Solving Use Of Variables

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Introduction for algebra 1 solving equations online:

Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. Solving Use of Variables are involved in the four basic operations such as addition, subtraction, multiplication and division. In algebra, the variables are related with coefficients, exponents, terms, expressions and equation. In Algebra, besides numerals we use symbols and alphabets in place of unknown numbers to make a statement. Hence, Solving Use of Variables may be regarded as an extension of Arithmetic.

Related terms for solving use of variables:

Variables

Algebraic variables are the alphabetical characters which are used for assigning the value. While solving the algebraic equation value of the variable will be changed. Widely used variables are x, y, z

Constant

An algebraic constants are the value whose value never change while solving the algebraic equation. In 2y+5, the value 5 is the constant.

Expressions

An algebraic Expression is the combination of variables, constant, coefficients, exponents, terms ...
... which are combined together by the following arithmetic operations, Addition, subtraction, multiplication and division.an example is given below

3x+4y+5

Term

Terms of the algebraic expression is concatenated to form the algebraic expression by the arithmetic operations such as addition, subtraction, multiplication and division. In the following example 3n2 + 2n the terms 3n2, 2n are combined to form the algebraic expression 3n2 + 2n by the addition operation ( + )

Coefficient

The coefficient of an algebraic expression is the value which is 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 equals the numbers or expressions. Most probably algebraic equation is used for finding the value of the variable.

example :y =3x2+4x+6

Example for solving use of variables:

Example 1: On my last birthday I weighed 125 pounds. One year later I have put on some pounds. Now the total weight is 175 pounds. Write the expression and solve it for x to find how many pounds increased?

Solution:

One year later some pounds are added, this is unknown weigh. So we have to consider it as x. so we get

The expression is 125+x=175

125-125+x=175-125 (add -125 on both sides)

X=50 pounds

The increased weight is 50 pounds.

Example 2: In three more years, Paul’s grandmother will be six times as old as Jack was last year. If Paul’s current age is added to his grandmother's current age, the total is 68. How old is each one now?

Let 'g' be Paul’s grandmother's current age

Let 'j' be Paul’s grandmother's current age

Those g and j are unknown values

If Paul’s current age is added to his grandmother's current age, the total is 68

j + g = 68

In six more years, Paul’s grandmother will be six times as old as Paul was last year

(g+3) = 6 (j-1)

If Paul’s present age is added to his grandmother's current age, the total is 68

j+g=68

Solving both equations we get Paul’s age (j) as 11 and Paul’s grandmother's age (g) as 57

Example 3:

Solve for the variable x, 2x-15=25

Solution

2x-15=25

2x-15 + 15=25 + 15(Add 15 on both sides)

2x = 40

2x/2=40/2(both sides divided by 2)

x=20

Example 4:

Simplify the following expression 5*(30÷2)+x=0

Solution:

5*(30÷2)+x=0 (evaluate the expression inside the parenthesis)

5*(15)+x=0

75+x=0

75-75+x=0-75 (add -75 on both sides)

X=-75

Practive problem for solving use of variables:

1. 2(x-2) - 3 = 5

Answer: x=6

2. 6x + 7 = -x – 7

Answer: x=-2

Comprehend more on about how to solve for inequalities and its Circumstances. Between, if you have problem on these topics graphing linear equations and inequalities Please share your views here by commenting.