High school algebra 2
Introduction :
Algebra is a practice for Solve Algebra Equations in step by step. Algebra is division of mathematics is that replacement letters for numbers. An algebraic equation is stand for the scale. In online, students get help with math algebra 2 topics. In online, algebra2 math questions are available with answers. Online learning is the one of the best method of learning. An algebra 2 topic includes system of equation, polynomials, factoring, probabilities, etc. In this article we shall discuss for high school algebra 2.
Sample problem for high school algebra 2:
High school algebra 2 problem 1:
Evaluate the given advance algebra 2 problem and find the values of given variable terms.
16p + 28q + 12r = 56
16p + 12q + 20r = 48,
32p + 12q + 8r = 52.
Solution:
Let as the given equations be identified as follows.
16p + 28q + 12r = 56------------- (i)
16p + 12q + 20r = 48------------- (ii)
32p + 12q + 8r = 52------------ (iii)
Consider the above given equations (i) and (ii). Subtracting the equation (i) and (ii) we get
16p + 28q + 12r = 56
16p + 12q + 20r = 48
16q – 8r = 8 ------------------ (iv)
Consider the above given equations (ii) and (iii). In the equation (ii) multiply with a numerical 2, we get
32p + 24q + 40r = 96
Subtracting the equation (ii) and (iii) we get
32p + 12q + 8r = 52
32p + 24q +40r = 96
–12q – 32r = –44
In the above equation multiply with -1 we get
12q + 32r = 44 ------------------- (v)
Consider the above equations (iv) and (v). In the equation (v) multiply with the numerical 4, we get
64q – 32r = 32
Add the equation (iv) and (v) we get
12q + 32r = 44
64q – 32r = 32
76q = 76
q = 1
Substitute q = 1 in an equation (v), we get
12p + 32(1) = 44
12p + 32 = 44
p = 1
Substitute p = 1, q = 1 in the equation (iii), we get
32(1) + 12(1) + 8r = 52
32 + 12 + 8r = 52
r = 1
Answers of a given algebraic variables are p = 1, q = 1, r = 1.
High school algebra 2 problem 2:
Solve the given algebraic expressions and simplify of 10p4 – 4p2 + 3p + 7 and 4p + 11p3 – 6p2 – 5.
Solution:
We are going to simplify the given two expressions.
Step 1: We using the algebraic properties, we get
= (10p4 – 4p2 + 3p + 7) + (11p3 – 6p2 + 4p – 5)
Step 2: In the next step we are combine the two equations, we get
= 10p4 + 11p3 – 4p2 – 6p2 + 3p + 4p + 7 – 5
Step 3: In the next step we are grouping the same values, we get
= 10p4 + 11p3 – (4 + 6) p2 + (3 + 4) p + 2
= 10p4 + 11p3 – 10p2 + 7p + 2.
High school algebra 2 problem 3:
Find the factoring value of the given function. 2p2- 12p + 18 = 0
Solution:
In the first stem we find factor value of the given numerical values of the ‘p’ coefficients.
36 (product)
/ \
- 6 - 6
\ /
-12 (sum)
2p2- 12p + 18 = 2p2 – 6p – 6p + 18
= 2p (p - 3) – 6p(p - 3)
= (2p - 6) (p – 3)
= (p - 3) (p – 3)
So the factor of the given function is 3 and 3.
Practice problem for high school algebra 2:
3p + q + 2r = 18
2r + 3q + r = 18
Answer: p = 3, q = 3, r = 3.
Algebra is a practice for Solve Algebra Equations in step by step. Algebra is division of mathematics is that replacement letters for numbers. An algebraic equation is stand for the scale. In online, students get help with math algebra 2 topics. In online, algebra2 math questions are available with answers. Online learning is the one of the best method of learning. An algebra 2 topic includes system of equation, polynomials, factoring, probabilities, etc. In this article we shall discuss for high school algebra 2.
Sample problem for high school algebra 2:
High school algebra 2 problem 1:
Evaluate the given advance algebra 2 problem and find the values of given variable terms.
16p + 28q + 12r = 56
16p + 12q + 20r = 48,
32p + 12q + 8r = 52.
Solution:
Let as the given equations be identified as follows.
16p + 28q + 12r = 56------------- (i)
16p + 12q + 20r = 48------------- (ii)
32p + 12q + 8r = 52------------ (iii)
Consider the above given equations (i) and (ii). Subtracting the equation (i) and (ii) we get
16p + 28q + 12r = 56
16p + 12q + 20r = 48
16q – 8r = 8 ------------------ (iv)
Consider the above given equations (ii) and (iii). In the equation (ii) multiply with a numerical 2, we get
32p + 24q + 40r = 96
Subtracting the equation (ii) and (iii) we get
32p + 12q + 8r = 52
32p + 24q +40r = 96
–12q – 32r = –44
In the above equation multiply with -1 we get
12q + 32r = 44 ------------------- (v)
Consider the above equations (iv) and (v). In the equation (v) multiply with the numerical 4, we get
64q – 32r = 32
Add the equation (iv) and (v) we get
12q + 32r = 44
64q – 32r = 32
76q = 76
q = 1
Substitute q = 1 in an equation (v), we get
12p + 32(1) = 44
12p + 32 = 44
p = 1
Substitute p = 1, q = 1 in the equation (iii), we get
32(1) + 12(1) + 8r = 52
32 + 12 + 8r = 52
r = 1
Answers of a given algebraic variables are p = 1, q = 1, r = 1.
High school algebra 2 problem 2:
Solve the given algebraic expressions and simplify of 10p4 – 4p2 + 3p + 7 and 4p + 11p3 – 6p2 – 5.
Solution:
We are going to simplify the given two expressions.
Step 1: We using the algebraic properties, we get
= (10p4 – 4p2 + 3p + 7) + (11p3 – 6p2 + 4p – 5)
Step 2: In the next step we are combine the two equations, we get
= 10p4 + 11p3 – 4p2 – 6p2 + 3p + 4p + 7 – 5
Step 3: In the next step we are grouping the same values, we get
= 10p4 + 11p3 – (4 + 6) p2 + (3 + 4) p + 2
= 10p4 + 11p3 – 10p2 + 7p + 2.
High school algebra 2 problem 3:
Find the factoring value of the given function. 2p2- 12p + 18 = 0
Solution:
In the first stem we find factor value of the given numerical values of the ‘p’ coefficients.
36 (product)
/ \
- 6 - 6
\ /
-12 (sum)
2p2- 12p + 18 = 2p2 – 6p – 6p + 18
= 2p (p - 3) – 6p(p - 3)
= (2p - 6) (p – 3)
= (p - 3) (p – 3)
So the factor of the given function is 3 and 3.
Practice problem for high school algebra 2:
- Evaluate the given advance algebra problem and find the values of given variable terms.
3p + q + 2r = 18
2r + 3q + r = 18
Answer: p = 3, q = 3, r = 3.
- Solve the given algebraic expressions and simplify of 4p4 – 6p2 + 3p + 2 and p + 7p3 – 5p2 – 4.
- Find the factoring value of the given function. p2- 5p + 6 = 0