Download SIMULTANEOUS EQUATIONS

SIMULTANEOUS EQUATIONS
Example 1.
Rhoda bought a fruit cake and two chocolate cakes at a cost of Sh.2, 000.On the
following day her brother Luther visited the same shop and bought two fruit cakes
and three chocolate cakes at a cost of Sh.3500.What was the cost of each cake?
To find the value of two unknowns in a problem, two different equations must be
given that relate the unknowns to each other. These two equations are called
simultaneous equations.
a)
Substitution method
From the situation above, let the cost of the fruit and chocolate cakes be x and y
respectively.
x + 2y = 2000
…(1)
2x + 3y = 3500
…(2)
a)
Label the equations so that the working is made clear.
b)
Write one variable in terms of the other using one of the equations. E.g
from (1) x = 2000 - 2y.
c)
Substitute this expression for x in equation (2) and solve to find y.
2(2000 – 2y) + 3y = 3500.
4000 – 4y +3y = 3500.
4000 – y = 3500.
y = 4000 – 3500
y = 500.
d)
Find x from any of the two equations using this value of y.
x + 2(500) = 2000.
x +1000 = 2000.
x = 2000 – 1000
x = 1000.
The fruit cake costs Sh.1000 and the chocolate cake costs Sh.500.
Exercise.
1.
Use the substitution method to solve the following.
2x + y = 5
2.
X + 3y = 5
3.
3x + y = 10
x + 2y = 8
2x + 3y = 14
4.
x–y=2
2x+ y = - 3
x – y = -3
b) Elimination method.
Example 2.
The cost of one litre of petrol and two litres of diesel is Sh.5800.If you buy two
litres of petrol and three litres of diesel, the cost would be Sh.9800.Find the unit
cost of petrol and diesel.
Let the cost of a litre of petrol be x and that of diesel y.
x + 2y = 5800
… (1)
2x + 3y = 9800
… (2)
a)
Label the equation so that the working is made clear.
b)
Choose an unknown in one of the equations and multiply the equations by
a factor or factors so that this unknown has the same coefficient in both
equations.
c)
Eliminate this unknown from the two equations by subtracting them, then
solve for the remaining unknown.
d)
substitute in the first equation and solve for the eliminated unknown.
(1) x 2
x + 2y =5800
…(1)
2x + 4y = 11,600
…(3)
2x + 3y = 9800
…(2)
Subtract (2) from (3)
y = 1800
Substituting in (1)
x + 2(1800) = 5800
x + 3600 = 5800
x = 5800 – 3600
x = 2200.
One litre of petrol costs Sh.2200 while diesel Sh.1800.
Exercise.
1.
Use the elimination method to solve the following:
2x + 5y = 24
2.
4x +3y = 20
3.
5x + 2y = 13
2x + 6y = 26
4x +y = 11
4.
9x + 2y = 28
x + 2y = 17
8x +3y = 45
C) Solution by Matrix method.
Example 3.
The wage bill for five men and six women workers is Sh.670, 000, while
the bill for eight men and three women is Sh.610,000.Find the wage for a
man and a woman.
Let the wage of a man be x and that of a woman y.
5x + 6y = 670,000.
8x + 3y = 610,000.
….(1)
…..(2)
(a) Re-write the equations in matrix form.
5
8
6
3
x
y
=
670,000
610,000
This shows the matrices of coefficients, variables and constants in the
order they appear.
(b) Find the inverse of the matrix of coefficients.
Note: In order to find an inverse of matrix requires us to find the
determinant of the matrix first.
The determinant of
5 6
8 3
is (5 x 3 ) – ( 6x 8) = - 33.
Thus the inverse of the above matrix would then be:
1
-33
.
3 -6
-8 5
(c) Pre-multiply through the matrix equation by this inverse.
1
3 -6
-33 -8
5
5 6
x
= 1
3 -6
670000
8 3
y
-33
-8 5
610000
When a matrix is multiplied by its inverse the result is an Identity
matrix. Thus the left hand side yields an identity matrix.
1 0
x
= 1
-1650000
0 1
y
-33
-2310000
=
50000
x
y
70000
Therefore a man earns Sh.50,000 while a woman Sh.70,000.
Exercise.
Formulate a pair of simultaneous equations and solve using the
matrix method.
1.
Find two numbers with a sum of 28 and a difference of 2.
2.
Twice one number added to three times another gives 26. Find the
number, if the difference between them is 3.
3.
The average of two numbers is 9, and three times the difference between
them is 18. Find the numbers.
D) Graphical solution of Simultaneous equations.
Example 4. Solve graphically
3x + 2y = 8
x – 4y = -2 .
The equations above are equations of straight lines. Hence we can draw the
two lines on the same graph page and find their point of intersection. The
co-ordinates of this point of intersection will provide the solutions to the
simultaneous equations.
For the line
3x + 2y = 8
x
0
2
4
y
4
1
-2
For the line x – 4y = -2.
x
2
6
10
y
1
2
3
Exercise: Draw the two lines on the same pair of axes and find the point of
intersection. State the x and y co-ordinates as the solution.
Example 5.
Solve the simultaneous equations below graphically.
x+y=3
y – x2 = 1.
One of the equations is a quadratic while the other is linear.
Draw the graph of x + y = 3 which is a straight line.
Table of values.
x
0
3
1
y
3
0
2
On the same graph page include the graph of y = x2 + 1 which is a curve.
Table of values.
x
-5
-4
-3
y
26 17 10
-2
-1
0
1
2
3
4
5
2
1
2
5
10
17
The point of intersection of the curve and the straight line provides the
solution to the equations.
Exercise: Draw the line and curve on the same pair of axes and read of the x
and y values as the solution.
Exercise. Solve the following simultaneous equations graphically.
1.
2x – 3y = - 12
3x + 2y = -5.
2.
4x + 5y = 1
7x – 3y = - 10.
Miscellaneous exercise.
1.
The line, with equation y + ax = c, passes through the points (1,5) and (3,1).
Find a and c.(Hint: For the point (1,5) put x = 1 and y = 5 into y +ax = c.)
2.
The curve y = ax2 + bx passes through (2,0) and (4,8). Find a and b.
3.
Given x2 – y2 = -5 and x + y = 5, find x and y.