**Introduction **

In grade7 algebra, beside numerals we use symbols and literals in placeof unknown numbers to make a statement. Grade7 Algebra consists of Arithmetic problems. Grade7 algebra also consisting of both numerals andliterals. The algebraic expression can be defined as is an expression involving numbers and letters, multiplied together.

## Topics under grade 7 algebra

- Literals
- Constant
- Variable
- Algebraic Expression
- Arithmetic expression

**Literals:**

The letters that is used to represent the numbers is called as Literals.

**Example:**

The sum of two variables is 12. It can be written as x + y = 12. Here, x,y are the literals.

**Constant:**

A variable that has fixed numerical value is called as constant.

**Example:**

4,7, 8/5 are constants.

**Variable:**

A term or quantity that has different numerical values is called as variable. It can be represented as x, y, z etc.

**Algebraic expression:**

The Algebraic expression is combination of variables, numerals and arithmetical expression .

**Example:**

The algebraic expressions are, **4x + 5y = 12**

**Arithmetic expression:**

The combination of numbers formed to using arithmetic operations is called as arithmetic expression.

**Example:**

7 - 5 = 2, 6 + 5 = 11. These are some of the arithmetic expression.

## Practice problem for grade 7 algebra

**Practice problem 1:**

**Classify into arithmetic expression and algebraic expression.**

**1) 2 + 3 = 5 2) 7x + 4y = 15 3) 4x + 3y = 10 4) 8 + 3 = 11**

**Solution:**

**Arithmetic expression:**

1) 2 + 3 = 5.

2) 8 + 3 = 11.

**Algebraic expression:**

1) 7x + 4y = 11.

2) 4x+ 3y = 10.

**Practice problem 2:**

Multiply the given two expressions (x + 3) (x – 4).

**Solution:**

First expression is (x + 3)

Second expression is (x – 4)

Multiply, (x + 3) (x - 4) = x^{2} + 3x - 4x – 12

= x^{2} + 3x - 4x – 12

**Answer:**

The final answer is x^{2} + 3x - 4x – 12.

**Practice problem 3:**

Add the given two algebraic expressions **(3x + 2) **and **(5x – 12)**

**Solution:**

Given expressions are **(3x + 2)** and **(5x – 12)**

Adding the two expressions, we get

= 3x + 2 + 5x – 12

= 8x – 10

**Answer:**

The final answer is** 8x – 10**

**Practice problem 4:**

Subtract the algebraic expressions** (2x + 12) **and **(8x – 14)**

**Solution:**

Given two expressions are (2x + 12) and (8x – 14)

Subtract the two algebraic expressions, we get

(2x + 12) – (8x – 14) = 2x + 12 – 8x + 14

= - 6x + 26

**Answer:**

The final answer is **– 6x + 26**