Mathematics Lesson Notes SS2 First Term
Mathematics Notes SS2 First Term – Edudelight.com
SCHEME OF WORK FOR SS2 MATHEMATICS SESSION
WEEKS TOPICS
- Revision of SS 1 Work
- Logarithm of numbers less than 1
- Approximations and percentage error
- Sequence and series
- Geometric Progression
- Revision of Factorization of Perfect Squares
- Review of the first half term’s work and test
- Construction of quadratic equation from sum and products of roots
- Simultaneous linear equations (revision)
- Use of the graphical methods to solve other related equations
- Revision of the 2nd half term is work and preparation for examination
- First term examination.
REFERENCE MATERIALS
THE MAN MATHEMATICS FOR SENIOR SECONDARY SCHOOL BOOK 2
ESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOL BOOK 2
WEEK ONE
REVISION OF SS1 WORK
WEEK TWO
LOGARITHM OF NUMBERS LESS THAN ONE
Here we use the negative power
EXAMPLES
- Use the logarithm tables to find the logarithms of (a) 0.005608 (b) 0.07896 (c) 0.0004968
Solution
- 0.005608 = 5.608 × 10-3in standard form, hence the characteristic is 3
To find the mantissa, look for 56 under 0 to obtain 7482 and continue until the column headed 8 in the difference part to find 6, which must be added to 7482 to give 7488
Therefore, log 0.005608 = 3 .7488-
Note: 3.7488 = –3+0.7488 and is not the same as -3.7488
- 0.07896 = 7.896 × 10-2 (b) 0.0004968 = 4.968 10-4
Characteristic is 2 characteristic is 4
The mantissa is 8974 the mantissa is 6962
Log 0.0004968 = 4.6962
- Use antilogarithm tables to find the numbers whose logarithms are; (a) 2.7903 (b) 4.0907(c) 3.5711
Solution
- the mantissa is 7903 in the antilogarithm table its correspond to 0.7903
The characteristic is 2, so the must be one zero between the decimal point and the 1sts.f
Antilog 2.7903 = 0.06170
- Mantissa is 0907 which correspond to 0.1232,Characteristic is 4,
Antilog 4.0907 = 0.0001232
- Mantissa is 5711 which correspond to 0.3725, characteristic is 3
Antilog 3.5711 = 0.003725
EXERCISE
- Using the logarithm tables, find the logarithm of (a)0.8702 (b) 0.0002312 (c) 0.00007008
- Using the antilog tables, the numbers whose are (a) 2.9805 (b) 3.5401 (c) 1. 9846
MULTIPLICATION AND DIVISION OF NUMBERS LESS THAN ONE
Recall (i) log A B = log A + log B (ii) log = log A – log B
EXAMPLES: Evaluate using Logarithm tables a). 0.08907 X 0.006792 (b). 0.00889 204.6
SOLUTION
a). 0.08907 X 0.006792
| NO | Log |
| 0.8907 | – 1.9497 |
| 0.006792 | – 3.8320 |
| 0.006049 | – 3.7817 |
0.08907 X 0.006792 = 0.006792
(b). 0.00889 204.6
| NO | Log |
| 0.00889 | – 3.9489 |
| 204.6 | – 2.3109 |
| 0.00004345 | – 5.6380 |
0.00889 204.6 = 0.00004345
EXERCISE:Evaluate the following, leaving your answer to 3.S.f
(a). 32.48 X 0.03467 X 0.00897 X 0.9458
(b). 8.75
POWER AND ROOTS OF NUMBERS LESS THAN
Example: Evaluate (a) (10.05872)2 (b)
Solution
| NO | Log | |
| 0.05872 | – 2. 7687 | |
| 0.058724 | – 2. 7687 x 4 | |
| 0.00001186 | – 5.0748 |
i.e 0.000012 to S.F
(b)
| NO | Log |
| 0.00894 | – 3. 9513 |
| – 3. 9513 3 | |
| 0.2075 | – 1.3171 |
Assignment: Work out the following given your answer in bar notation (a) 2. 98 x 3
(b) 3. 7 4 (c) 4. 2351 3
Mathematics Notes SS2 First Term – Edudelight.com
WEEK 3
APPROXIMATION AND PERCENTAGE ERROR
ROUNDING OFF OF NUMBERS
Example: Round off 78539 to the nearest (a) ten (b) Hundred (c) thousand
Solution
- 78539 = 78540 (b) 78500 (c) 79000
Decimal places
Example: Round off 567.5684 to the (a) nearest whole number (b) 1 d.p (c) 2d.p (d) 3 d.p
Solution
(a) 568 (b) 567.6 (c) 567.57 (d) 567.568
Example 2: Round off 26.5274 to the nearest (a) tenth (b) hundredth (c) thousandth
Solution:
- 26.5 (b) 26.53 (c) 26.567
Significant figures (s.f)
Example: Round 0.0006729 to (a) 1sf (b) 2s.f (C) 3 s.f
Solution
- 0.0007 (b) 0.00067 (c) 0.000673
Percentage error
All measurements are not exact. Therefore whenever we use measured or values or round off value. We introduce error into the figure. Example if Ajide is 1.6 tall to 1 d.p. then the actual height could be between 1.55m and 1.65m. the error can be calculated as 1.55 – 1.6 or 1.65 – 1.6 i.e -0.05 or + 0.05 i.e 0.05
Absolute error: it is the difference between the measured value and the actual value.
Relative error: it is obtain by dividing the absolute error by the actual value
Percentage error: Percentage error = x
That is x = 3.125 = 3.1% to 1s.f
Example
- The weight of a book is given as 1.1kg but the actual weight is 1.24 kg. What is the percentage error?
Solution
Error = 1.24kg – 1.11kg = 0.14kg
Percentage error = x 100%
x 100%
- Instead of expressing 2/21 as a decimal number to 2 s.f a student expressed it to 2 d.p. Find the percentage error.
Solution
2/21 = 0.095238 = 0.095 to 2 s.f,Error= 0.10 – 0.095 = 0.005
Percentage error =
Exercise
- A carpenter was told to make a rectangular desk with a top which has dimensions 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.
- Calculate the percentage error in the
- Length and breadth (ii) area of the table top
- Find the product of the two errors in (ai)
- A man bought 5 reams of duplication paper, each of which was supposed to contain 480 sheets. The actual number of sheets in the packets were: 435, 420, 405, 415 and 440.
- Calculate, correct to the nearest whole number the average percentage error for the packets of papers.
- If the agreed price for a full ream was N35.00 find correct to the nearest naira, the amount by which the buyer was cheated.
Assignment
- A sale girl have a change of N1.15 to a customer instead of N1.25 calculate her percentage error.
- A boy measures the length and bread of a rectangular lawn as 59.6m and 40.3m respectively instead of 60m and 40m. What is the percentage error in his calculation of the perimeter of the lawn?
Week 4
SEQUENCE AND SERIES
A sequence is a list of numbers that follows a rule.
Examples: Find the next two terms of the sequence (a) 3, 7, 11, 15, 19 (b) 1, 4, 9, 16, 25 (c) 1, 1, 2, 3, 5, 8 (d) 81, 27, 9, 3
The nth term of a sequence (Tn or Un)
The nth term of a sequence is the term from which all other term can be generated using natural 1, 2, 3, 4 …………n
Example: The nth term of a certain sequence is given as Tn = 3X2n-1
- Write down the first three terms of the sequence
- What is the 10th tern of the sequence
Solution
Tn= 3 x 2n-1
- The first term T1 = 3 x 2o = 3
The second term T2 = 3 x 21 = 6
The Second term T3= 3 x 22 = 12
The first three terms are 3, 6, 12
- The 10th T10 = 3 x 29= 1536
2. Find the nth term of the sequence 2, 6, 10, 14, 18
Solution
Since the rule is add 4 each time. So we try 4n – 2 or 4n + 2 when n=1, 4n-2 = 4-2 =2.
When n = 2, 4n-2=8-2=6
When n=3 4n-2=12-2 = 10
When n=4, 4n – 2 = 16 – 2 = 14
When n = 5, 4n-2 = 20-2 = 18. Tn = 4n-2
Exercise: Find the first three terms of the sequence with nth term Un=3n2
State the nth term of the sequence 2, 5, 8, 11, 14
A Series
A series is formed by addition the terms of a sequence e.g 3+6+9+12+………….is an infinite series while 1+2+4+8+16 is finite series.
Arithmetic Progression (A.P)
A sequence or series is called an A.P if each term is obtained by adding a constant number to the previous term. The constant is called the common difference
| S/N | SEQUENCE | FIRST TERM (A) | COMMON DIFFERENCE (D) |
| 1 | 1, 3, 5, 7,9 | 1 | 2 |
| 2 | 16, 11, 6, 1, 1, -4 | 16 | -5 |
| 3 | -5, -3, -1, 1, 3 | -5 | 2 |
The nth term of an A.P
If the first term of an A.P is a and the common difference (d) Then the first four terms are U1=a, U2=a+d, U3=a+2d and U4 = a+3d in general Un=a+ (n-1) d
Example
- Find the (a) 10th (b) 15th (c) 100th (d) nth term of A.P 5, 10, 15, 20……..
Solution
5, 10, 15, 20………… (c) U100 = a+99d
= 5 + 495
= 500
- U10 = a +9d (d) Un = a + (n-1) d
= 5 + 45 = 5 + (n-1)5
= 50 = 5 + 5n – 5
Un = 5n
- U15 = a+14d
= 5 + 70
= 75
- Determine the number of the term which is 83 in the A.P 3, 8, 13, 18,…………
Solution
a=3, d=5, n=? Un = 83
Un = a+(n-1)d
83 = 3 + 5(n-1)
5n – 2 = 83
5n = 85
n=17
Exercise:
In an AP the difference between the 8th and 4th term is 20 and 8th term is times the 4th term. Find the (i) common difference (ii) the first term of the sequence.
Arithmetic mean
| b = |
If a, b and c are three consecutive terms of an AP then b is the arithmetic mean of a and c.
Example:
If x+4, 2x+8 and 12 are three consecutive terms of an A.P. Find the value of x.
Solution
2x+8 = OR
4x + 16 = x + 16 2X + 8 – (x +4) = 12 –(2x +8)
3x = 0 x + 4 = 4 -2x
X = 0 3x = 0
x = 0 we have 4, 8, 12 in the A.p
Exercise: If 8, x, y, -4 are in an A.P. Find x and y
Sum of terms of an A.P
| Sn = (a + l) |
| Sn = (2a + (n-1)d |
If the first term of an A.P is a, the common difference (d) then the sum of the first nth term denoted Sn is given as Or
Where l is the Last term or nth term. L = a+(n-1)d
Examples: (1) Find the sum of the first 20 terms of the A.P 17 + 12 + 7 + 2 +……………
(2) Find the sum of the A.P 2 + 4 + 6 + …………. + 48
Solution
- a = 17, d = 15 a = 2, l = 48, n = 24
Sn = (2a + (n-1) d] Sn = (a + l)
S20 = 10(34 + 191 – 5) S24 = 12(2 + 48)
S20 = -610 S24 = 600
Exercise:
A man is able to save N50.00 of his salary in a particular year. After, every year he saved N20.00 more than the preceding year. How long does it take him to save N4370? JAMB
Assignment
- The first and last terms of an A.P are 21 and -47 respectively. If the sum of the series is given as -234. Calculate (a) the number of terms in the A.P (b) the sum of the first 18 terms
- Yusuf has just secured a job in a big company as a clerical officer. His starting salary is N180, 000 per annum with annual investment of N2, 000. Determine his salary in the 20th year and find his total salary he would have and in the first 20 yrs.
WEEK 5
Geometric Progression (G.P)
A sequence in which each term is obtained from the proceeding term by multiplying or dividing by a constant factor is called the G.P
Examples: 16, 8, 4, 2, 1 & 1, 3, 9, 27, 81
Nth term of a G.P
| Un = arn-1 |
If the first term of a G.P is a and the common ratio r. Then the first three terms are U1= a, U2= ar and U3 = ar2. In general
Example: Find (a) the 8th (b) 34th and (c) nth term of the G.P 10, 20, 40, 80
2. if the first, second and last term of a G.P are 41, -20 and 12500 respectively. Find the number of terms in the G.P.
Solution
- a = 10, r = = 2 2. a = 4, ar = -20, arn-1 = -12500
- U8 = ar9 = 1280 r = = -5
- U34 = ar33 = 5 x 234 Un = arn-1
- Un = arn-1 = 10 x 2n-1 -12500 = 4(-5)n-1
2 x 5 x 2n-1 (-5)2 = (-5)n-1
5 x 2n n-1 = 5, n = 6
Exercise: The third and fifth term of a G.P are and respectively Find the (a) Common ratio (b) First term
| b = |
Geometric Mean
If a, b and c are three consecutive term of a G.P b is called the geometric mean of a and b.
Example: if x + 3, x +8 and x + 18 are 3 consecutive terms of a G.P. Find the value of x.
Solution
(x + 8)2 = (x + 8) (x + 18), 5x = 10, x =2
Exercise: If , x, , y ……. are in G.P. Find the product of x and y.
Solution
, = , 16y = 8, xy =
X = = y =
The sum of nth terms to a G.P
The sum of a geometric progression with first term (a) and Common ratio(r) is given as
| Sn = |
| If /r/ < 1 |
| 1 |
| Sn = |
| 2 |
| If /r/ > 1 |
Examples: Find the sum of the first 6 terms of the following G.P
- 90 + 30 + 10 + + ………………
- 2 + 4 + 8 + 16 + ………….
Solution
- a = 90, r = , /r/ < 1 b. a = 2, r = 2, /r/ > 1
Sn = S6=
S6 =
= S6 = 126
| S = |
Sum to infinity of a G.P
From Sn = as n x /r/ < 1
Examples: (1) Find the sum to infinity of the G.P
- 1 + + …….. (b) 1 + + 2 + 2 + ………
Solution
a = 1, r = a = 1, r =
S = S =
= = = 10
Assignment: Exercise E Page 35, No 7 – 9
WEEK 6
Completing Quadratic Expressions into Perfect square
Examples of perfect squares include (i) x2 + 2x + 1, x2 – 6x + 9, 4x2– 4x + 1 etc.
Example: Find the term when added will change the expression x2 + 4x into perfect square.
Solution
Let the constant term be n
X2 + 4x + n = (x + a)2
X2 + 4x + n = x2 + 2ax + a2
2a = 4
a = 2
n = a2
n = 22 = 4
OR
Step 1: Find half the coefficient of x i.e 2
Step 2: square this result i.e 22 = 4
Step 3: add this square to the original expression to obtain a perfect square. X2 + 4x + 4
Exercise: Change the following expressions into a perfect square
- 2x2 – 3x b. 3x2 – 2m c. 2v2 – 5v d. 2x2 + x
e. 5x2 + 4x f. 3x2 – 10x g. – 2t
Solving quadratic Equations by completing the Square method
Recall if x2 + bx + 2 = 0 then 2 = 0 .
The RHS of x2 + bx + 2 = 0 is called a perfect square
Example: Solve the following equations using the completing the square method.
(a) 4x2 – 4x – 3 = 0 (b) x2 + 7x + 10 = 0 (c) -6t -2 = -5t2 (d) x2 -10x + 8 = 0
Solution
Step 1: Add 3 on both sides 4x2 – 4x = 3
Step 2: divide both sides by 4 x2 – x =
Step 3: add (half the coefficient of x)2 x2 – x + 2 = + 2
Step 4: Factorize LHS 2 = 1
Step 5: Take square root on both sides =
X = +1 + or X = -1 +
X = or x =
- X2 + 7x + 10 = 0 (c) 5t2 – 6t – 2 = 0
X2 + 7x = -10 t2 =
X2 + 7x + 2 = -10 + 2 t2 + 2 = + 2
2 = 2 = +
= t =
X = or x = t – 0.6 =
X = -2 or x = -5 t = 1.47 or t = -0.27
Solving quadratic equations by Formula
Given the equation ax2 + bx + c = 0
X2 + x + = 0, dividing both sides by a
X2 + x = , subtracting from both sides
X2+ + 2 = + 2 make the LHS a perfect square
2 = Factorizing the LHS
= , Take square root of both sides
This is known as the quadratic formula
Examples: Solve the equation 2x2 + 4x – 9 = 0 by using the quadratic formula. Give your answer to 2d.p
Solutions
2x2 + 7x – 11 = 0 x =
a = 2, b=7, c = -11
x = x = 1.176 or x = -4.676
x = x = 1.18 or x = -4.68
Assignment
Solve correct to 2dps the quadratic equation
2x2 + 7x – 11 = 0. Did you know x =
Mathematics Notes SS2 First Term – Edudelight.com
WEEK SEVEN
MID TERM BREAK
WEEK 8
Construction of Quadratic Equations from sum and products of roots
If are the roots of the equation ax2 + bx + c = 0.
Then we have x2– ( ) x + =0 OR X2 – (sum of roots)x + (Product of roots) = 0
Examples: Find the quadratic equations whose roots are;
- 2, 3 (b) -3, 2 (c) , (d) 1, (e) 3, -3
Solution
- = 2 + 3 = 5 (b) = -3 + 2 = -1
(c) = + = 2 (d) = 1 + =
= x = = 1 x =
X2 – 2x + – 0 x2 +
4x2 – 8x + 3 = 0 3x2 – 4x + 1 = 0
(e) = -3 + 3 = 0
= -3 x 3 = -9
X2 – 9 = 0
Word Problems leading to Quadratic Equations
Examples: (1) The product of two numbers is 40 and their sum of 13. Find the numbers.
(2) The breadth of a rectangle is 3cm less than the length, If its area is 88cm2. Find breadth.
Solution
| 1 |
Let the no be x and y
x + y = 13 (2) Let length = l
| 2 |
Breadth = l -3
xy = 4 Area = l(l-3) =88
| 3 |
| 1 |
L2 – 3L = 88
From y = 13 – x L2 – 3L – 88 = 0
| 2 |
| 3 |
(L – 11) (L + 8) = 0
We substitute into L = 11 or L = -8
x (13 – x) = 40 L=11 since L > 0
13x – x2 = 40 and bread = L – 3
X2 – 13x + 40 = 0 = 11 – 3
(x – 5) (x – 8) = 0 = 8cm
X = 5 or x = 8 length = 11cm
y = 5 when x = 8
and y = 8 when x = 5
Exercise: (1) The sum of two numbers is 44 and their product is 483. What are the numbers?
(2) The area of a rectangle is 35cm2 and the perimeter is 24cm. find the length and breadth.
Assignment
- The sum of two numbers is 20 and the sum of their sequence is 218. Find the numbers.
- A child is 15 years and his mother is 38 years in what year was the product of their ages 78?
- A father is twice as old as his son. The product of their ages 10 years ago was 532. Find the age of the father now.
Mathematics Notes SS2 First Term – Edudelight.com
WEEK 9
Simultaneous Equations
Two Linear equations in two variables are called simultaneous linear equations in two variables. We solve a simultaneous equations by looking for the values of two unknowns.
Examples are: 3x – 2y = 5 (ii) x + y = 9 etc
2x + 4y = 6 x – y = 7
Method of Solving linear Simultaneous Equations
- Elimination Method:
Examples: solve the equations 5x-2y=16, 3x+4y+6=0
| 1 |
Solution
5x-2y=16 Sub y = -3 into (1) to have
| 2 |
5x – 2(-3) = 16
3x+4y-6 5x = 16 – 6
We eliminate x as follows
3x (1) 15x – 6y = 48 iii x = 2
5x (2) 15x + 20y = -30 iv y = -3 and x = 2
-26y = 78
-26 -26
Y = -3
Exercise: Solve x + y = 12 and x – y = 6 (ii) 2x + y = 10 and x – y = 2
(iii) 2x + 3y = 11, 4x + y = 12Using elimination method
Substitution method
| 1 |
Example: 5x-2y=16
| 2 |
3x+4y=-6
From (1) 2y=5x – 16
| 3 |
Y= 5x-6
2
Sub (3) into (2) gives 3x + 4 = -6, multiply by 2
6x + 20x-64=-12
26-x=52
X=2
From (3) Y= = , y=-3
Exercise: Solve the following pairs simultaneous equations by substitution method. (a)2x+5y=6 , 5x-2y=9 (b) y=3x, 4y=5x+14.
Assignment: ExB2. No 24-28, page 55
Graphical method
Linear equations gives straight lines. While quadratic equations give curves.
| 2 |
| 1 |
Examples: 5x-2y=16 3x+4y =6
Step: For each equation, express one of the variable, in terms of the other variable. The obtain two tables of values.
| x | 0 | 1 | 2 | 3 | 4 |
| y | -8 | -55 | -3 | 2 |
From (1) 2y=5x-16
Y=
From (2) 4y=-6-3x
| x | 0 | 1 | 2 | 3 | 4 |
| y | -1-5 | -2.25 | -3 | -3.75 |
Y=
Step2: From the table of values plot the two straight line graphs
| 2, 3 |
| 5 |
| 4 |
| 3.5 |
| 2.5 |
| 4.5 |
| 3 |
| 2 |
| 1.5 |
| 1 |
| 0.5 |
| 1 |
| 2 |
| 3 |
| -10 |
| -9 |
| -8 |
| -7 |
| -6 |
| -5 |
| -4 |
| -3 |
| -2 |
| -1 |
| -2 |
| -3 |
| -4 |
| -5 |
| -1 |
| Y = |
| Y = |
Step 3: The Coordinator of the point where the two lines meet give the required solution.
Therefore x = 2, and y = -3 that is (2, -3).
Exercise: Solve graphically the following simultaneous equations (a) 2x – y = 0, 3x + 2y = 5
- 2x + 3y + 5 = 0, x + y + 2 + 0.
Graphical Solution of linear and Quadratic Equation
Example: (a) Using a scale of 1cm for 1 unit of the x-axis and 1 cm for 5 unit on the y – axis and draw the graph of y = 2x2 – 2x – 1 in the interval -5
(b) Using the same scale and axes draw the graph of 2x – 3y + 5 = 0
(c) Use the graph to solve the simultaneous equations 2x – 3y + 5 = 0 and y = 2x2 – 2x – 1
Solution
(b) 2x – 3y + 5 = 0
y =
| X | -1 | 2 | 5 |
| y = | 1 | 3 | 5 |
| X | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y = 2x2 – 2x – 1 | 59 | 39 | 23 | 11 | 3 | -1 | -1 | 3 | 11 | 23 | 39 |
- The x-coordinates of the points where the straight line meets the curve gives the required solution that is x = 2 or x = -0.7
Exercise: Solving graphically (a) 2x – y = 0, 2x2– y = 0
- Y = 3x – 6, y = (x – 2) (x – 3). Using a scale of 1cm for 1unit on each axes in the interval -2 x 6.
Graphical Solution of quadratic Equations
Example: Using a scale of 1 cm for 1 unit on the x axis 1cm for 10 units on the y-axis draw the graph of y = 2x2 – 9x + 4 from x = -5 to x = 5
Use your graph to solve the following equations
- 2x2 – 9x + 4 = 0 (b) 2x2 – 9x + 7 = 0
Solution
| X | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| Y | 99 | 72 | 49 | 30 | 15 | 4 | -3 | -6 | -5 | 0 | 9 |
- The roots of the equation 2x2 – 9x + 4 = 0 are the x-coordinates of the points where the straight line y=0 (x – axis) meets the curve. The roots are therefore x = 0.5 or x = 4.
- 2x2– 9x + 7 = 0 can be written as 2x2 – 9x + 4 = -3. Where line y= -3 meets the curve. The required roots are x = 1, x = 3.5
Exercise ExD No 12, 16 and 17 page 60
Word Problems involving Simultaneous Linear Equations
Examples: (1) A two-digit number is six times the sum of its digits by 9. Find the number
(2) The perimeter of a rectangle is 102m. The difference between the length of the rectangle and its breadth is 9m. Find the length and breadth of the rectangle.
Solution
- Let the unit digit be x (2) Let length = b, breadth = b
| 1 |
and tens by y Perimeter = 2(L + b) = 102
yx = 10y + x = 6(x + y) 2L + 2b = 102
| 1 |
| 1 |
| 2 |
10y + x = 6x + 6y also L – b = 9
| 2 |
5x – 4y = 0 1 x 2L + 2b = 102
Also when the digits are reversed 2 x 2L – 2b = 18
Xy = 10x + y = 6(x + y) – 9 =
| 2 |
10x + y = 6x + 6y – 9
| 1 |
| 1 |
4x – 5y = -9 b = 21m
4 x 20x – 16y = 0 from L – b = 9
| 2 |
L = 9 + 21
5 x 20x – 20y = -45 L = 30m
=
| 1 |
y = 5
from 5x = 4y
x =
x = 4
Assignment: Page 65, No 14, 15 and 16
Mathematics Notes SS2 First Term – Edudelight.com
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