Algebra 1 and 2 study guide
Introduction :
Algebra is one of the major part of mathematics. The topics are geometry, analysis, topology, combinatory, and number theory, algebra is one of the main branches of pure math. Algebra 2 deals with the solving equation, inequalities, graph and functions, system of equations and inequalities, polynomials and factoring, fractional expression, powers and roots, complex numbers and quadratic functions. In this article we are going to see some problems on algebra 1 and algebra 2.
Example problems on algebra 1 and 2 study guide:
Algebra 1 and 2 study guide – Example: 1
Adding Polynomials f(x) = (19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42)
Solution:
(19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42)
= 19x3 + 11x2 + 7x – 15 + 6x3 - 8x2 - 21x + 11 + 4x2 - 3x - 42
= 19x3 + 6x3 + 11x2 - 8x2 + 4x2 + 7x - 21x - 3x – 15 + 11 - 42
= 25x3 + 7x2 - 17x - 46
Answer: (19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42) = 25x3 + 7x2 - 17x - 46
Algebra 1 and 2 study guide – Example: 2
Solve the given equation 16c + d = 26 and 4c + d = 122. Find the values of c and d?
Given:
16c + d = 26
4c + d = 122
Solution:
16c + d = 26------ (1)
4c + d = 122------ (2)
To solve the first equation for the d
d = 26 – 16c
Now, we can substitute the 26 – 16c for h in the second equation.
4c + (26 – 16c) = 122
-12c + 26 = 122
-12c = 122 - 26
-12c = 96
c = -96/12
c = -8
Substitute the c = -8 in the first equation
16c + d = 26
16(-8) + d = 26
-128 + d = 26
d = 26 + 128
d = 154
Answer: c = -8 and d = 154
Verification:
16c + d = 26
16(-8) + 154 = 26
-128 + 154 = 26
26 = 26
Algebra 1 and 2 study guide – Example: 3
Solve the equation by using the factorization method x2 + 37x+322= 0
Solution:
x2 + 37x+322= 0
x2 + 14x+23x+322=0
x(x+14) +23(x+14) = 0
(x+14)(x+23)=0
x+14=0 and x+23=0
x+14-14=0-14 and x+23-23=0-23
x=-14 and x=-23
Therefore, x = {-14,-23}
Answer: x1 = -14 and x2 = -23
Practice problems on algebra 1 and 2 study guide:
1. Solve the given equation 12c + d = 133 and 5c + d = 77. Find the values of c and d?
Answer: c = 8 and d = 37
2. Solve the equation by using the factorization method x2 + 40x+396= 0
Answer: x1 = -18 and x2 = -22
3. Add the given polynomial f(x) = (12x3 + 19x2 + 3x – 21) + (5x3 – 18x2 – 12x + 18) + (6x2 – 11x – 44)
Answer: (12x3 + 19x2 + 3x – 21) + (5x3 – 18x2 – 12x + 18) + (6x2 – 11x – 44) = 17x3 + 7x2 - 20x - 47
Algebra is one of the major part of mathematics. The topics are geometry, analysis, topology, combinatory, and number theory, algebra is one of the main branches of pure math. Algebra 2 deals with the solving equation, inequalities, graph and functions, system of equations and inequalities, polynomials and factoring, fractional expression, powers and roots, complex numbers and quadratic functions. In this article we are going to see some problems on algebra 1 and algebra 2.
Example problems on algebra 1 and 2 study guide:
Algebra 1 and 2 study guide – Example: 1
Adding Polynomials f(x) = (19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42)
Solution:
(19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42)
= 19x3 + 11x2 + 7x – 15 + 6x3 - 8x2 - 21x + 11 + 4x2 - 3x - 42
= 19x3 + 6x3 + 11x2 - 8x2 + 4x2 + 7x - 21x - 3x – 15 + 11 - 42
= 25x3 + 7x2 - 17x - 46
Answer: (19x3 + 11x2 + 7x – 15) + (6x3 – 8x2 – 21x + 11) + (4x2 – 3x – 42) = 25x3 + 7x2 - 17x - 46
Algebra 1 and 2 study guide – Example: 2
Solve the given equation 16c + d = 26 and 4c + d = 122. Find the values of c and d?
Given:
16c + d = 26
4c + d = 122
Solution:
16c + d = 26------ (1)
4c + d = 122------ (2)
To solve the first equation for the d
d = 26 – 16c
Now, we can substitute the 26 – 16c for h in the second equation.
4c + (26 – 16c) = 122
-12c + 26 = 122
-12c = 122 - 26
-12c = 96
c = -96/12
c = -8
Substitute the c = -8 in the first equation
16c + d = 26
16(-8) + d = 26
-128 + d = 26
d = 26 + 128
d = 154
Answer: c = -8 and d = 154
Verification:
16c + d = 26
16(-8) + 154 = 26
-128 + 154 = 26
26 = 26
Algebra 1 and 2 study guide – Example: 3
Solve the equation by using the factorization method x2 + 37x+322= 0
Solution:
x2 + 37x+322= 0
x2 + 14x+23x+322=0
x(x+14) +23(x+14) = 0
(x+14)(x+23)=0
x+14=0 and x+23=0
x+14-14=0-14 and x+23-23=0-23
x=-14 and x=-23
Therefore, x = {-14,-23}
Answer: x1 = -14 and x2 = -23
Practice problems on algebra 1 and 2 study guide:
1. Solve the given equation 12c + d = 133 and 5c + d = 77. Find the values of c and d?
Answer: c = 8 and d = 37
2. Solve the equation by using the factorization method x2 + 40x+396= 0
Answer: x1 = -18 and x2 = -22
3. Add the given polynomial f(x) = (12x3 + 19x2 + 3x – 21) + (5x3 – 18x2 – 12x + 18) + (6x2 – 11x – 44)
Answer: (12x3 + 19x2 + 3x – 21) + (5x3 – 18x2 – 12x + 18) + (6x2 – 11x – 44) = 17x3 + 7x2 - 20x - 47