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Answers For Algebra Systems

Introduction:
Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. Answer for algebra systems deal with the following basic operations such as addition, subtraction, multiplication and division. Answer for algebra systems use variables, constant, coefficients, exponents, terms and expressions.. In answer for algebra systems, we also use the following properties such as commutative, associative, identities and inverse.
Most important term in free algebra online:
“Answer for algebra systems” describes the terms such as variables, constant, coefficients, exponents, terms and expressions.
Variables:
Algebraic variables are the alphabets where we are assigning the values. While solving the algebraic equation value of the variable will be changed. Widely used variables are x, y, z
Constant:
Algebraic constants are the value whose values never changes while solving the algebraic equation. In 43y + 33, the value 33 is the constant.
Expressions:
An algebraic Expression is the mixture form of variables, constant, coefficients, ...
... exponents, terms which are combined together by the following arithmetic operations such as Addition, subtraction, multiplication, and division. The example of an algebraic expression is given below
51y + 61
Term:
Terms of the algebraic expression is used to form the algebraic expression by the arithmetic operations such as addition, subtraction, multiplication and division. In the following example 2n2 + 3n the terms 2n2, 3n are combined to form the algebraic expression 2n2 + 3n by the addition operation ( + )
Coefficient:
The coefficient of an algebraic expression is the value present just before the terms. From the following example, 3n2 + 2n the coefficient of 3n2 is 3 and 2n is 2
Equations:
An algebraic equation balances the numbers or expressions. Most probably algebraic equation is used for the value of the variable. The example of the equation is given below
3n +3 = 6
Order of the operation in answer for algebra systems:
1. First, Reduce the algebraic expression whatever inside the parentheses.
2. Next, Reduce the exponents.
3. Next, Reduce the multiplication or division operations.
4. Finally, Reduce the addition or subtraction operations.
Examples of answer for algebra systems:
Example 1:
2(a-2)+4a-2(a-4)+10
Solution:
2(a-2)+4a-2(a-4)+10 = 2(a-2)+4a-2(a-4)+10
= 2a–4+4a–2a+8+10
= 2a+4a-2a–4+8+10
= 4a+14 (divide both terms by 2)
= 2a+7
Example 2:
4x - 2 = 2x - 8
Solution:
4x - 2 = 2x – 8
4x – 2 + 2 =2x -8 + 2 (Add 2 on both sides)
4x = 2x -6
4x – 2x =2x -2x - 6 (Add -2x on both sides)
2x = -6
2x /2 = -6 / 2 (Divided both sides by 2)
X = -3
Example 3:
Solve the equation 15x + 10 = -50
Solution
15x + 10 = -50
15x + 10 – 10 = -50 – 10 (Add -10 on both sides)
15x = -60
15x / 15 = - 60 / 15 (Divided both sides by 15)
x = - 4
Example 4:
Solve the equation |-5x + 5| -8 = -8
Solution:
|-5x + 5| -8 = -8
|-5x + 5| -8 + 8 = -8 + 8 (Add 8 on both sides)
|-5x + 5| = 0
|-5x + 5| is same as -5x + 5, now solve for x
-5x + 5 = 0
-5x + 5 - 5= 0 - 5 (Now add -5 on both sides)
-5x=-5
-5x / 5 = -5 / 5 (Now divide both sides by -5)
-x = - 1 are equal to x = 1
Example 5:
x+y=9
-x+2y=0
Substitution method for linear equation:
x+y=9 ---------------------- equation 1
-x+2y=0---------------------- equation 2
If we add the equations 1 and 2, we will get
3y=9
3y/3 = 9/3 ( both sides are divided by 3 )
y = 3
Substitute y = 3 in the equation 1, so we will get
x + 3 = 9
x+3-3=9-3 ( -3 is added on both sides)
x=6
Elimination method for linear equation:
x+y=9 ---------------------- equation 1
-x+2y=0---------------------- equation 2
Take the equation 1
x+y=9
x+y-y=9-y ( -y is added on the both sides )
x=9-y
Substitute x=9-y in the equation 2, we will get
-(9-y ) +y=0
-9+y+2y=0
-9+3y=0
-9+9+3y=0+9
3y=9
3y/3=9/-3 ( both sides are divided by 3)
Y=3
Substitute y=3 in the equation 1
X+3=9
X+3-3=9-3 ( Add -3 on the both sides)
X=6
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