Algebra Problems – Solutions and Identities

Algebra Problems with Solutions

1. Solve for x in the equation 2x + 3 = 5

x = 2

2. Solve for x in the equation 3x + 2 = 5

x = 1

3. Solve for x in the equation 2x – 1 = 3

x = 2

Algebra Identities

An algebra identity is a statement that is always true in mathematics. Algebra identities are usually written as equations, and they can be used to solve problems. Some common algebra identities include the associative and distributive laws, the identity and inverse laws, and the commutative law.

6th Grade Algebra Problems

6th grade algebra problems involve solving equations and inequalities, manipulating algebraic expressions, and working with fractions and decimals.

One common type of 6th grade algebra problem is solving equations. An equation is a mathematical statement that two things are equal. In algebra, equations are usually written with an equal sign (=) between two algebraic expressions. To solve an equation, you must find a value for the variable that makes the expressions on either side of the equal sign equal.

Another common type of 6th grade algebra problem is solving inequalities. Inequalities are mathematical statements that two things are not equal. In algebra, inequalities are usually written with a less-than or greater-than sign (< or >). To solve an inequality, you must find a value for the variable that makes the expressions on either side of the inequality equal.

Algebra Mathematics Problems and Solutions

Problem:

A carpenter wants to make a rectangular box with a volume of 64 cubic feet. What are the dimensions of the box if the length is 8 feet and the width is 4 feet?

Solution:

The dimensions of the box are 8 feet by 4 feet.

Algebra Word Problems

1. A plumber charges $25 for a service call plus $50 per hour of service. Write an equation in slope-intercept form for the cost, C, after h hours of service.

C = 25 + 50h

Algebra Identities

Given below are the algebra identities that you must learn to solve Algebra problems.

1. (a + b)2 = a2 + b2 + 2ab

2. (a – b)2 = a2 – b2 – 2ab

3. a2 – b2 = (a + b) (a – b)

4. a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab

5. a3 + b3 = (a + b) (a2 – ab + b2)

6. a3 – b3 = (a – b) (a2 + ab + b2)

7. (a + b)3 = a3 + 3ab (a + b) + b3

8. (a – b)3 = a3 – 3ab (a – b) – b3

9. Consider the following example:

   Six less than a number is equal to two. Determine the number

   Solution:

   Let the given number be denoted as ‘x’.

   According to condition, you have x – 6 = 2

   By solving it algebraically for x, you get,

   x = 2 + 6 = 8

   Hence, your required number is 8.