**Introduction to 7th grade Algebra:-**

In Algebra, beside numeral we use symbol and literal in place of unknown number to make a statement. 7th grade Algebra is a cluster of mathematics that uses mathematical statements to describe relationships linking things that vary over time. For example: the relationship connecting price and number of item.Here we learning lot of 7th grade Algebra based problems.

**7th grade algebra topics:-**

• algebra Exponentials and Logarithms

• algebra Absolute Value Equations and Inequalities

• algebra Sequences and Series Combinatory

• algebra Advanced Graphing

• Graphing Polynomials for algebra

• Graphing Rational Functions

• Exponents for algebra

• Radicals for algebra

• Polynomials for algebra

• Factoring for algebra

• Division of Polynomials for algebra

• Solving Equations for algebra

• Solving Inequalities for algebra

Here we learning lot of 7th grade Algebra topics.

**7th grade algebra formulas:-**

• Difference of squares a^{2}-b^{2}=(a-b)(a+b)

• Difference of cubes a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})

• Sum of cubes a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})

• Formula for (a+b)^{2} and (a-b)^{2},(a+b)^{2}=a^{2}+2ab+b^{2},(a-b)^{2}=a^{2}-2ab+b^{2}

• Formula for (a+b)^{3} and
(a-b)^{3},(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3},(a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}

Here we learning lot of 7th grade Algebra based formulas.

**7th grade algebra Problem 1:**

Solve - 5 (x + 2) = - 16

**Solution:**

- 5 ( x + 2 ) = - 16

Step 1 : Remove the Brakets, since there is a - sign before the bracket, + before the quantity becomes - and - becomes +.

-5x – 10 = -16

Step 2 : Take -10 to the other side, when switching the sides, the sign switches, so -10 is +10 on the other side

-5x = - 16 + 10

-5x = - 6

Step 3: Take 5 to the other side, here 5 is a multiple, so when it changes sides, it multiplies as 1/5

x = -6/5

Here we learning lot of 7th grade Algebra based problems.

**7th grade algebra Problem 2:**

Solve: 8 (x + 3) + 5 (x – 3) = 4

**Solution:**

8 ( x + 3 ) + 5 ( x – 3 ) = 4 (Remove the brakets, here 8 will multiply both x and 3, and 5 will multiply x and -3

8x + 24 + 5x – 15 = 4

8x + 5x + 24 - 15= 4 (rearrange the terms and simplify like terms)

13x + 9 = 4

13x = 4 - 9

13x = - 5

x = - 5/13

Here we learning lot of 7th grade Algebra based problems.

**7th grade algebra Problem 3:**

Solve for x and y for the following equations

5x + y = 8

6x + y = 7

**Solution:**

5x + y = 8 ------- (1)

6x + y = 7 ------- (2)

Solve the 1st equation for the y

y = 8 - 5x

Now, we substitute 8 - 5x for y in the 2nd equation.

6x + (8 - 5x) = 7

x + 8 = 7

x = 7 – 8

x = -1.

5x + y = 8

5(-1) + y = 8

y = 8 + 5

y = 13

Therefore the value of x = -1 and y = 13.

Here we learning lot of 7th grade Algebra based problems.