Algebraic Expressions Definition The algebraic expressions have the variables and the constants. The algebraic expressions are the finite combination of the symbols that are formed according to the rules of the context. The algebra is an expression which is used to designate the value for the given values in the expression. The expression might be depending on the values assigned to the values assigned in the expression. The expression is the syntactic concept in the algebra. The online provides the connectivity between the tutors and the students. This article has the information about learn online algebraic expressions. Simplifying Algebraic Expressions Below are the examples on Simplifying algebraic expressions Know More About Order Of Operations Calculator

Example 1 Compute the factors for the expression x2+ 56x+ 768. Solution: The given expression is x2+ 56x+ 768. Step 1: x2+ 56x+ 768 = x2+24x+ 32x+ (24x 32) Step 2: x2+ 56x+ 768 = x(x+24) + 32(x +24) Step 3: x2+ 56x+ 768 = (x+24) (x+32) Step 4: x+24 and x+32 The factors for the given expression x2+ 56x+ 768 are (x +24) and (x +32). Example 2 Compute the factors for the expression x2+ 76x+ 1440. Solution: The given expression is x2+ 76x+ 1440. Step 1: x2+ 76x+ 1440 = x2+36x+ 40x+ (36x 40) Learn More About Solve For X Calculator

Step 2: x2+ 76x+ 1440 = x(x+36) + 40(x +36) Step 3: x2+ 76x+ 1440 = (x+36) (x+40) Step 4: x+36 and x+40 The factors for the given expression x2+ 76x+ 1440 are (x +36) and (x +40). Example 3 Simplify the expression 24xy + 10 x2y + 14 x y + 16 x2y + 10 x y. Solution: The given expression is 24xy + 10 x2y + 14 x y + 16 x2y + 10 x y. Step 1: 24xy + 10 x2y + 14 x y + 16 x2y + 10 x y = x y (24 +14 +10) + x2y (10+ 16) Step 2: 24xy + 10 x2y + 14 x y + 16 x2y + 10 x y = 48 x y + x2y (10+ 16) Step 3: 24xy + 10 x2y + 14 x y + 16 x2y + 10 x y = 48 x y + 26 x2y The value for the given expression 12xy + 13 x2y + 16 x y + 20 x2y is 48 x y + 26 x2y

Solving Linear Equations with Fractions Learn about solving linear equations with fraction. In a linear equation the degree of the equation will be always one. It contains variables and constants and contain a first power of variables, also does not contain product of two variables. A linear equation may contain one or more variables in it. Linear equation is a simple equation. It has lots of application in mathematics and many other fields. Solving a equation is finding the value of unknown variables in the equation. The equation involving fraction contains terms of the form `a/b`. Here the method to solve is by taking the fraction term to one side and taking least common multiple(LCM) and them cross multiply the equations bring terms with variables to one side and constants to other side. Take variable common and then take remaining term to other side and simplify them.Now we will see some examples of solving equations with fraction below. Solving Equations with Fractions

Below are the examples on solving linear equations with fractions Example 1: Solve the linear equation `y/4` = 3 Solution: Step 1: Given equation `y/4` = 3 Step 2: Multiply by 4 on both sides `y/4` Ă— 4 = 3 Ă— 4 y = 12 Example 2: Solve the linear equation `x/3` + 5 = 9 Solution: Step 1: Given equation Read More About Factoring Trinomials Calculator

`x/3` + 5 = 9 Step 2: Subtract 5 on both sides of the equation `x/3 ` + 5 - 5 = 9 - 5 `x/3` = 4 Step 3: Multiply by 3 on both sides of the equation `x/3` × 3 = 4 × 3 x = 12 Example 3: Solve the linear equation `x/3` + `(5x)/3` = 4 Solution: Step 1: Given equation `x/3` + `(5x)/3` = 4 Step 2: Add the fractions `x/3` and ` (5x)/3`

Since the denominators of both the fractions are common, we can add the numerators directly as follows. `(x + 5x)/3` = 4 `(6x)/3` = 4 Step 3: Multiply by 3 on both sides of the equation 6x = 12 Step 4: Divide by 6 on both sides x=2 Example 4: Solve the linear equation `m/3` + `(3m)/5` - 4= 6 Solution: Step 1: Given equation `m/3` + `(3m)/5` - 4 = 6 Step 2: Add 4 on both sides of the equation

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