Intersection of Two Straight Lines

Two lines will intersect at a point. The point will have a pair of values as (x1, y1). The straight lines are represented by equation with two or one variable in x and y.  If those equations are solved to get the value of x and y, will represent the point intersection of those two lines. To solve the set of lines to get the value of x and y , we can use either the method of elimination or substitution. Now let us discuss few problems on this topic intersection of two straight lines.

Example Problems on Intersection of Two Straight Lines

Ex 1: Find the point of intersection of the following two lines

2x + 3y = 10; 2x + y = 6.

Sol: Given: 2x + 3y = 10 --------------(1)

2x + y = 6  --------------(2)

The point intersection of above two lines can be obtained by solving them as follows:

(1) – (2) implies: 2x + 3y = 10

2x + y = 6

We get, 2y = 4

y = 2.

From (2), 2x + (2) = 6

2x = 6 – 2 = 4

2x = 4

x = 2.

Therefore, the point of intersection is (2, 2).

Ex 2: Find the point of intersection of the following two lines

x + 2y = 1; 5x + 4y = -7.

Sol: Given: x + 2y = 1 ------------------(1)

5x + 4y = -7 -----------------(2)

The point of intersection of above two lines can be obtained by solving the as follows:

(1) x 5 – (2) implies: 5x + 10y = 5

5x + 4y = -7

We get, 6y = 12

y = 2.

Therefore, from (1), x + 2(2) = 1

Implies, x = 1 – 4 = -3.

Therefore the point of intersection is (-3, 2).

Ex 3: Find the point of intersection of the following two lines

3x + y = 10; y = 7.

Sol: Given: 3x + y = 10 -------------(1)

y = 7 -----------(2)

Since, y = 7 is one of the line, the value of the y coordinate will be 7.

Therefore, from (1), we get , 3x + 7 = 10

3x = 10 – 7 = 3

x = 1.

Therefore, the point of intersection is ( 1, 7).

Practice Problems on Intersection of Two Straight Lines

1. Find the point of intersection of lines 3x + 2y = 7 and x + y = 3.

[ Answer: (1, 2)]

2. Find the point of intersection of lines 3x - 2y = -2 and x - 2y = 6.

[ Answer: (-4, -5)]

3. Find the point of intersection of lines 4x - 3y = -10 and 3x + y = -1.

[ Answer: (-1, 2)]

Solving Binomials by Factoring

Introduction: 

In algebra, the polynomials which have two terms are called binomials. To factor binomials, we need to follow the following methods:

(i) 2a + ab = a(2 + b ) [ Here the given expression has two terms, where a is the common value]    
     = a( 2 + b)         
(ii) a² – b² = ( a + b) ( a – b)   [ This is the standard form]
(iii) (a + b)² = ( a + b)(a +b)
(iv) (a - b)² = ( a - b)(a - b)

Product of two polynomials will give three terms.
( a + b)² = ) a² +b² + 2ab
( a - b)² = ) a² +b² - 2ab

Let us see few problems on this topic solving binomials by factoring.

Example Problems on Solving Binomials by Factoring

Ex 1: Solve (x + 3) (x – 4) = 0

Soln: Given: (x + 3) (x – 4) = 0
This implies: x + 3 = 0 or x – 4 = 0
That is : x = -3, 4.
Therefore the solution is { -3, 4}.

Ex 2: Solve x + y = 7 and xy = 12, find x and y.

Soln: Given : x + y = 7 -----------(1)
                            xy = 12 ---------(2)
Therefore, x – y = sqrt [(x+y)² – 4xy]
                             = sqrt[ 7² – 4(12)]
                             = 1
Therefore, x – y = 1 ---------------(3)
From (1) and (3), we get:
x + y = 7 -----------(1)
x – y = 1 -----------(3)
2x = 8, this implies that x = 4.
Therefore, (1) implies 4 + y = 7
Hence y =3.
Therefore the solution is {4,3}.

Ex 3: Solve x - y = 5 and xy = 24, find the value of x + y.

Soln: x + y = sqrt [(x-y)² + 4xy]
                   = sqrt[ 5² – 4(24)]
                   = 11.
Therefore from, x + y = 11
                             x – y = 5,
We get 2x = 16.
Therefore, x = 8.
Hence from x + y = 11. y = 3.
Therefore the solution is { 8, 3}.

Ex 4: Solve x + y = 11 and xy = 24, find the value of x² – y².

Soln: x – y = sqrt [(x+y)² – 4xy]
                   = sqrt[ 11² – 4(24)]
                   = 5
Therefore, x² – y² = ( x + y )( x – y)
                             = 11 x 5 = 55.

Practice Problems on Solving Binomials by Factoring

1. If a + b = 9  and ab = 36, find a - b
[Ans: a – b = 5]

2. If a – b = 4 and ab = 12, find a² – b².
[Ans: a² – b² = 32]

Graph of Sinx

 The coordinate graph is called the Cartesian coordinate plane. The graph contains a couple of the vertical lines are called coordinate axes. The vertical axis of the y axis value and the horizontal axis value is the x axis value. The points of the intersection of those two axes values are called the origin of coordinate graphing pictures.  The trigonometry graph is a sin or cos waves. In this graph equation is in the form of y = mx + c. m is nothing but a sin or cos. In this article we shall discuss graph of sin x.

Sample Problem for Graph of Sin X:

Graph of sin x problem 1:

Solve the given trigonometry functions 3sin 5x - y = 0 and draw the graph for the given function.

Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                                      3sin 5x - y = 0
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                                  3sin 5x – y = 0
                                                y = 3sin 5x
In the next step we are find out the plotting points of the above equation.
In the above equation we put x = -5 we get
           y = 3sin 5(-5)
           y = -0.4
In the above equation we put x = -4 we get
            y = 3sin 5(-4)    
           y = -2.7
In the above equation we put x = 0 we get
            y = 3sin 5(0)
           y = 0
In the above equation we put x = 2 we get
            y = 3sin 5(2)
           y = -1.6
From equation (1) we get the following value

X	-5	-4	0	2
y	0.4	-2.7	0	-1.6

Graph:

Graph of Sin X Problem 2:

Solve the given trigonometry functions y = 2sin 3x and the draw graph for the given function.
Solution:
In the first step we find the plotting point of the given trigonometry functions. The given function is
                      y = 2sin 3x
We are going to find out the plotting points for a given equation. In the first step we are going to change equation in the form of y = mx + c, we get the following term
                        2sin 3x – y = 0
                              2sin 3x = y                                   
In the above equation we put x = -4 we get
           y = 2sin 3(-4)
           y = 1
In the above equation we put x = -3 we get
            y = 2sin 3(-3)   
           y = -0.82
In the above equation we put x = 0 we get
            y = 2sin 0
           y = 0
In the above equation we put x = 3 we get
            y = 2sin 3(3)
           y = 0.82
From equation (1) we get the following value

x	-4	-3	0	3
y	1	-0.82	0	0.82

Graph:

Multiply Complex Numbers

The complex number consists of imaginary part and real part is defined as the complex number.

The complex number is written the form x + yi where x and y are real numbers and i is defined as the imaginary unit.

The normal numbers in the complex numbers is extended by using extra numbers.

The complex numbers are used in quantum physics, engineering and applied mathematics.

Multiply Complex Number:

Multiplication: (a + bi)(c + di) = (ac - bd) + (bc + ad)i

Examples on Multiply Complex Numbers:

Example 1:

Solve (2 + 2i)(5 + 3i)

Solution:

The multiplication of the complex numbers in the arithmetic form is given as,

(2 + 2i)(5 + 3i) = 10 + 6i + 10i + 6i^2

= 10 + 16i - 6   { Since i^2  = -1 }

= 4 + 16i.

The complex number multiplication answer is 4 + 16i.

Example 2:

Solve

(3 + 7i)(4 + 3i)

Multiplication of the complex number in algebraic form is given as,

(3 + 7i)(4 + 3i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 3:

Solve

(4 + 7i)(3 + 3i)

Multiplication of the complex number in algebraic form is given as,

(4 + 7i)(3 + 3i) = 12 + 21i + 12i + 21i^2

= 12 + 33i - 21

= -9 + 33i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+33i.

Example 4:

Solve

(4 + 3i)(3 + 7i)

Multiplication of the complex number in algebraic form is given as,

(4 + 3i)(3 + 7i) = 12 + 9i + 28i + 21i^2

= 12 + 37i - 21

= -9 + 37i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -9+37i.

Example 5:

Solve

(2 + 5i)(4 + 6i)

Multiplication of the complex number in algebraic form is given as,

(2 + 5i)(4 + 6i) = 8 + 20i + 12i + 30i^2

= 8 + 32i - 30

= -28 + 32i.

These are examples of multiplication of complex numbers.

Hence the complex number multiplications answer is -28+32i.

Practice Algebra Calculus

Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus may be seen as introducing new ways of thinking about algebra problems.

A branch of mathematics, algebra calculus  mainly deals with derivatives, integrals, limits, infinite series, differential equations, and functions, is called calculus. It is used for computing rate of change of velocity and rate of change of acceleration, the slope of a curve, and optimization

Types of Calculus and Problems in Algebra Problems:


Calculus can be classified into two major types as follows

Differential calculus
Integral calculus

Ex 1: Compute the values of Δy and dy if y = f(x) = x^3 + x^2 − 2x + 1  where x changes (i) from 2 to 2.05 and (ii) from 2 to 2.01

Sol :    (i) We have f(2) = 2^3 + 2^2 − 2(2) + 1 = 9

f(2.05) = (2.05)^3 + (2.05)^2 − 2(2.05) + 1 = 9.717625.

And Δy = f(2.05) − f(2) = 0.717625.

In general dy = f ′(x) dx = (3x^2 + 2x − 2)dx

When x = 2, dx = Δx = 0.05 and dy = [(3(2)^2+2(2)−2] 0.05 = 0.7

(ii) f(2.01) = (2.01)^3 − (2.01)^2 − 2(2.01) + 1 = 9.140701

Δy = f(2.01) − f(2) = 0.140701

When dx = Δx = 0.01, dy = [3(2)^2 + 2(2) − 2]0.01 = 0.14

Remark :   The approximation Δy ≈ dy becomes better as Δx becomes smaller . Also dy was easier than to compute Δy. For more complicatedfunctions it may be impossible to compute Δy exactly. In such cases theapproximation by calculus is especially useful.
Example 2 Algebra Calculus Problems:

Ex 2 :  Use differentials to find an approximate value for

`root(3)65` .

Let y = f(x) =`3sqrtx`=x.^1/3 Then dy =1/3x ^-2/3 dx

Since f(64) = 4. We take x = 64 and dx = Δx = 1

This gives dy =1/3 (64) ^−2/3 (1)

=1/3(16)       = 1/48

365      = f(64 + 1) ≈ f(64) + dy = 4 +1/48

≈ 4.021

Practice Problems

Pro 1:    If f(x) = x^4, then find f'(x)

Sol :              f(x) = x^4

f'(x) = 4x^3

Pro 2:    If f(x) = 5x3^ + 2x^2, then find f'(x).

Sol :              f(x) = 5x^3 + 2x^2

f'(x) = 15x^2 + 4x

Pro 3:   x7  dx  =  (x^7+1) / (7 + 1) + c

Sol :                   = x^8/8 + c

Algebra with Negative Numbers

The algebra is a division of mathematics, which can be explained as calculation of arithmetic operations. It is also referred to that character type of abstract algebra pattern. The negative number is indicating the symbol "-". For example the number -6, is the negative number. The elementary algebra is the division of the algebra.  In the subsequent we see in detail about algebra with negative numbers.

Example Problem for Algebra with Negative Number

Example 1:

Given two number -23x and -43x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -23x + (-43x)

= - (23x+43x)

= - (66x)

Answer for this problem is -66x.

Example 2:

Given two number -120x and -56x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -102x - (-56x)

= -102x+56x

= - 46x

Answer for this problem is -46x.


More over Problem for Algebra with Negative Number

Example 3:

Given two number -430x and -123x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -430x + (-123x)

= - (430x+123x)

= - (553x)

Answer for this problem is -553x.

Example 4:

Given two number -1230x and -1405x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -1230x - (-1405x)

= - 1230x + 1405x

= 175x

Answer for this problem is 175x.

Example 5:

Given two number -2345x and -450x, adding these negative number

Solution:

In the problem given two numbers are the negative numbers.

Adding these two number

= -2345x + (-450x)

= - (2345x+450x)

= - (2795x)

Answer for this problem is -2795x.

Example 6:

Given two number -5623x and -2345x, subtracting these negative number

Solution:

In the problem given two numbers are the negative numbers.

Subtracting these two number

= -5623x - (-2345x)

= - 5623x + 2345x

= - 3278x

Answer for this problem is -3278x.

Multiplication Problems

Introduction to multiplication problems:
Multiplication – It is a repeated addition.
For example 2 * 3 = 6 this is nothing but 2+2+2 = 6 or 3+ 3 =6
The order of multiplication problems does not affect the answer.The answer will not be changed even though the performing order changed.
Distribution in multiplication problems 
            To multiply 4(10+5), we multiply 4*10 and 4*5, adding the result. Thus,
                             4(10+5) = (4*10) + (4*5)
                                           =40+20
                                           =60
This is nothing but the distributive property of multiplication.
It will be easy to do multiplication if we are well known with multiplication tables.
When we multiply two numbers, there is no matter which number is first or second but the answer will be same.

More on Multiplication Problems

Example 1:

Multiply 46 * 3
Solution:
First multiply the ones digit with 3 and next tens digit with 3
If we have any remainder carry to next digit and add with it
46 * 3 =138

Example 2:

Multiply 45 * 5
Solution:
First multiply the ones digit with 5 and next tens digit with 5
If we have any remainder carry to next digit and add with it
45 * 5 = 225

Example 3:

Multiply  100 * 39
Solution:
This multiplication problem is very simple
Just multiply 1*39 and add two zero at the end
So 3900 is the answer

Example 4:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 5:

Multiply:   66 1* 23
Solution:
      6 6 1
(*)    2 3
-------------
     1983
     1322
-------------
  1 5 203
------------

Example 6:

Multiply:  2546 * 121
Solution:
                      2 5 4 6
              (*)       1 2 1
          -------------------------
                  2 5 4 6
               5 0 9 2
           2 5 4 6
--------------------------------
          1 0 1 8 4
----------------------------------

Example 7:

Multiply 45 * 9
Solution:
First multiply the ones digit with 9 and next tens digit with 9
If we have any remainder carry to next digit and add with it
45 * 9 = 405

Example 8:

Multiply  100 * 65
Solution:
This multiplication problem is very simple
Just multiply 1*65 and add two zero at the end
So 6500 is the answer.

Example 9:

Multiply 83 * 6
Solution:
First multiply the ones digit with 6 and next tens digit with 6
If we have any remainder carry to next digit and add with it
83 * 6 = 498

Example 10:

Multiply  1000 * 869
Solution:
This multiplication problem is very simple
Just multiply 1*869 and add three zero at the end
So 869000 is the answer.