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Simultaneous
equations
Clement
Radcliffe, Contributor
We
will complete our review of simultaneous
equations today by looking at the
solutions to some of the practice
examples that were given for homework.
3x
+ 2y = 1 . . . (1)
4x
- y = 16 . . . (2)
Multiply
equation (2) by 2
...
8x - 2y = 32. (3)
Add
equations (1) and (3)
...
11x = 33
...
x = 33/11 = 3.
Substituting
x = 3 into equation (1):
...
3 x 3 + 2y = 1
...
9 + 2y = 1
...
2y = 1 - 9 = - 8
...
y = - 4
Answer:
x = 3, y = -4.
2x
= 11 + 3y . . . (1)
x
+ 2y + 12 = 0 . . . (2)
Using
the substitution method: from equation
(1), x = (11 + 3y)/2
Substituting
into equation (2):
(11
+ 3y)/2 + 2y +12 = 0.
Multiply
all terms by 2:
...
2 x (11 + 3y)/2 + 2 x 2y + 2 x 12
= 0.
...
11 + 3y + 4y + 24 = 0
...
7y = - 24 -11 = - 35
...
y = - 35/7 = - 5
Substituting
into equation (1):
...
x = (11 + 3y)/2 = (11 + 3 x - 5)/2
= (11 - 15)/2
...
x = - 4/2 = -2
Answer:
x = -2, y = -5.
...
2x + 3y = -1 . . . (1)
5x
- 2y = 18 . . . (2)
Multiply
equation (1) by 2 and equation (2)
by 3.
...
4x + 6y = -2 . . . (3)
15x
- 6y = 54 . . . (4)
Adding
equations (3) and (4)
...
19x = 52
...
x = 52/19
Substituting
into equation (3)
=
4 x 52/19 + 6y = -2
...
208/19 + 6y = -2
...
6y = -2 -208/19 = -246/19
...
y = -246/19 ÷ 6
...
y = -41/19
Answer:
x = 52/19 , y = -41/19.
Now
to a new topic:
SOLUTION
OF QUADRATIC EQUATIONS
The
following are the methods which are
commonly used at this level.
- Factorisation
- Graphs
- Completing
the squares
- Formula
method
We
will now begin with the factorisation
method.
POINTS
TO NOTE
- Quadratic
equations are expressed in the form
ax2 + bx + c = 0, where a, b and
c are constants.
- The
factorisation method is used if,
and only if, the expression ax2
+ bx + c can be factorised.
- Given
the equation x2 + 7x + 10 = 0, then
by factorising the left hand side,
you get (x
+ 2 )( x + 5 ) = 0.
If
(x + 2 )( x + 5 ) = 0
then
(x + 2) = 0, that is x = - 2
OR
x + 5 = 0, that is x = -5.
Solutions
are x = -2 and x = -5.
- Be
reminded that the solutions of the
equation are the values which satisfy
the equation. These can be checked
by substitution as follows:
If
x2 + 7x + 10 = 0, then
if x = -2, 4 -14 + 10 = 0.
Similarly,
where x = -5, then 25 -35 + 10 = 0.
The equation is satisfied by both
solutions.
We
shall look at some examples.
1.
Solve 3x2 - 7x -6 = 0
Using
factorisation:
3x2
- 7x -6 = 0
...(3x
+ 2) (x - 3) =0
...3x
+ 2 = 0, that is, 3x = - 2
...x
= -2/3
When
x - 3 = 0,
...x
= 3
Answer:
x = - 2/3 and 3
2.
Solve the equation:
1
- 9x2 = 0
Factorising
using difference of two squares:
1
- 9x2 = (1 - 3x)(1 + 3x)
...(1
- 3x)(1 + 3x) = 0
...1
- 3x = 0 or x = 1/3
1
+ 3x = 0 ... 3x = -1
...x
= - 1/3.
Answer:
x = 1/3 or - 1/3.
ALTERNATIVELY
1
- 9x2 = 0
...9x2
= 1
x2
= 1
9
...x
= ± 1/3.
3.
Solve the equation: 3(x +2)2
= 7(x + 2) (June 1990, no. 3a)
3(x
+2)2 = 7(x + 2) Clearing
the brackets:
...3(x2
+ 4x + 4) = 7x + 14.
...3x2
+ 12x + 12 = 7x + 14.
...3x2
+ 12x -7x + 12 - 14 = 0.
...3x2
+ 5x - 2 = 0. Factorising:
...(3x
- 1)(x + 2) = 0
...3x
-1 = 0, that is, x = 1/3.
OR
x + 2 = 0, that is, x = -2.
Answers
are x = 1/3 and -2.
NB
You may also solve the above using
the factorisation method.
Now
that you are comfortable with solving
simultaneous linear equations and
some quadratic equations, you can
now attempt the following for homework.
x2
- 3x - 4 = 0
6x2
- x - 15 = 0
2x2
- x - 3 = 0
x2
+ x = 6
Solve
the simultaneous equations: 3a - 1/2b
= 4
9a
+ 2b = -2
I
urge you to find other examples in
your textbooks and past papers and
do them. After all, practise becomes
perfect.
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Andre
McDaniel, of Cornwall College,
receives the champion school
award from Janet Sharp, executive
vice-president and resident
actuary of Sagicor Life Jamaica,
at the Insurance Association
of Jamaica's Vivian Rochester
Memorial Mathematics Competition
prize-giving ceremony, recently.
-Photos by Rudolph Brown/Chief
Photographer
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Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
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