Subject: Compulsory Maths

An equation that can be written in the form ax + b =0, where a and b are constants, is called a linear equation in one variable.

For example, x - 4 =10 is a linear equation in x.

A value, such that, when you replace the variable with it, makes the equation true is called the solution of the equation.

If a = b, then a + c = b + c

For example:

Solve: x - 10 = -2

Solution:

x - 10 = -2

or, x - 10 + 10 = -2 + 10, addition axiom

\(\therefore\) x = 8

If a = b, then a-c = b-c

For example:

Solve for x: x+2 = 6

Solution:

x+2 =8

or, x = 8 - 2

\(\therefore\) x =6

If a = b, then ac = bc

For example:

Solve for x: \(\frac{x}{2}\) = 5

Solution:

\(\frac{x}{2}\) = 5

or, x = 5 \(\times\) 2

\(\therefore\) x = 10.

To solve word problems on linear equations apply the following steps:

- Read the problem carefully to find out what is given and what is to be given.
- Represent the unknown quantity by x or by some other letter.
- Write the relation between known and unknown.
- Solve the equation to obtain the value of unknown linear Inequality.

**Examples**

- Two numbers are in the ratio 3:4. If the sum of the numbers is 49, Find the numbers.

Solution:

Let the numbers be 3x and 4x

Then, 3x + 4x = 49

or, 7x = 49

or, x = \(\frac{49}{7}\)

\(\therefore\) x = 7

Now, 3x = 3 x 7 =21 and 4x = 4 x 7 =28

Hence, the required numbers are 21 and 28. - Two- third of the number is less than 10 than the original number. Find the number.

Solution:

Let the number be x.

Then, \(\frac{2}{3}\) x = x - 10

or, 2x = 3x - 30

or, 2x - 3x = -30

or, -x = -30

\(\therefore\) x = 30

Hence, the required number is 30.

An inequality is a relationship between two quantities that are not equal. In equations, one side is equal to the other side. In linear inequalities, one side is bigger than or smaller than or equal to the other side.

A linear equation in one variable has only one solution. An equality in one variable has a set of possible solutions.

The symbols used in inequality are:

- > means 'greater than'.
- < means 'less than'.
- ≥ means 'greater than or equal to'
- ≤ means ' less than or equal to'

**Example:**

Given that x is an integer. State the possible integer values of x in the following inequalities.

- x > 7
- x ≤ -3

Solution

- x is greater than 7

Solution set = { 8, 9, 10 } - x is less than or equal to -3

Solution set = { -3, -4, -5, -6 }

**Solving linear inequalities in one variable**

Solving linear inequalities are the same as solving linear equations with one important exception.

When you multiply or divide an inequality by a negative value, it changes the direction of the inequality.

Look at this statement: 5> 2

Suppose we multiply both sides by -1.

( -1 ) (5 )> (- 1 ) ( 2 )

- 5 <-2

-5 less than -2 because it is further to the left on the number line.

Thus, -5 < -2

The two equations having two unknowns are called Simultaneous equations. They are called simultaneous because they must be solved at the same time. For example, x+y = 10 and x-y = 2 are simultaneous equations. To solve a simultaneous equation we must use graph. For this, we have to draw the lines of the equations, The point where the lines cross is the solution.

**Example:**

Solve graphically: x + y = 6, 2x - y = 8

Solution:

Given equations are x + y = 6 and 2x - y = 8

x + y = 6

y = 6 - x ...........(i)

Table for y = 6 - x:

x | 3 | 0 | 2 |

y | 3 | 6 | 4 |

Again, 2x -y = 8

2x - 8 = y

y = 2x - 8 ..........(ii)

Table for y = 2x - 8

x | 4 | 6 | 3 |

y | 0 | 4 | -2 |

Draw the lines of the equations on a graph. Since, both lines intersect at P(6, 2), solution of the given equations is x = 6 and y = 2

A general quadratic equation can be written in the form ax^{2}+ bx + c = 0, where x represents a variable or an unknown and a, b and c are constants with a≠ 0.(If a = 0, the equation is the linear equation.). For example: 2x^{2} = 5x +2 = 0, is a quadratic equation where a = 2, b = 5 and c=2

**Solution of quadratic equations by factoring**

To make the equation true, the solution of a quadratic equation consists of all numbers (roots). All quadratic equations have 2 solutions (i.e 2 roots).

Sometimes it is easy to find solutions or roots of a quadratic equation by factoring. Indeed, the basic principle to be used is: If a and b are real numbers such that ab = 0, then a = 0 or b = 0.

**Examples**

- Solve: x
^{2}= 49

Solution:

x^{2}- 49 =0

x^{2}-7^{2}= 0

(x+7) (x-7) = 0

x +7 = 0 or x - 7 =0

∴ x = -7 or x = 7 - Solve: x
^{2}+ 7x - 8 = 0

x^{2}+ (8 -1) x + 8 = 0

x^{2}+ 8x -1x +8 =0

x (x + 8) -1(x + 8) = 0

(x + 8) (x - 1) = 0

x + 8 =0 or x - 1 = 0

∴ x = 8 or x = 1

- An equation can be written in the form ax + b = 0. It can also be represented in a graph.
- An inequality is a relationship between two quantities that are not equal.
- A linear equation in one variable has only one solution.
- A general quadratic equation can be written in the form ax
^{2}+ bx + c = 0, where x represents a variable or an unknown and a, b and c are constants with a≠ 0.(If a = 0, the equation is the linear equation).

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Solve:

7x=21

Solution:

Here given, 7x=21

Dividing by 7 in both side

\(\frac{7x}{7}\)=\(\frac{21}{7}\)

∴x=3

Solve:

x-8=9

Solution:

Here given, x-8=9

Adding 8 on both side

x-8 +8=9+8

∴x=17

Solve:

3x+4=13

Solution:

Here given,

3x+4=13

Subtracting 4 from both sides

3x+4-4=13-4

or, 3x=9

Dividing 3 on both side

\(\frac{3x}{3}\)=\(\frac{9}{3}\)

∴x=3

Solve:

5x-14=8

Solution:

5x-14=8

or, 5x-14+14=8+14 [Adding 14 on both side]

or, 5x=22

Dividing by 5 on both side

\(\frac{5x}{5}\)=\(\frac{22}{5}\)

∴ x = \(\frac{22}{5}\)

Solve:

\(\frac{10x+8}{6}\)=17

Solution:

\(\frac{10x+8}{6}\)=17

or, \(\frac{6×(10x+8)}{6}\)=17×6 [Multiplying by 6 on both side]

or, 10x+8=102

or, 10x+8-8=102-8 [Subtracting 8 from both side]

or, 10x=94

or, \(\frac{10x}{10}\)=\(\frac{94}{10}\) [Dividing by 10 on both side]

∴ x=\(\frac{47}{5}\)

Solve:

8x+9=10

Solution:

8x+9=10

or, 8x+9-9=10-9 [Subtracting 9 from both side]

or, 8x=1

or, \(\frac{8x}{8}\)=\(\frac{1}{8}\) [Dividing by 8 on both side]

∴ x = \(\frac{1}{8}\)

Solve:

13x-14=12

Solution:

13x-14 = 12

or, 13x-14+14 = 12+14 [Adding 14 on both side]

or, 13x = 26 [Dividing by 13 on both side]

or, \(\frac{13x}{13}\) = \(\frac{26}{13}\)

∴ x = 2

Solve:

\(\frac{x+1}{6}\)=7

Solution:

\(\frac{x+1}{6}\)=7

or, 6×\(\frac{(x+1)}{6}\)=7×6 [Multiplying by 6 on both side]

or, x+1= 42

or, x+1-1= 42-1 [Subtracting 1 from both side]

∴ x = 41

Solve and check:

5x+3=2x+6

Solution:

5x+3=2x+6

Subtract 3 from both side

or, 5x+3-3=2x+6-3

or, 5x=2x+3

Subtract 2x from both side

5x-2x=2x-2x+3

or, 3x=3

Divide by 3 on both side

\(\frac{3x}{3}\)=\(\frac{3}{3}\)

∴ x = 1

By checking 5x+3=3x+6 or, 5×1+3=2×1+6 or, 5+3=2+6 ∴ 8 = 8 proved |

Solve and check:

4x+7=3x+10

Solution:

4x+7=3x+10

or, 4x+7-7=3x+10-7 [subtracting 7 from both side]

or, 4x=3x+3

or, 4x-3x=3x-3x+3 [Subtracting 3x from both side ]

or, x=3

∴ x = 3

By checking 4x+7=3x+10 or, 4×3+7=3×3+10 or, 12+7=9+10 ∴ 19 = 19 proved. |

Solve and check:

9+14x=27-11x

Solution:

9+14x=27-11x

or, 9+14x-9=27-11x-9 (subtracting 9 from both side)

or, 14x=18-11x

or, 14x-11x=18-11x-11x (subtracting 11x from both side)

or, 25x=18

or, \(\frac{25x}{25}\)=\(\frac{28}{25}\)

∴ x=\(\frac{18}{25}\)

By Checking 9+14x=27-11x or, 9+14×\(\frac{18}{25}\)=27-11×\(\frac{18}{25}\) or, 9+\(\frac{252}{25}\)=27-\(\frac{198}{25}\) or, \(\frac{9×25+252}{25}\)=\(\frac{27×25-198}{25}\) =\(\frac{225+252}{25}\)=\(\frac{675-198}{25}\) ∴\(\frac{477}{25}\)=\(\frac{477}{25}\) Proved. |

Solve and check:

4(x+4)=3(x-1)

Solution:

4(x+4)=3(x-1)

or, 4x+16=3x-3

or, 4x+16-16=3x-3-16 ( Subtracting 16 on both side)

or, 4x=3x-19

or, 4x-3x=3x-3x-19( Subtracting 3x on both side)

∴x=-19

4x+16=3x-3 or, 4×(-19)+16=(-3)×(-19) or, -76+16=-57-3 ∴ -60=-60 Proved. |

Solve:

\(\frac{x-2}{x+2}\)=\(\frac{4}{3}\)

Solution:

\(\frac{x-2}{x+2}\) = \(\frac{4}{3}\)

or, 3(x-2 )= 4(x+2) (By cross multiplication)

or, 3x-6 = 4x+8

or, 3x-6+6 = 4x+8+6) (Adding 6 on both side)

or, 3x-4x = 4x-4x+14 (Subtracting 4x from both side)

or, -x = 14

∴ x = -14

Solve;

\(\frac{3-4x}{5-4x}\) \(\times\) \(\frac{7}{2}\)

Solution:

Here given, 2(3-4x) = 7(5-4x) (By cross multiplication)

or, 6-8x = 35-28 x

or, 6-8x-6 = 35-28x-6 (Subtracting 6 from both side)

or, 9x =10x+8

or, 9x-10x = 10x+8-10x (Subtracting 10x from both side)

or, -x = 8

∴ x = -8

Solve:

10 % of x= 35

Solution:

10 % of x = 35

or, x×\(\frac{10}{100}\) = 35

or, \(\frac{x}{10}\) = 35

or, x = 35×10 (By cross multiplying)

∴ x = 350

Solve:

2\(\frac{1}{2}\)% of 500 = x

Solution:

2\(\frac{1}{2}\)% of 500 = x

or, 500×\(\frac{5}{2}\) = 100x( By cross multiplication)

or, \(\frac{500×5}{2×100}\) = x

or, \(\frac{25}{2}\) = x

∴ x = 12.5

Solve:

13% of x = 6.5

Solution:

13% of x =6.5

or, x×\(\frac{13}{100}\) = 6.5

or, x×13 6.5×100

or, x = \(\frac{6.5×100}{13}\)

or, x =\(\frac{65×100}{130}\)

or, x = 5×10

∴ x = 50

Solve:

33% of x+x = 266

Solution:

33% of x+x = 266

or, x+x×\(\frac{33}{100}\) = 266

or, \(\frac{100x+33x}{100}\) = 266

or, \(\frac{133x}{100}\) = 266

or, x = \(\frac{266×100}{133}\)

or, x = 2×100

∴ x = 200

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