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Simultaneous
equations
(cont'd)
ClementRadcliffe, 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.
Solve
simultaneously:
x
+ y = 7 ... (1)
2x
+ y = 10 ... (2)
Equation
(2) - Equation (1),
...
x = 3. Substituting x = 3 into equation
(1):
...3
+ y = 7
...y
= 4.
Answer:
x = 3, y = 4
Solve
simultaneously:
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.
Multiply
all terms by 2:
...
11 + 3y + 4y + 24 = 0
...
7y = - 24 -11 = - 35
...
y = - 35/7 = - 5
Substituting
into equation (1):
...
x = 11 + 3y = 11 + 3 x - 5 = 11 -
15
2
2 2
...
x = - 4/2 = -2
Answer:
x = -2, y = -5
Solve
simultaneously:
...
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 ÷ 6
19
...
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
- Formula
method
- Graphs
- Completing
the square
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
...
(3 x + 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
+ 3x = 0 ... 3x = -1
...
x = - 1/3.
Answer:
x = ? 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
Now
that you are comfortable with solving
simultaneous linear equations and
some quadratic equations, you can
now attempt the following for homework:
- x2
- 9x + 14 = 0
- 2x2
- x - 15 = 0
- 2x2
- x - 3 = 0
- x2
+ x = 6
- Solve
the equation: y = 2x2 - 3x - 2 when
y = 0
This
is the final lesson for this term,
so let me use the opportunity to wish
for you all a happy holiday. May you
use the time to be fully charged for
the challenges that await us in the
new year.
Clement
Radcliffe is the principal of Glenmuir
High School in May Pen.
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