Download GCSE Mathematics

Analytical Toolbox
Simultaneous Equations
By
Dr J.P.M. Whitty
Learning objectives
•
After the session you will be able to:
•
•
•
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Find the solution to two simultaneous equations
using graphical means.
Algebraic solutions to systems of two equations.
Appreciate the solution protocol required for
systems of three equations.
Use math software to solve systems of equations.
2
Learning Check: Finding the Line
• Here’s the StarGate bit:
• A point in three dimensions needs six pieces of
information to be fully described.
• A course therefore seven
• Since a line exists essentially in 2D then only two
pieces of information.
•
Two points
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A gradient and a point
3
Example
 Find the equation of the line given that it
passes though the points (-2,1) and (6,5)
y
1 5
1
Find the Gradient m 


x  2  6 2
1
Find the intercept y  x  c
2
1
1
Any point will do 6  (5)  c or 1  (2)  c
2
2
4
Examples
Find the equation of the line given the
points (-2,4) and (4,1) expressing your
answer in the form ax+by=c.
2. A line has a gradient of –0.75 and passes
through a point (3,-4), state the equation of
the line.
3. Find the equation of a line with a gradient of
unity given that it passes through the point
(-1,-2).
1.
5
Learning Check
 Solve the following
using a graphical
method
 y=2x -1 &
 y= 8-x
 Hence, use MATLAB to
verify your results.
 Q. Is there a way to solve
such problems algebraically?
6
Class Examples Time
• Solve the following problems using a
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graphical method (via MATLAB) and verify
the results using substitution
y=x+3 & y=7-x
y=x-4 & 2x+y=5
x+y =3 & y=1-2x
y=x+4 & y=3x
y=2x-1 & y=3x+2
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Algebraic Solution
 One way to obtain an algebraic solution is to
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substitute one equation into the other and obtain
the solution of the resulting linear equation. Thus:
y=x+3 (1) & y=7-x (2)
x+3=7-x
2x=4; x=2
y=x+3=5
Also 7-2=5 (which satisfies the second)
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Algebraic Solution Continued
 Solve the following simultaneously
 x+y =3 & y=1-2x
• x+(1-2x)=3
• -1x=2
• x=-2
• -2+y=3; y=5
•
Does the other equation fire also?
9
Class Examples Time
• Use the method of substitution
• y=x-4 & 2x+y=5
• y=x+4 & y=3x
• y=2x-1 & y=3x+2
10
Method of Elimination
Let us consider the class examples again
written in a slightly different form, i.e.
equating the variables to a number thus
• y-x=-4 & y+2x=5
• y-x=4 & y-3x=0
• y-2x=1 & y-3x=2
• Since in each of these examples that
coefficients of y are identical than we are just
able to subtract the equations
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Method of Elimination
 Write on equation under the other thus
 y - x=4
 y+2x=5
 0-3x=-9
 x=3; y-3=4; y=1
 Does the other equation also fire?
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Method of Elimination
 This method can be used when ever the either the x
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or the y coefficients are the same: consider:
3x+y=7 & x+y=3
5x+4y=17 & 3x+4y=15
Additionally if the signs are different we add instead
of subtracting
5x-2y=5 & 3x+2y=19
4x+3y=19 & 2x-3y=5
13
Class Examples Time
• Solve the following simultaneous equations using the
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•
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method of elimination verity your answers using
MATLAB.
2x+y=8 & x-y=1
2x+3y=14 & 2x-y=6
3x+5y=5 & x–5y =15
3x+2y=13 & 3x+y=9
6x+5y=3 & 2x+5y=9
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Equations with unlike coefficients
 So far we have only considered sets of
equations where the coefficients of either the
x or the y values are the same. A problem
occurs when this is NOT the case, consider:
• 2x+ y= 9
• 5x+2y=22
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General Method of Ellimination
(Cross Multiplication)

A.
B.
C.
The method of attack is ALWAYS the same
regardless and is as follows:
Multiply the first equation by the coefficient of y in
the second equation.
Multiply the second equation by the coefficient of y
in the first equation.
The coefficients of y are now equal so they can
now be solved as previously. (i.e. if they have the
same sign subtract otherwise add)
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Cross multiplication
 Let us now consider
how this works when
applying this to the
last example.
• 2x+ y= 9 (1)
• 5x+2y=22 (2)
• 2(2x+ y= 9) (1)
• 1(5x+2y=22) (2)
4x+2y=18 (1)
5x+2y=22 (2)
-1x+0=-4
x=4
2(4)+y=9
y=1
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Discussion Examples

1.
2.
3.
4.
5.
Use the general method of elimination to
solve the following:
2x+y=18 & 2x-3y=1
5x+2y=18 & 7x-8y=9
8x+7y=22 & 12x-5y=2
3x-4y=1 & 9x-7y=13
2x-3y=8 & 8x+15y=5
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Summary
 Simultaneous equations can be solve via the method
substitution or elimination generally the following
rules apply to which method should be used


Method of substitution if one of the equations y=mx+c
of x=ay+b then substitute this equation into the other.
If the equations are both un the form ax+by=c then
use the elimination method
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Solution of three simultaneous equations
Problems occur in scientific and engineering problems
where more unknowns are required. Hence more
equations are required. We will only concern ourselves
here with three unknowns as in principle the methods
shown in this lecture can easily be extended.
2 I1  3I 2  4 I 3  26
I1  5I 2  3I 3  87
 7 I1  2 I 2  6 I 3  12
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Discussion Examples
How would we approach these problems:
1)
3x  4 y  z  10
2 x  3 y  5 z  9
x  2y  z  6
2)
x yz 4
2 x  3 y  4 z  33
3x  2 y  2 z  2
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Use of mathematic Software
 Once you have more than four equations even with
the higher math methods of Matrices and
Determinates it is still cumbersome and due to the
amount of computations errors inevitably creep in
to calculation. Thus the need for computer
solutions of such systems of equations is self
evident. To solve such systems we can use a
number of math software packages inc. MatLab &
Derive
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Computer Solutions of Equations
Let us consider the
systems of equations as
before:
The solution protocol is
the same each time using
the math package
#1 3x  4 y  z  10
#2 2 x  3 y  5 z  9
#3 x  2 y  z  6
Go to SOLVE in the top
menu
Select system and enter
the appropriate expression
numbers (in this case
#1,#2 and #3)
The package then
returns the solution vector
for the system
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MatLab Computer Solution
1)
2 I1  3I 2  4 I 3  26
Use the following method to
solve the following systems of
equations:
I1  5I 2  3I 3  87
I. Type each of the coefficients on
the RHS into a matrix A=[2, 4,  7 I1  2 I 2  6 I 3  12
-4;1 -5 -3; -7 2 6] (Important to
remember the semicolons)
2)
II. Type the RHS as a column
x yz 4
vector [26; -87; 12]
III. Type in x=A\b and the system
2 x  3 y  4 z  33
will return the solution (NB a
3x  2 y  2 z  2
backslash \ is required)
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Summary
• Have we met our learning objectives?
• Specifically: are you able to:
• Find the solution to two simultaneous equations using
graphical means.
• Algebraic solutions to systems of two equations.
• Appreciate the solution protocol required for systems of
three equations.
• Use math software to solve systems of equations.
25
Derive Solution of Equations
Let us consider the
systems of equations as
before:
The solution protocol is
the same each time using
the math package
#1 3x  4 y  z  10
#2 2 x  3 y  5 z  9
#3 x  2 y  z  6
Go to SOLVE in the top
menu
Select system and enter
the appropriate expression
numbers (in this case
#1,#2 and #3)
The package then
returns the solution vector
for the system
26
Homework
1.
2.
3.
Find the equation of the line given the points (-1,6)
and (4,1) expressing your answer in the form
ax+by=c.
A line has a gradient of 1/2 and passes through a
point (-2,4), state the equation of the line.
Find the equation of a line with a gradient of unity
given that it passes through the point (0,-4).
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Homework

1.
2.
3.
4.
5.
Use the most suitable method to solve the
following simultaneous equations:
7x-10y=16 & 4x-15y=37
y=3x-1 & 2y-6x=2
3x-5y=13 & 7x+3y=1
3x+5y=5 & x-5y=15
3x+2y=8 & 7x+10y=24
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