Algebra 11 grade
Introduction :
Statements based on this variable are manipulated using the rules of operations that apply to numbers, such as addition. This can be done for a variety of the reasons, including equation solving. Algebra is much broader than the elementary algebra and studies what happens when different rules of operations are used and when operations are devised for things other than numbers. Addition and multiplication can be generalized and their precise definition lead to structures such as groups rings and fields. (Source.Wikipedia)
Theorem of algebra 11 grade:
Remainder Theorem: Let P(x) is any Degree of Polynomial greater than or equal to one and let a be any real number. When P(x) is divided by the binomial (x–a) the remainder is P (a).
Example 1:
Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).
Solution:
P(x) = x3 – 5x2 + 7x – 4. By the remainder theorem of algebra grade when P(x) is divided by
(x – 1), the remainder is P(1).
∴ The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.
Example 2:
Find the remainder when 3x3 + 4x2 – 5x + 8 is divided by x + 2.
Solution:
When P(x) is divided by (x + 2), the remainder is P(–2).
∴ The remainder P(–2) = 3(–2)3 + 4(–2)2 – 5(–2) + 8 = 3(–8) + 4(4) + 10 + 8 = 10
Algebra 11 grade to application simultaneous linear equations:
Example 1:
A bag contains five, ten and two rupees coins. The total number of coins is 20 and the total value of money is Rs.125. If the second and third sorts of the coins are interchanged the value will be decreased by Rs.6. Find the number of coins in each sort.
Solution:
Let x, y, z be the number variable of Rs.10, Rs.5 and Rs.2 respectively. The total number of coins is 20 x + y + z = 20------------- (1)
If the total value of money is Rs.125 10x + 5y + 2z = 125--------------------- (2)
If the II and III sorts of coins are interchanged the value will be decreased by Rs.6
10x + 5z + 2y = 125 – 6 = 119
(3) 10x + 2y + 5z = 119
10 × (1) – (2) 5y + 8z = 75 (4)
------------------------------------------------
(2) – (3) 3y – 3z = 6
y – z = 2 (5)
Let us solve (4) and (5)
(4) 5y + 8z = 75
8 × (5) 8y – 8z = 16
13y = 91 y = 7
Substituting y = 7 in (5) we get
z = 5
Substituting z = 5, y = 7 in (1) we get
x + 7 + 5 = 20 20 – 7 – 5 or x = 8
Number of Rs.10, Rs.5 and Rs.2 respectively 8, 7 and 5.
Practice problem for algebra 11 grade:
1. Find m if 5x5 – 9x3 + 3x + m leaves a remainder 7 when divided by (x+1).
Answer: 6.
2. The sum of three packets is 24. Among them one packet is equal to half of the sum of other two packets but four times the difference of them. Find the packets are calculated.
Answer: The packets are 8, 9 and 7.
I like to share this Algebraic Expression Definition with you all through my blog.
Statements based on this variable are manipulated using the rules of operations that apply to numbers, such as addition. This can be done for a variety of the reasons, including equation solving. Algebra is much broader than the elementary algebra and studies what happens when different rules of operations are used and when operations are devised for things other than numbers. Addition and multiplication can be generalized and their precise definition lead to structures such as groups rings and fields. (Source.Wikipedia)
Theorem of algebra 11 grade:
Remainder Theorem: Let P(x) is any Degree of Polynomial greater than or equal to one and let a be any real number. When P(x) is divided by the binomial (x–a) the remainder is P (a).
Example 1:
Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).
Solution:
P(x) = x3 – 5x2 + 7x – 4. By the remainder theorem of algebra grade when P(x) is divided by
(x – 1), the remainder is P(1).
∴ The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.
Example 2:
Find the remainder when 3x3 + 4x2 – 5x + 8 is divided by x + 2.
Solution:
When P(x) is divided by (x + 2), the remainder is P(–2).
∴ The remainder P(–2) = 3(–2)3 + 4(–2)2 – 5(–2) + 8 = 3(–8) + 4(4) + 10 + 8 = 10
Algebra 11 grade to application simultaneous linear equations:
Example 1:
A bag contains five, ten and two rupees coins. The total number of coins is 20 and the total value of money is Rs.125. If the second and third sorts of the coins are interchanged the value will be decreased by Rs.6. Find the number of coins in each sort.
Solution:
Let x, y, z be the number variable of Rs.10, Rs.5 and Rs.2 respectively. The total number of coins is 20 x + y + z = 20------------- (1)
If the total value of money is Rs.125 10x + 5y + 2z = 125--------------------- (2)
If the II and III sorts of coins are interchanged the value will be decreased by Rs.6
10x + 5z + 2y = 125 – 6 = 119
(3) 10x + 2y + 5z = 119
10 × (1) – (2) 5y + 8z = 75 (4)
------------------------------------------------
(2) – (3) 3y – 3z = 6
y – z = 2 (5)
Let us solve (4) and (5)
(4) 5y + 8z = 75
8 × (5) 8y – 8z = 16
13y = 91 y = 7
Substituting y = 7 in (5) we get
z = 5
Substituting z = 5, y = 7 in (1) we get
x + 7 + 5 = 20 20 – 7 – 5 or x = 8
Number of Rs.10, Rs.5 and Rs.2 respectively 8, 7 and 5.
Practice problem for algebra 11 grade:
1. Find m if 5x5 – 9x3 + 3x + m leaves a remainder 7 when divided by (x+1).
Answer: 6.
2. The sum of three packets is 24. Among them one packet is equal to half of the sum of other two packets but four times the difference of them. Find the packets are calculated.
Answer: The packets are 8, 9 and 7.
I like to share this Algebraic Expression Definition with you all through my blog.