algebra (2).docx
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ALGEBRA (2).docxTRANSCRIPT
5 x 3Coefficient
Exponent/ Power/ Index
Variable/ Literal/ Base
ALGEBRAThe science of equations/ expression.
The letters used in algebra is known as literal or variable.
Note: 1x1 can be written as x
–1xy can be written as –xy
Exercise1.Form Algebraic expressions
1. 7 times a number x2. 4 times a number x minus 5
3. The sum of two numbers x and y divided by another number p
4. Half of a number y5. Double the product of the numbers x,y and z.
6. Double the product of a and b and divide by 5
7. 3 times a number minus 4 times a number y.
8. Ali bought p pencils for Rf. 3 each and q pens for Rf. 7 each. What is the total cost?
9. 7ab times twice xy
SIMPLIFICATION OF ALGEBRAIC EXPRESSION. Simplifying like terms To multiply and divide algebraic expression follow the indices rule. You can add and subtract only like terms. That is, the variables and their exponents should be
same.
Eg: a) 3ab + 2ab = 5ab but 3ab +2b cannot be added togetherb) 3a2 – 2a2 = a2
3a – 2a2 cannot be added together
EXERCISE 1Simplify
1. 9ab – 5ab2. 6xy – 12xy + 2xy3. 4gh + 5jk – 2gh +74. 2x2y – x2y + 3xy2 –2 y2x5. 7cd – 8dc + 3cd6. 4ab + 10bc – 2ab – 5cd7. 3ba – ab + 3ab – 5ab8. 2p2 – 5p2 + 2p – 4p9. 3a2b – 2ab +4 a2b – ba
10.x5 – 5x3 + 2 – 2x3
11. h3+5h – 3 – 4h2 – 2h +7 +5h2 12. 23z2 + 17k2 – 3z2 +813. x5+ x5+ x5+x5
14. x2+ 5x +4 – x2 + 6x –315. 2 + 3x2+ x4 – 3x2 +116. 7x × 2y2 ×(2y)2
17. (3x)2 × 3x2 × 5y18. 2xy2 × 3x2y + 4x3y3
Expand and simplify [Single bracket]
Expand the bracket by multiplying the term out side the bracket with each term inside the bracket.
8 x−2 (3 x+5 )8 x−6 x−10
And then simplify by collecting all like terms.8 x−6 x−10
2 x−10
EXERCISE 2Expand and Simplify
1. 4( 2x – 5) 2. 5x (3x +5y)3. 3x + 2( x+1)4. 7+ 3(2x – 3)5. 3x – 4(2x – 5)6. 3ab – 2a(b – 2) 7. x(x – 2) + 3x(x – 3)8. y(3y – 1) – (3y – 1)9. 7b(a + 2) – a(3b +3)10. 3(x – 2) – (x – 2)11. 7(2x +2) – (x – 2)
Expand and simplify [Multiplying two expressions in brackets]
Multiply each term in the first bracket with each term in the second bracket.[Remember to take the terms with the sign]
(3+2 y )( 4 y−6 )
3 (4 y−6 )+2 y (4 y−6)Remove both the brackets
3 (4 y−6 )+2 y (4 y−6 )
12 y−18+8 y2−12 y
Simplify the like terms
12 y−18+8 y2−12 y
8 y2−18
EXERCISE 3Expand and Simplify
1. (x +1) (x +3)2. (y +4) (y+5)3. (x – 2) (x +2)4. (x+5) (x – 2)5. (x – 3) (x – 2)6. (3x +y) (x + 2y)
7. (5x – y) (3y – x)8. 2(x– 1) (x+2)9. 4(2y – 3) (3y +2)10. 2x(2x – 1) (2x+1) 11. (x + 4)2
12. (2x +y)2
13. (x+ 2)2 + (2x + 1)2 14. (x – 3)2 – (x + 1)2
Expanding brackets using IDENTITIES(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
EXERCISE 4 Expand using identities
1. (3 + x)2
2. (5 + y)2
3. (2p + 5)2
4. (3x + 2y)2
5. (3 – x)2
6. (7 – y)2
7. (x – 6)2
8. (2x – 2y)2
9. (5m – 4n)2
10. (x + 2)2 + (x + 4)2
11. (2y – 1)2 + (y – 3)2
12. (5– x)2 + (6 + x)2
13. (4 + 2x)2 – (x + 1)2
14. (2g – 6)2 – (g + 1)2
15. (x + 7)2 – (2x + 3)2
16. (y + 3)2 – (y – 3)2
Find the unknown p and/or q of the following using identities.
1. (x + p)2 = x2 + 8x + 16
2. (x + q)2 = x2 +12x + 36
3. (x – p)2 = x2 – 10x + 25
4. (px + q)2 = 4x2 +12x +9
5. (px – 1)2 = 16x2 – qx + 1
6. (px – q)2 = 4x2 – 20x + 25
SUBSTITUTION
EXERCISE 1. Evaluate the following given that x = –3 and y = 2.
a) x2 b) x2 – 2y c) xy2 d) (xy)2 e) x2
yf) 2y2 g) x + 3y
h) (2x)2 – (3y)2 i) 3x2 + 3 j) x2 – y2 k) 2x + 3y4
2. Evaluate the following given that a = 4, b = –2 and c = –3.
a) a(b + c) b) a2 (b – c) c) 2c(a – c) d) b2(2a +3c)
e) b2 + 2b + af) √(a2+c2) g) b2
a+ 2 c
b h) √ab+c2
3. Evaluate the following given that k = –3, m = 1 and n = –4.
a) 2 k+mk−n
b) 5m√k2+n2 c) kn−k
2md)
k+m+n
k2+m2+n2 e) m√k−n
f) m2(2k2 – 3n2) g) k2m2( m – n)
CHANGING THE SUBJECT OF THE FORMULA
EXERCISE 1
1. Make x the subject of the following.
a) 2x = 5
b) Ax = B
c) 9x = T + N
d) x + B = T
e) N = x + D
f) N2 + x = P + Q
g) x – A = E + K
h) F = B – x
2. Make y the subject of the following
a) Ay + C = N
b) Ny – F = H
c) Z – Py = T
d) j = my + c
e) A(y + B) = C
f) b(y – d) = q
g) n = r(y + t)
h) g = m(y + n)
EXERCISE 2
1. Make a the subject of the following.
a) aD
= B
b) a−A
B = T
c) g = a−R
D
d) Aa+B
D = B
e) n = ea−f
h
f) M (a+B)
N = T
g) ra
= B
h) pa
+ q = M
i) C – da
= B
j) Y
a−C – T = 0
2. Make x the subject of the following
a) x2 = B
b) x2 + A = B
c) b = a+ x2
d) C– x2 = m
e) mx2 = n
f) ax2 – t = m
g) m
x2 = a + b
h) x2
a + b = c
i) √ x = 2
j) √ ( x+1 ) = 5
k) √ ( x−D ) = A2
l) g = √ (c−x )
m) √ ( M−N x ) = P
n) a√ ( x2−k ) = b
o) √ Mx
= N
p) √ ( x2−4 ) = 6
L.C.M AND H.C.F OF ALGEBRAIC EXPRESSIONS
L.C.M. of algebraic expressions should include all the terms with their highest power.
H.C.F. of algebraic expressions should include the common terms with their lowest power
EXERCISE
Find the LCM and HCF of the following algebraic expressions.
1. a2 , a3 , a3 ,a5
2. 10a3b2c3, 8a2b2c2
3. a2 , b , c3
4. p3qr2, pq2r3 , 5p2q3r5. p3q2 , p2 q3 , p2q6. a2b3c3, a3b3 ab2c2
7. 3mn2, 6mnp, 12m2np8. 2ab, 5b, 7ab2
9. 3x2yz, 6xy2z3, 3xyz2
10. x2, x3, x5
11. a3b, a2b2, ab3
12. 6x2, 9x3, 15x
13. 2y4, 8y, 10y2
14. 5p2, 3q3, 7pq15. 5xy, 10x2y, 15xy2
16. 10a2b, 15ab2, 20a2b17. 3x, 618. (x+7)2, (x+7)2, (x+7)19. x, (x−1)20. (m−n3), (m−n)21. (x+1)2, (x−1), (x+1) (x−1)2
22. (2x+3)2, (x+4), (2x+3)(x+4)23. (x+1)(x−3), (x−3)(x+4), (x+4)(x−2)24. (y + 2)2, (y + 2)(y – 3), 3(y + 2)
FACTORIZATION
SIMPLE FACTORIZATION
6 x2+14 x
Step 1:
Take out the highest common factors [HCF] of the variables and the coefficients
2 ×3 x× x2+2 ×7 x
HCF =2 x
Step 2:
Write the remaining terms inside the bracket
2 ×3 x× x2+2 ×7 x
2 x (3 x2+7)
EXERCISE
Factorize the following
1) 4x – 6
2) 8m +12n +16r
3) 3pqr – 9pqs
4) 8x2y – 4xy2
5) 8pq + 6pr – 4ps
6) 5pq – 10qr + 15qs
7) 4ut – 16t + 20rt
8) m3 – m2n + mn2
9) 56x2y – 28xy2 10) 72m2n + 36mn2 – 18m2n2
FACTORIZATION BY GROUPING
x2−xy−2 x+2 y
Step 1:
Group the expression into two parts
x2−xy−2 x+2 y
Step 2:
Apply simple factorization to the groups separately
x2−xy−2 x+2 y
x (x− y )−2(x− y)
Step 3:
Apply simple factorization to the answer
x (x− y )−2(x− y)
( x− y )(x−2)
EXERCISE
1) 3m + 3n + mx + nx
2) 6x + xy + 6z + yz
3) rs – 2ts + rt – 2t2
4) ab – 4bc + ac – 4c2
5) mn – 2mr – 3rn – 6r2
6) sx – tx + sy – ty
7) 2mh – 2mk + nh – nk
8) 2ax – 2ay – bx + by
9) ms +2mt2 – ns – 2nt2
10) pr – 4p – 4qr + 16q
FACTORIZATION OF TRINOMIALS
3 x2+7 x−6
Step 1:
Multiply the x2 term and constant
3 x2+7 x−6
3 x2×−6=−18 x2
Step 2:
Find factors of the product such that they add up to the x term.
Step 3:
Split the middle term using the factors
3 x2+7 x−6
= – 18x2
9x – 2x7
x9
3 x2+9 x−2x−6
Step 4: Factorize by Grouping
3 x2+9 x−2x−6
3x(x + 3) – 2 (x + 3)
(x + 3)(3x – 2)
EXERCISE
1) x2 + 8x + 12
2) x2 + 6x + 5
3) x2 + 13x + 12
4) x2 + 6x + 9
5) x2 – 7x + 12
6) x2 – 8x + 12
7) x2 – 12x + 36
8) x2 – 15x + 36
9) x2 + 2x – 15
10) x2 + x – 12
11) x2 + 3x – 54
12) x2 + 2x – 8
13) x2 – 2x – 8
14) x2 – x – 20
15) x2 – 2x – 15
16) x2 – x – 12
17) 2x2 + 4x + 2
18) 2x2 + 7x + 6
19) 2x2 + x – 6
20) 2x2 – 7x + 6
21) 9x2 – 6x + 1
22) 6x2 – x – 1
23) 3x2 + 8x + 4
24) 2x2 – 3x – 5
FACTORIZATIONS BY DIFFERENCE IN TWO SQUARE
25 y2−144
Using the identity: a2 – b2 = (a + b)(a – b)
(5y)2 – (12)2
(5y + 12)(5y – 12)
EXERCISE
1) x2 – 16
2) x2 – y2
3) 4x2 – 1
4) 16x2 – y2
5) 25p2 – 36q2
6) 9m2 – 4n2
7) m2n2 – 9 p2
8) 9
16 a2 – 25
9) 144x2 – 1
10) 4a4 – 16b2
11) x2 – x
12) 9x3 – 36x
13) 3y3 – 12y
14) a3 – ab2
15) 18m3 – 8mn2