 
Simultaneous Equations and Intersections
Sometimes the equations first need to be rearranged:
e.g. 2
y  2 x  4       (1)
3 x  y  11
      (2)
Solution: Equation (2) can be written as
y  11 3 x       (2a)
Now, eliminating y between (1) and (2a) gives:
2 x  4  11  3 x

5 x  15

Substituting into (1): y  2 x  4
x3

y2
The point of intersection is ( 3, 2)
Simultaneous Equations and Intersections
Exercises
1.
Solution:
Eliminate y:
Find the point of intersection of the
following pairs of lines:
y  4  2 x       (1)
y  x  5       (2)
4  2x  x  5
 9  3x 
 y  2
Point of intersection is (3,2)
2 x  y  7       (1)
2.
y  3  x       (2)
y  7  2 x       (1a)
Rearrange
Solution: (1):
Eliminate y: 3  x  7  2 x
 x  4  y  1
Point of intersection is (4,1)
x3
Simultaneous Equations and Intersections
e.g. 3 Find the points of intersection of y  x 2 and
y  3  2x
y  x2
There are 2 points
of intersection
y  3  2x
We again solve the equations simultaneously but this
time there will be 2 pairs of x- and y-values
Simultaneous Equations and Intersections
e.g. 1
y  x2
      (1)
y  3  2 x       (2)
Since the y-values are equal we can eliminate y by
equating the right hand sides of the equations:
x 2  3  2 x This is a quadratic equation, so
we get zero on one side and try
2

x  2x  3  0
to factorise:
 ( x  1)( x  3)  0
 x  1 or x  3
To find the y-values, we use the linear equation,
which in this example is equation (2)
x 1 
y  3  2(1)
 y 1
x  3 
y  3  2(3)  y  9
The points of intersection are (1, 1) and (-3, 9)
Simultaneous Equations and Intersections
Sometimes we need to rearrange the linear equation
before eliminating y
y  x2  3
e.g. 2
      (1)
y  3 x  1       (2)
Rearranging (2) gives y  3 x  1       (2a )
x 2  3  3x  1
x 2  3x  4  0
Eliminating y:
y  3 x  1 (4, 13)

 ( x  1)( x  4)  0

x  1
or
x4
( 1,  2)
Substituting in (2a): x  1  y  2
x4
 y  13
y  x2  3
Simultaneous Equations and Intersections
Exercise
Find the points of intersections of the following
curve and line
y  x 2  2       (1)
x y8
      ( 2)
The solution is on the next slide
Simultaneous Equations and Intersections
y  x 2  2       (1)
x y8
      ( 2)
y  8  x       (2a )
Rearrange (2):
Solution:
x2  2  8  x
x2  x  6  0

 ( x  3)( x  2)  0
 x  3
Eliminate y:
or
Substitute in (2a):
y  8 x
x  3 
y  8  ( 3 )

y  11
x2 
y  8  (2)

y6
The points of intersection are
(3, 11)
and
x2
(2, 6)
Simultaneous Equations and Intersections
Special Cases
e.g. 1 Consider the following equations:
y  x 2  2       (1)
y   x  1       (2)
y  x2  2
y  x  1
The line and the
curve don’t meet.
Solving the equations simultaneously will not give
any real solutions
Simultaneous Equations and Intersections
Suppose we try to solve the equations:
y  x 2  2       (1)
y   x  1       (2)
Eliminate y:

x2  2  x  1
x2  x  1  0
Calculating the discriminant, b 2  4ac we get:
b 2  4ac  (1) 2  4(1)(1)
 1 4
 3  0
b  4ac  0  The quadratic equation has no real roots.
2
Simultaneous Equations and Intersections
y  x 2  3       (1)
y  4 x  1       (2)
e.g. 2
Eliminate y:
x 2  3  4 x  1
 x 2  4x  4  0
The discriminant, b 2  4ac  4 2  4(1)( 4)  0
The quadratic equation
has equal roots.
Solving x  4 x  4  0
2
y  x2  3
y  4 x  1
 ( x  2)( x  2)  0
 x  2 (twice)
x  2  y  7
The line is a tangent to the curve.
Simultaneous Equations and Intersections
SUMMARY
 A linear and a quadratic equation represent a line
and a curve.
 To solve a linear and a quadratic equation
simultaneously:
• Eliminate one unknown to give a quadratic
equation in the 2nd unknown, e.g. ax 2  bx  c  0
•
•
b 2  4ac  0  2 points of intersection
b 2  4ac  0  the line is a tangent to the curve
Substitute into the linear equation to find the
values of the 1st unknown.
Solve for the 2nd unknown
b 2  4ac  0  the line and curve do not meet and
the equations have no real solutions.
Simultaneous Equations and Intersections
Exercises
Decide whether the following pairs of lines and curves
meet. If they do, find the point(s) of intersection.
For each pair, sketch the curve and line.
1.
y  x2  3
y  2x  2



2.
y  x2  3
y  7x  7



3.
y  x2  3
y  x 1 0



Simultaneous Equations and Intersections
Solutions
1.
y  x2  3
y  2x  2




x 2  3  2 x  2

x 2  2x  1  0
b 2  4ac  4  4(1)(1)  0
b 2  4ac  0  the line is a tangent to the curve
x 2  2x  1  0
 ( x  1)( x  1)  0
x  1

y4

y  x2  3
y  2x  2
Simultaneous Equations and Intersections
Solutions
2.
y  x2  3
y  7x  7




x2  3  7x  7
 x 2  7 x  10  0
b 2  4ac  49  4(1)(10)  9
b 2  4ac  0  there are 2 points of intersection
x 2  7 x  10  0
 ( x  2)( x  5)  0
 x  2, 5
x2  y7
x  5  y  28
y  x2  3
y  7x  7
Simultaneous Equations and Intersections
Solutions
3.
y  x2  3
y  x 1 0





x2  3  x  1
x2  x  4  0
b 2  4ac  ( 1) 2  4(1)( 4)  15
b 2  4ac  0  there are NO points of intersection
y  x2  3
y  x 1  0
Simultaneous Equations and Intersections
Simultaneous Equations and Intersections
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Simultaneous Equations and Intersections
Two Lines At the point of intersection, we
notice that the x-values on both lines are the
same and the y-values are the same.
y  x  3
y  2x  5
y  2x  5
y  x  3
As the y-values are
the same, the righthand sides of the
equations must also be
the same.
  x  3  2x  5

 2  3x
2

x

3
Substituting into one of the original equations, we
can find y: y   x  3

y   23   3  y  11
3
The point of intersection is
 23 , 113 
Simultaneous Equations and Intersections
1 quadratic equation and 1 linear equation
e.g.
y  x2
      (1)
y  3  2 x       (2)
Since the y-values are equal we can eliminate y by
equating the right hand sides of the equations:
x 2  3  2 x This is a quadratic equation,

x  2x  3  0
 ( x  1)( x  3)  0
so we get zero on one side
and try to factorise:
2

x 1
or
x  3
To find the y-values, we use the linear equation,
which in this example is equation (2)
x  1  y  3  2(1)  y  1
x  3 
y  3  2(3)  y  9
The points of intersection are (1, 1) and (-3, 9)
Simultaneous Equations and Intersections
Sometimes we need to rearrange the linear equation
before eliminating y
y  x2  3
e.g.
      (1)
y  3 x  1       (2)
Rearranging (2) gives y  3 x  1       (2a )
x 2  3  3x  1
x 2  3x  4  0
Eliminating y:

 ( x  1)( x  4)  0

x  1
or
x4
Substituting in (2a): x  1  y  2
x4
 y  13
Simultaneous Equations and Intersections
Special Cases
e.g. 1 Consider the following equations:
y  x 2  2       (1)
y   x  1       (2)
The line and the
curve don’t meet.
Solving the equations simultaneously will not give
any real solutions.
The discriminant b 2  4ac  0
Simultaneous Equations and Intersections
y  x 2  3       (1)
y  4 x  1       (2)
e.g. 2
Eliminate y:
x 2  3  4 x  1
 x 2  4x  4  0
The discriminant, b 2  4ac  4 2  4(1)( 4)  0
The quadratic equation
has equal roots.
Solving x 2  4 x  4  0
 ( x  2)( x  2)  0
 x  2 (twice)
x  2  y  7
The line is a tangent to the curve.
Simultaneous Equations and Intersections
SUMMARY
 A linear and a quadratic equation represent a line
and a curve.
 To solve a linear and a quadratic equation
simultaneously:
• Eliminate one unknown to give a quadratic
equation in the 2nd unknown, e.g. ax 2  bx  c  0
•
•
b 2  4ac  0  2 points of intersection
b 2  4ac  0  the line is a tangent to the curve
Substitute into the linear equation to find the
values of the 1st unknown.
Solve for the 2nd unknown
b 2  4ac  0  the line and curve do not meet and
the equations have no real solutions.
Download
Related flashcards

Algebra

20 cards

Polynomials

21 cards

Ring theory

15 cards

Category theory

14 cards

Elementary algebra

12 cards

Create Flashcards