# 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 plug-in 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 plug-in 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.

1. x + 1
2. 12`x^2` + 6x
3. 48 `x^10` `y^8`
4. x – intercept = 1; y – intercept = 1
5. (x + 1) (9x + 5)

Basic Algebraic Laws
 Commutative Law For Addition:a + b = b + aex. 5 + 7 = 7 + 5 =12Commutative Law For Multiplication:a x b = b x aex. 3 x 6 = 6 x 3 = 18Associative Law For Addition:(a + b) + c = a + (b + c)ex. (4 + 6) + 5 = 4 + (6 + 5) = 15Associative Law For Multiplication:(a x b) x c = a x (b x c) ex. (2 x 3) x 4 = 2 x (3 x 4) = 24Distributive Law:a x (b +c) = (a x b) + (a x c)ex. 3 x (2 + 5) = (3 x 2) + (3 x 5)=21

 Even though the above laws apply to addition and multiplication, they do not apply to subtraction and division. For example:7 - 5 = 2, which is not the same as 5 - 7 = -2.

## 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

 Rules for areas: Rectangle or square = width x height circle = 2 x pi x radius x radius triangle = 1/2 x base x height Polynomial Solutions for y = ax2 + bx + c Sqrt is the same as square root here. x = -b +/- sqrt(b^2 - 4ac) 2a

 Distance Formula between two points on a cartesian graph Let the first point be (x1,y1) and the second point be (x2,y2). 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 = (x2 - x1)^2 + (y2 - y1)^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[(x2 - x1)^2 + [(y2 - y1)^2 ]

 Rules For Exponential Forms Under Multiplication And Division xm times xn = x m + n ex. 23 x 24 = 23 + 4 = x7 xmdivided by xn = xm - n ex. 27 / 24 = 27 - 4 = 23

 Rules For Exponentials Being Raised To A Power (xm)n = xm^n = xmn ex. (x5)3 = x 5^3 = x15

 Rules For Exponentials and 1/N Roots (xm)1/n = x m/n ex. (x15)1/3 = x5

 Steps For Solving For A Single Variable Equation 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.

 How To Reduce Rational Expressions x5 - 2x4 - 15x3 x6 + x5 + 12x4 (x3)(x2 - 2x - 15)Factor out the common factorof x3 from the numerator and (x4)(x + 7x + 12)factor out the common factor of x4.

 (x3)(x + 3)(x - 5) Factor out the remaining polynomials in the numerator (x4)(x + 4)(x + 3) and the denominator. (x3)(x + 3)(x - 5)Cancel out any common (x4)(x + 4)(x + 3)factors.x - 5 We get the answer. x(x + 4)