**What are Algebraic Expressions?**

Algebraic expressions are the mathematical expressions we get when mathematical operations such as addition, subtraction, multiplication, division, etc., are operated upon any variable. The mathematical expression 2x + 5 is an example of an algebraic expression. The figure below shows the different components of an algebraic expression.

- A symbol that does not have a fixed value is called a
*variable*. Variables can take any value. In the figure above,**x**is the variable. - The numbers that are multiplied by the variables are called
*coefficients*. In the figure above, the coefficient of**x**is**2**. - Any symbol not joined to a variable is called a
*constant*because its numerical value does not change. In the figure above, the constant is**5**. - A
*term*can be a variable alone, a constant alone, or a combination of variables and constants by multiplication, division, or exponentiation. In the figure above,**2x**and**5**are the terms of the algebraic expression. - Terms are separated by the plus (+) sign or the minus (-) sign. One or more terms added together make an expression. In the figure above,
**2x**+**5**is an expression with two terms.

**Like Terms and Unlike Terms**

In adding or subtracting algebraic expressions, it is crucial to have a knowledge of like and unlike terms because the operation of addition and subtraction can only be performed on like terms. Like terms are terms that have the same variables raised to the same powers. Like terms can have a different coefficient, but the variables and their exponents must be the same. For example, **6x**, **2x**, and **– 8x** are like terms, but **7y ^{2}**,

**4xy**,

**– 2xy**

^{2},

**x**, and

**5**are

**unlike terms**

**.**

In combining like terms, simply add the numerical coefficients and just copy the variable part. For example, the terms **2y** and **5y** are like terms so that they can be combined. To combine, add the numerical coefficients **2** and **5**, that is, **2 + 5, **then attach the variable** y. **Thus, **2y **and **5y **combined are **7y**.

On the other hand, we cannot combine unlike terms into a single term. For example, the terms

**– 3x**and

**6y**are unlike terms because they have different variables. So, combining them will result in an expression with two terms such as

**– 3x + 3y**.

**How to simplify algebraic expressions?**

Simplifying algebraic expressions is the process of writing an expression in the most efficient and compact structure possible while maintaining the value of the original expression. It is a useful mathematical skill because it converts complex or difficult-to-read expressions into simpler ones. The process entails collecting like terms, which involves adding or subtracting terms in an expression.

Below are some basic rules and steps in simplifying an expression:

- If any, eliminate the grouping symbols, such as braces, brackets, and parentheses by multiplying factors.
- Use the laws of exponents to remove any grouping symbols if the terms are being raised to powers.
- Combine the like terms but keep the unlike terms as they are.
- Combine the constants.

**Example #1**

Simplify 5*x* + 2(x – 3)

**Solution**

We cannot combine terms when they are in any grouping sign. Thus, we need to eliminate the parenthesis by multiplying any factor outside the grouping to all the terms inside it.

5x + 2(x – 3) = 5x + 2x – 6

= **7x – 6**

If a minus sign is placed in front of a grouping, it affects all the operators contained within the grouping symbols. Putting a minus sign in front of a group converts the addition operation to subtraction and vice versa.

**Example #2**

Simplify 6x – (4 – x).

**Solution**

6x – (4 – x) = 6x + (–1) [4 + (–x)]

= 6x + (–1) (4) + (–1) (–x)

= 6x – 4 + x

= **7x – 4**

However, if a plus (+) sign comes before the grouping symbols, the parentheses are simply erased without affecting the operators contained within the grouping.

**How to add algebraic expressions?**

In adding algebraic expressions, we combine the like terms, and the unlike terms are written as they are. When combining like terms, make sure to group them and keep the signs of all the terms in the algebraic expression the same.

There are two ways in performing the addition of algebraic expressions.

**Horizontal Method –**In this method, you must write all expressions horizontally and then arrange the terms such that the like terms are positioned beside each other. These like terms are then added or combined.**Vertical Method –**In this method, you need to write each algebraic expression in a separate row such that the like terms are lined up vertically or in a column. The like terms are then added or combined column-wise.

**Example #1**

What is the sum of 2x^{2}, – 3x^{2}, and 5x^{2}?

**Solution (Horizontal Method)**

Algebraic Expression Addition Process | Step- by-step Explanation |

2x^{2} – 3x^{2} + 5x^{2} | Set up the addition horizontally. Notice that 2x^{2}, – 3x^{2}, and 5x^{2} are like terms because they all have x^{2}as their common variable.^{ } |

4x^{2} | Combine the like terms into a single term by adding their numerical coefficients. |

Therefore, the sum of 2x^{2}, – 3x^{2}, and 5x^{2} is 4x.^{2} |

**Solution (Vertical Method)**

Algebraic Expression Addition Process | Step-by-step Explanation |

2x^{2} – 3x ^{2} + 5x ^{2} ——– | Set up the addition vertically. Line up the like terms in the same column. |

2x^{2} – 3x ^{2} + 5x ^{2}——– 4x^{2} | Combine the like terms into a single term by adding their numerical coefficients vertically. Put the result below the bar. |

Therefore, the sum of 2x^{2}, – 3x^{2}, and 5x^{2} is 4x.^{2} |

**Example #2**

Simplify (p + 2q – 3r + 4) + (2p + 4q + 6r – 2).

**Solution (Horizontal Method)**

Algebraic Expression Addition Process | Step- by-step Explanation |

(p + 2q – 3r + 4) + (2p + 4q + 6r – 2) | Set up the addition horizontally. |

p + 2q – 3r + 4 + 2p + 4q + 6r – 2 | Remove the grouping symbols. The sign of the terms in the polynomials remains the same. |

p + 2p + 2q + 4q – 3r + 6r + 4 – 2 | Arrange the terms of the new polynomial in a standard form such that the like terms are beside each other. The like terms are p and 2p, 2q and 4q, – 3r and 6r, and 4 and – 2. |

3p + 6q + 3r + 2 | Combine the like terms into a single term by adding their numerical coefficients. |

Therefore, (p + 2q – 3r + 4) + (2p + 4q + 6r – 2) is equal to 3p + 6q + 3r + 2. |

**Solution (Vertical Method)**

Algebraic Expression Addition Process | Step-by-step Explanation |

p + 2q – 3r + 4 + 2p + 4q + 6r – 2———————- | Set up the addition vertically. Line up the like terms in the same column. The like terms are p and 2p, 2q and 4q, – 3r and 6r, and 4 and – 2. |

p + 2q – 3r + 4+ 2p + 4q + 6r – 2———————– 3p + 6q + 9r + 6 | Combine the like terms into a single term by adding their numerical coefficients vertically. Put the answers below the bar. |

Therefore, (p + 2q – 3r + 4) + (2p + 4q + 6r – 2) is equal to 3p + 6q + 3r – 2. |

**How to subtract algebraic expressions?**

The process for subtracting algebraic expressions is the same as the process for adding them. It can be done horizontally or vertically. The only difference is that when subtracting one algebraic expression from another, you must change the signs of each term in the expression being subtracted and then combine like terms.

But why do we have to change the signs of an algebraic expression when being subject to subtraction? Let us take an example of two numbers, **3** and **−5**. Suppose if we must subtract **−5** from **3**, we write it as 3−(−5). We know that the product of two positive signs or two negative signs is positive, and the product of two unlike signs is negative. In the example: 3−(−5) = 3 + 5 = 8. If we do not change the sign, we will have 3 – 5 = – 2, which is completely different. So, changing the signs of the subtrahend when subtracting algebraic expressions is necessary.

In other words, the subtraction symbol (−) must be distributed to each term of the subtrahends before combining like terms. The most common error in subtracting algebraic expressions is failing to change the sign of EVERY term in the subtrahend or the expressions being subtracted.

**Example #1**

What is the difference between 2x and –7x?

**Solution (Horizontal Method)**

Algebraic Expression Subtraction Process | Step-by-step Explanation |

2x – (–7x) | Set up the subtraction horizontally. |

2x + 7x | Distribute the minus sign to the term in the subtrahend. Thus, the –7x will become 7x. Then, proceed like in the process of addition. |

9x | Combine the like terms into a single term by adding their numerical coefficients. |

Therefore, the difference between 2x and –7x is 9x. |

**Solution (Vertical Method)**

Algebraic Expression Subtraction Process | Step-by-step Explanation |

2x– – 7x————– | Set up the addition vertically. |

2x + 7x————– | Distribute the minus sign to the term in the subtrahend. Thus, the –7x will become 7x. Then, proceed just like in the process of addition. |

2x + 7x————– 9x | Combine the like terms into a single term by adding their numerical coefficients vertically. Put the result below the bar. |

Therefore, the difference between 2x and –7x is 9x. |

**Example #2**

Subtract** **(a^{2}b –2a^{2} – bc + 5) and (– 3a^{2} +3a^{3}) from (b^{3} +3a^{2}b – 6a^{2} – 6bc +7a^{3}).

**Solution (Horizontal Method)**

Algebraic Expression Subtraction Process | Step-by-step Explanation |

(b^{3} + 3a^{2}b – 6a^{2} – 6bc + 7a^{3}) – (a^{2}b –2a^{2} – bc + 5) – (– 3a^{2} + 3a^{3}) | Set up the subtraction horizontally. |

b^{3} + 3a^{2}b – 6a^{2} – 6bc + 7a^{3} – a^{2}b + 2a^{2} + bc – 5 + 3a^{2} – 3a^{3} | Distribute the minus sign to the terms in the subtrahends. Thus, a^{2}b will become – a^{2}b, –2a^{2} will become 2a^{2}, – bc will become bc,^{ }5 will become –5, – 3a^{2} will become 3a^{2}, and 3a^{3} will become – 3a^{3}. Then, proceed just like in the process of addition. |

b^{3} + 3a^{2}b – a^{2}b – 6a^{2} + 2a^{2} + 3a^{2} – 6bc + bc + 7a^{3} – 3a^{3} – 5 | Arrange the terms of the new algebraic expression such that the like terms are beside each other. |

b^{3} + 2a^{2}b – a^{2 }– 5bc + 4a^{3 }– 5 | Combine the like terms into a single term by adding their numerical coefficients. |

Therefore, (b^{3} +3a^{2}b – 6a^{2} – 6bc +7a^{3}) – (a^{2}b –2a^{2} – bc + 5) – (– 3a^{2} +3a^{3}) is equal to b.^{3} + 2a^{2}b – a^{2 }– 5bc + 4a^{3 }– 5 |

**Solution (Vertical Method)**

Algebraic Expression Subtraction Process | Step-by-step Explanation |

Set up the subtraction vertically. Line up the like terms in the same column. | |

Distribute the minus sign to the terms in the subtrahends. Thus, a^{2}b will become – a^{2}b, –2a^{2} will become 2a, ^{2}– bc will become bc,^{ }5 will become –5, – 3a^{2} will become 3a^{2}, and 3a^{3} will become – 3a.^{3} Then, proceed just like in the process of addition. | |

Combine the like terms into a single term by adding their numerical coefficients vertically. Put the result below the bar. | |

Therefore, (b^{3} +3a^{2}b – 6a^{2} – 6bc +7a^{3}) – (a^{2}b –2a^{2} – bc + 5) – (– 3a^{2} +3a^{3}) is equal to b.^{3} + 2a^{2}b – a^{2 }– 5bc + 4a^{3 }– 5 |

**How to solve problems involving adding and subtracting algebraic expressions?**

To solve problems involving algebraic expressions, follow these steps:

- Analyze the problem.
- List all the given information.
- Set up the addition or subtraction process.

**Problem #1**

What should be added to 2m^{2 }+ 3mn^{ }+ n^{2} to obtain 6mn + n^{2}?

**Solution**

The required expression is equal to the subtraction of 2m^{2 }+ 3mn^{ }+ n^{2} from 6mn + n^{2}. Thus, required expression is

= (6mn + n^{2}) – (2m^{2 }+ 3mn^{ }+ n^{2})

= 6mn + n^{2} – 2m^{2 }– 3mn^{ }– n^{2}

= 6mn – 3mn^{ }+ n^{2 }– n^{2 }– 2m^{2}

= 3mn – 2m^{2}

Therefore, 3mn – 2m^{2} must be added to 2m^{2 }+ 3mn^{ }+ n^{2} to obtain **6mn + n**^{2}**.**

**Problem #2**

Jim and Kate get paid per project. Jim is paid a project fee of \$20 plus \$8 per hour, while Kim is paid a project fee of \$15 plus \$10 per hour. Write an algebraic expression representing how much a company will pay to hire both to work the same number of hours on a project.

**Solution**

Step 1: Write the algebraic expressions to represent how much the company will pay each person: Project Fee + Hourly rate × Number of hours worked. Let h represents the number of hours they will work on the project. So, Jim will be paid \$20 + \$8h, and Kate will be paid \$15 + \$10h.

Step 2: Add the algebraic expressions to represent the total amount a company will pay to hire Jim and Kate. Thus, (\$20 + \$8h) + (\$15 + \$10h).

Step 3: Perform addition of algebraic expressions and simplify the result by combining like terms. So, \$20 + \$8h + \$15 + \$10h = \$35 + \$18h.

Therefore, a company will pay **35 + 18h dollars** to hire both Jim and Kate.

**Problem #3**

In a Mathematics test, Sarah scored p marks. Nina scored 7 marks more than Sarah, while Faye scored 4 marks less than Nina. Work out the total mark of the three girls together.

**Solution**

Step 1: Write the algebraic expressions that will represent the score of each girl in the mathematics test. Sarah got a score of p, Nina got a score of p + 7, and Faye got a score of (p + 7) – 4.

Step 2: Add the algebraic expressions to represent the total mark that the three girls got. Thus, p + (p + 7) + [(p + 7) – 4].

Step 3: Perform addition of algebraic expressions and simplify the result by combining like terms. So, p + p + 7 + p + 7 – 4 = 3p + 10

Therefore, the total mark of the three girls together is **3p + 10**.

**Problem #4**

What is the length of the third side of the triangle if its perimeter is 8x^{2 }+ 34xy + 10 units and the length of its other two sides are 10xy – 6x^{2} and 2x^{2 }+ 10xy – 4?

**Solution**

Step 1: The perimeter of a triangle is equal to the sum of the lengths of all its three sides. So, to determine the length of the third side, we must subtract the length of the other two sides, which are 10xy – 6x^{2} and 2x^{2 }+ 10xy – 4, from the given perimeter, which is 8x^{2 }+ 34xy + 10. That is, (8x^{2 }+ 34xy + 10) – (10xy – 6x^{2}) – (2x^{2 }+ 10xy – 4).

Step 2: Perform subtraction of algebraic expressions to solve for the length of the third side and simplify the result by combining like terms. Hence,

(8x^{2}+34xy+10) –(10xy–6x^{2}) –(2x^{2}+10xy–4) =8x^{2}+34xy+10–10xy+6x^{2}–2x^{2}–10xy+4

=8x^{2}+6x^{2}–2x^{2}+34xy–10xy–10xy +10+4

=12x^{2 }+ 14xy + 14

Therefore, the length of the third side of the triangle is **12x**^{2 }**+ 14xy + 14** units.

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