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4
a = 1 and b = 4
CHAPTER 4: SIMULTANEOUS EQUATIONS
To solve between one linear and one non-linear equation.
Method : Substitution Example : Solve
x + 2y = 4 --------1
2 2
5 x
y y
x
-----2 from 2, × xy
2x
2
+ 2y
2
= 5xy ------------3 from 1, x = 4
– 2y substitute in 3
24 – 2y
2
+ 2y
2
= 54 – 2yy
216 – 16y + 4y
2
+ 2y
2
= 20y – 10y
2
8y
2
+ 10y
2
+ 2y
2
– 32y – 20y + 32 = 0 20y
2
– 52y + 32 = 0 4
5y
2
– 13y + 8 = 0 5y
– 8y – 1 = 0 y =
8 5
or 1 y =
8 5
, x = 4 – 2
8 5
= 4
16 5
=
4 5
y = 1, x = 4 – 2 = 2
Thus, x = 2, y = 1 and x =
4 5
, y =
8 5
.
Note Be careful not to make the
mistake 4
– 2y
2
=16 + 4y
2
wrong If the equations are joined, you have to
separate them.
Solve x
2
+ y
2
= x + 2y = 3 x
2
+ y
2
= 3 and
x + 2y = 3
CHAPTER 5: INDEX AND LOGARTHM Index form:
b = a
x
Logarithm form log
a
b = x Logarithm Law :
1. log
a
x + log
a
y = log
a
xy 2.
log
a
x – log
a
y = log
a
x y
3. log
a
x
n
= nlog
a
x 4.
log
a
b =
c c
log log
a b
5. log
a
a = 1 6.
log
a
1 = 0 Example: Find the value of
5 3
log
4
8 – 2log
4
3 + log
4
18
5 3
log
4
8 – 2log
4
3 + log
4
18 = log
4
5 3
2
8 18
3
= log
4
32 18 9
= log
4
64 = log
4
4
3
= 3log
4
4 = 3 × 1 = 3 To solve index equations, change to the same
base if possible. If not possible to change to the same base take logarithm on both sides of the
equation. Example: Solve 3.27
x-1
= 9
3x
3.27
x-1
= 9
3x
3 × 3
3x-1
= 3
23x
3
1 + 3x – 3
= 3
6x
1 + 3x – 3 = 6x
2 = 3x x =
2 3
Example: Solve 5
x+3
– 7 = 0 5
x+3
– 7 = 0 5
x+3
= 7 log 5
x+3
= log 7 x + 3log 5 = log 7
x + 3 =
log 7 log 5
= 1.209 x = 1.209
– 3 = 1.791
Example: Solve
a a
log 384 log
144 log 6
a
= 4
384 6 log
144
a
= 4
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5
a
log 16
= 4 16 =
4
a
= a
2
a =
4
CHAPTER 6: COORDINATE GEOMETRY 1.
Distance between Ax
1
, y
1
and Bx
2
, y
2
AB =
2 2
2 1
2 1
x x
y y
Example: If M2k, k and N2k + 1, k – 3 are
two points equidistant from the origin O. Find the value of k.
MO = ON
2 2
2 2
2 2
1 3
k k
k k
Square, 4k
2
+ k
2
= 4k
2
+ 4k + 1 + k
2
– 6k + 9 0 =
2k + 9 2k = 9
k =
9 2
2. Point which divides a line segment in
the ratio m : n
1 2
1 2
,
nx mx
ny my
n m
n m
Example: Given Q2, k divides the line which joins P1, 1 and R5, 9 in the ratio m : n. Find
a the ratio m : n b the value of k
a
5
n m
n m
= 2 n + 5m = 2n + 2m
5m – 2m = 2n – n
3m = n
1 3
m n
thus, m : n = 1 : 3 b
3 1 1 9 1 3
= k
12 3
4
= k 2.
Equation of a straight line Gradient form: y = mx + c
Intercept form:
1
x y
a b
Graident = m =
int ercept int ercept
y x
=
b a
General form: ax + by + c = 0 The equation of straight line given the
gradient, m, and passes through the point x
1
, y
1
: y
– y
1
= mx – x
1
Equation of a straight line passing throug two points x
1
, y
1
and x
2
, y
2
is
1 2
1 1
2 1
y y
y y
x x
x x
Example: Find the equatioon of the straight line a with gradient 3 and passes through
1, 2
b passes through 2, 5 and 4, 8 a Equation of straight line
y 2 = 3x – 1
y + 2 = 3x – 3
y = 3x – 5
b Equation of straight line
5 8 5
2 4
2
y x
5
3 2
2
y x
2y – 5 = 3x – 2
2y – 10 = 3x – 6
2y = 3x + 4 3.
Parallel and Perpendicular Line Parallel lines,
m
1
= m
2
Perpendicular lines, m
1
× m
2
= 1
Example: Find the equation of the straight line which is parallel to the line 2y = 3x
– 5 and passes through 1, 4
2y = 3x – 5 , y =
3 2
x -
5 2
m =
3 2
, passes through 1, 4 Persamaan garis lurus ialah
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6
y – 4 =
3 2
x – 1
2y – 8 = 3x – 3
2y = 3x + 5 Example: Find the equation of the straight line
which is perpendicular to the line
1 3
4
x y
and passes through 2, 3
1 3
4
x y
, m
1
=
4 3
=
4 3
4 3
× m
2
= 1
m
2
=
3 4
, passes through 2, 3 The equation of the straight line is
y – 3 =
3 4
x – 2
4y – 12 =
3x + 6 4y + 3x = 18
4. Equation of Locus
Example: Find the equation of the locus for P which moves such that its distance from Q1, 2
and R 2, 3 is in the ratio 1 : 2
Let Px, y, Q1, 2, R 2, 3
PQ : PR = 1 : 2
1 2
PQ PR
PR = 2PQ
2 2
2 2
2 3
2 1
2
x y
x y
Square, x
2
+ 4x + 4 + y
2
– 6y + 9 = 4x
2
– 2x + 1 + y
2
– 4y + 4 = 4x
2
+ 4y
2
– 8x – 16y + 20 0 = 4x
2
– x
2
+ 4y
2
– y
2
– 12x – 10y + 7 3x
2
+ 3y
2
– 12x – 10y + 7 = 0
CHAPTER 7: STATISTICS 1.
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