Algebra 1 Mathematics Assignment Help
homework assignment help .com is most useful online help portal for the students that providing all online Algebra help Service. Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. Algebra 1 notes cover the four basic operations in algebra such as addition, subtraction, multiplication and division. The most important terms of algebra, variables, constant, coefficients, exponents, terms and expressions are covered algebra 1 notes. We will know the symbols and alphabets in the place unknown values for algebra 1 notes. Therefore, students are getting help for algebra 1 notes.
Examples for Algebra 1 Notes:
Example 1:
Simplify the expression. (5x + 3)  (2x + 3)
Solution:
(5x + 3)  (2x + 3) Now we have to multiply “–“ with the expression 2x + 3. So we get,
= 5x + 2 2x  3 5now we have to group the terms. So we get,
= 5x – 2x + 3 – 3
= 3x
Example 2:
Multiply the following terms (4x) ( 2x + 4).
Solution:
(4x) ( 2x + 4)Now we have to multiply 4x with 2x +4 like below.
= (4x `xx` 2x) + (4x `xx` 4) Note: (`a^m` ) (`a^n` ) = `a^(m+n)`
= 8`x^2` + 16x
Example 3:
Multiply the following terms (4`x^6` ) ( 5`x^8` `y^21` ).
Solution:
(4`x^6` ) ( 5`x^8` `y^21` ).we have to multiply 4`x^6` with 5`x^8` `y^21` like below.
= 4`x^6` `xx` 5`x^8` `y^21.`
= 4 `xx` 5 (`x^6` ) (`x^8` ) (`y^21` ) Note: (a^m) (a^n) = a^(m+n)
= 20 `x^(6+8)` `y^21`
= 20 `x^14 ` `y^21`
Example 4:
Find the intercepts for the following equation 6x + 3y = 7
Solution:
6x + 3y = 7
X – Intercept:
For finding the x – intercept, we have to plugin y=0.
6x + 3(0) = 7
6x = 7 now we have to divide both sides by 6.
`(6x)/6` = 7/6
X = 1.17
Y intercept:
For finding y intercept, we have to plugin x=0.
6(0) + 3y = 7
3y = 7 now we have to divide both sides by 3.
`(3y)/3 ` = 7/3
Y= 2.33
Example 5:
Factor the following expression 8`x^2` + 12x+4.
Solution:
8x^2 + 12x+4
Now we have to multiply the coefficient of the term 8`x^2` with the constant term 4. So we get
8 × 4 =32
Now we have to find the factor to 32. The factors are 4,8. By these factors, we can split out the term 11x as 7x + 4x. Therefore, the given equation become like below.
(8`x^2` + 8x) + (4x+4) Now we have to take out the common terms like below.
8x(x + 1) + 4(x +1)
(x + 1) (8x +4)
Practice Problems for Algebra 1 Notes:
1. Subtract the following expression. (3x + 4) – (2x +3)
2. Multiply the following terms (3x) ( 4x + 2).
3. Multiply the following terms (6`x^5` `y^4` ) ( 8 `x^5` `y^4` ).
4. Find the intercepts for the following equation 7x + 7y = 7
5. Factor the following expression 9`x^2` + 14x+5.
Answer key:
 x + 1
 12`x^2` + 6x
 48 `x^10` `y^8`
 x – intercept = 1; y – intercept = 1
 (x + 1) (9x + 5)
Commutative Law For Addition: a + b = b + aex. 5 + 7 = 7 + 5 =12 Commutative Law For Multiplication: a x b = b x aex. 3 x 6 = 6 x 3 = 18 Associative Law For Addition: (a + b) + c = a + (b + c)ex. (4 + 6) + 5 = 4 + (6 + 5) = 15 Associative Law For Multiplication: (a x b) x c = a x (b x c) ex. (2 x 3) x 4 = 2 x (3 x 4) = 24 Distributive Law: a x (b +c) = (a x b) + (a x c)ex. 3 x (2 + 5) = (3 x 2) + (3 x 5)=21 

Some Basic Algebra Rules
The additive inverse for a is (a) for any constant.or variable. a + (a) = 0
ex. 5 + (5)= 0
The mulitplicative inverse for a is 1/a for any constant or variable. a x 1/a = 1
ex, 5 x 1/5 = 1
The identity element for addition is 0 for any constant or variable. a + 0 = a
ex, 5 + 0 = 5
The identity element for multiplication is 1 for any constant or variable. a x 1 = a
ex, 5 x 1 = 5
The additive inverse for a is (a) for any constant.or variable. a + (a) = 0 ex. 5 + (5)= 0 The mulitplicative inverse for a is 1/a for any constant or variable. a x 1/a = 1 ex, 5 x 1/5 = 1 The identity element for addition is 0 for any constant or variable. a + 0 = a ex, 5 + 0 = 5 The identity element for multiplication is 1 for any constant or variable. a x 1 = a ex, 5 x 1 = 5 

Distance Formula between two points on a cartesian graph Let the first point be (x_{1},y_{1}) and the second point be (x_{2},y_{2}). We use the Pythagorean thereom for a right triangle, using the length between the two x values as one side and the length between the y values as another side of a right triangle. The diagonal line, or shortest distance between the two points would be the hypotenuse of that right triangle. The distance, which we'll call z, between those two points would be: z^2 = (x_{2}  x_{1})^2 + (y_{2}  y_{1})^2 We solve for z by taking the square root of both sides. Sqrt is the same as square root here. We get z = sqrt[(x_{2}  x_{1})^2 + [(y_{2}  y_{1})^2 ] 
x^{m } times x^{n} = x ^{m + n} ex. 2^{3} x 2^{4} = 2^{3 + 4} = x^{7} x^{m}divided by x^{n} = x^{m  n} ex. 2^{7} / 2^{4} = 2^{7  4} = 2^{3} 
(x^{m})^{n} = x^{m^n} = x^{mn} ex. (x^{5})^{3} = x ^{5^3} = x^{15} 
(x^{m})^{1/n} = x ^{m/n} ex. (x^{15})^{1/3} = x^{5} 
ex. 4x  9 = 7 + 9x + 19 4x  9 = 26 + 9x Combine Like Terms. In the case, +9 = +9 add the 7 and the19. Add 9 to both sides to get rid of the 9 on the left hand side. 9 is the additive inverse of 9 and will cancel that term to zero. ^{} 4x + 0 = 35 + 9x 9x =  9x We need to get rid of the 9x on the right side now, so we'll subtract both sides by 9x. ^{} 5x = 35 (1)(5x) = (1)(20) Multiply both sides by 1 to get rid of the minus sign in front of the 5x. ^{} 5x = 35 (1/5)(5x) = (1/5)(35) Multiply both sides by 1/5, which is the same as dividing both sides by 5. We want to get rid of the 5 in front of the x to solve for x. x = 7 Our solution to the equation is x = 7, If you plug 7 into the original equation, the right side will equal the left side of the equation. 
x^{5}  2x^{4}  15x^{3} ^{} x^{6} + x^{5} + 12x^{4} (x^{3})(x^{2}  2x  15)Factor out the common factor^{}of x^{3} from the numerator and (x^{4})(x x^{4}. 
(x^{3})(x + 3)(x  5) Factor out the remaining polynomials in the numerator (x^{4})(x + 4)(x + 3) and the denominator. (x^{3})(x + 3)(x  5)Cancel out any common ^{}(x^{4})(x + 4)(x + 3)factors. x  5 We get the answer. ^{} x(x + 4) 
AssignmentHelp
 Mathematics Assignment
 Physics Assignment
 Chemistry Assignment
 Biology Assignment
 English Assignment
 Finance Assignment
 Accounts Assignment
 Statistics Assignment
 Economics Assignment
 Management Assignment
 Electrical Assignment
 Mechanical Assignment
 Programming Assignment
 Other Assignment
 Others Services
 Terms condition
 Sitemap
 Packages
 AboutUs
OnlineTutoring
 Mathematics Online Tutoring
 Physics Online Tutoring
 Chemistry Online Tutoring
 Biology Online Tutoring
 English Online Tutoring
 Finance Online Tutoring
 Accounts Online Tutoring
 Statistics Online Tutoring
 Economics Online Tutoring
 Operations Online Tutoring
 Electrical Online Tutoring
 Mechanical Online Tutoring
 Programming Online Tutoring