Algebra is an important branch of Mathematics, which deals with several types of problems. These problems may either be related to constants or variables we mainly concentrate on "how to solve numbers.” When we solve any algebra problem there are several games using which you can easily learn the concept of algebra like kids have a great fun when they solve the puzzles, play some computer games by finding secret doors, etc. |

Elementary algebra deals with the basic concepts of algebra, which is one of the main concept of mathematics. It is a course for the students, who are starting from the beginning level of algebra.

Elementary algebra covers numbers, linear equations, polynomials, factoring, square roots, quadratic equations and graphing.

These topics also come under intermediate algebra or algebra 1. But, here in elementary algebra, we will study only the basic pace of those topics.

Given below are few problems based on elementary algebra topics.

### Solved Examples

**Question 1:**Solve the linear equations given below:

7x - y = 5

5x - 10y = 21

**Solution:**

**Given:**

7x - y = 5 --> (1)

5x - 10y = 21 --> (2)

Solve equation (1) for y,

y = 7x - 5

Now, substitute the value of y in the equation (2)

5x - 10(7x - 5) = 21

5x - 70 + 50 = 21

- 65x = 21 - 50

- 65x = -29

x = $\frac{-29}{-65}$

= $\frac{29}{65}$

By subsitituting the x value in the equation y = 7x - 5, we get

y = 7($\frac{29}{65}$) - 5

y = $\frac{203}{65}$ - 5

y = $\frac{203 - 325}{65}$

y = $\frac{-122}{65}$

$\therefore$, x = $\frac{29}{65}$ and y = $\frac{-122}{65}$

**Question 2:**Draw a graph for the equation y = 2x + 4

**Solution:**

**Given:**y = 2x + 4

At, x = 0

y = 2(0) + 4

y = 4

At, x = 1

y = 2(1) + 4

y = 6

At, x = - 2

y = 2(-2) + 4

y = 0

At, x = -3

y = 2(-3) + 4

y = -2

$\therefore$, co-ordinates are (0, 4), (1, 6), (-2, 0) and (-3, -2)

**Question 3:**Solve the given quadratic equation x

^{2}- 3x - 72 = 0, using the quadratic formula.

**Solution:**

**Given:**x

^{2}- 3x - 72 = 0

We can solve this above equation using the quadratic formula: x = $\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

Here, a = 1, b = -3 and c = -72

Now, substitute the above values in the formula,

x

_{(1, 2) }= $\frac{-(-3)\pm\sqrt{(-3)^{2}-4(-72)} }{2}$

x

_{(1, 2)}

_{ }= $\frac{-(-3)\pm\sqrt{9-4(-72)} }{2}$

x

_{(1, 2) }= $\frac{3\pm \sqrt{279}}{2}$

$\therefore$ x

_{1}= $\frac{3 + \sqrt{279}}{2}$ and x

_{2}= $\frac{3 - \sqrt{279}}{2}$

Given below are few problems based on abstract algebra topics:

### Solved Examples

**Question 1:**Find the 7

^{th}term for the following sequences:

8, 12, 16, .......

**Solution:**

**Given:**First term (a) = 8

Difference between next term and previous term, 12 - 8 = 4

and 16 - 12 = 4

$\therefore$ Common difference = 4

The general form of arithmetic progression is

$a_n$ = a + (n - 1)d

Now, $a_7$ = a + 6d

= 8 + 6 x 4

= 8 + 24

= 32

Therefore, the 7

^{th}term of the sequence is 32.

**Question 2:**Solve the following linear equations with fractions:

$\frac{5x}{3} - \frac{6}{7}$ = $\frac{x}{8}$

**Solution:**

**Given:**

$\frac{5x}{3} - \frac{6}{7}$ = $\frac{x}{8}$

Multiply both sides by LCD of the denominator.

LCD of 3, 7, 8 = 168

168$(\frac{5x}{3}-\frac{6}{7}$ = $\frac{x}{8})$

After dividing and multiplying, we get

280x - 144 = 21x

Subtract 21x both the sides

280x - 21x = 21x - 21x + 144

259x = 144

x = $\frac{259}{144}$