Elimination Using Addition and Subtraction

In systems of equations where the coefficients of terms contain the same variable are opposites, the elimination method can be applied by adding the equations. If the coefficients of those terms are the similar, then the elimination method can be done by subtracting the equations. Let us some example problems for elimination using addition and subtraction.

Example Problems of Elimination Using Addition and Subtraction:

Example 1:

Solve: a-b=3 and 3a+b=1

Solution:

Step 1:  Here we are going to solve these equations.

Step 2: We need to add the two equations. Here the coefficients of the b terms,(-b+b)are opposite. So we can easily cancel the terms.

Step 3: When we simplify we get 4a=4.therefore a= 1.

Now we need to calculate the value of b.

Step 6: so, we plug in a value in any equation to find b.

Step 7: here the second equation is 3a+b=1.

Step 8: So,3(1)+b=1.Therefore b = -2

How to check the solution:

We know that a and b value now. So substitute a and b value in any equation. Let us take the first equation.

Step 1: a-b=3.The value of a is 1, and the value of b is -2.

Step 2: Therefore, 1- (-2)=3 .

Step 3: When we add both the sides, we get 3=3.

Suppose if we get a different answer on both the side, the answer must be wrong.



Example 2:

Solve: u+v=3 and –u+2v=6.

Solution:

Step 1:  Here we are going to solve these equations.

Step 2: We need to add the two equations. Here the coefficients of the b terms,(-u+u)are opposite. So we can easily cancel the terms.

Step 3: When we simplify we get 3v=9.therefore v= 3.

Now we need to calculate the value of u.

Step 6: so, we plug in v value in any equation to find u.

Step 7: here the first equation is u+v=3.

Step 8: So,u + 3 = 3.Therefore u = 0.



How to check the solution:

We know that a and b value now. So substitute a and b value in any equation. Let us take the first equation.

Step 1: u+v=3 The value of u is 0, and the value of v is 3.

Step 2: Therefore,0+3=3. 3 = 3

Step 3: When we add both the sides, we get 3=3.

Suppose if we get a different answer on both the side, the answer must be wrong.

These are the example problems of elimination using addition and subtraction.

Practise Problems of Elimination Using Addition and Subtraction:

1) X+y = 7 and x-y =9

2) 2s-r =12 and s+r = -27



Answer key :

1) X = 8 and y = -1

2) S = -22 and r = -5.

Number Divisible by 4

A number divisible by 4 means nothing but a Division operation. Each and every number should divided by 4,that is called number divisible by 4.Four is a even number, When the number is even we can get the even integer numbers ,When we divided the odd number mean We cannot get the real number ,Only got the fraction numbers. Division is a one of the arithmetic operation. Arithmetic operations 1) Addition  2) Subtraction 3) Multiplication 4) Division

Step by Step Number Divisible by 4:


Form of Manual division method,

a / b = c ,where

Here number divide by 4 so we can use the constant of divisor 4

a = dividend.

b = 4 is called as divisor.

c = quotient.

Example:

12 / 4 = 3

Example problem 1: 32 divide by 4

32 divide by 4

Solution:

From the problem

32 is a dividend

4 is a divisor

In numerically it can written as 32/4

Step 1:

First we find the how many 4’s are available in the dividend

Step 2:

After find the  no of multiples in the dividend, We got the answer

32/4=8

Four is the quotient of 8

Remaining should be zero

Using Algebra Division Number Divisible by 4

Example 2: Using algebra division Number divisible by 4

Algebra division example using polynomial:

4x2+4x+4 / 4

4x2+4x+4 is dividend

4 is divisor

x2+x+1   is quotient of (4x2+4x+4) / 4

Example algebra division problem:

Division of: (42+4x+4)/4

Step 1:

(4X2+4x+4) is dividend

4 is the divisor

First we can arrange the terms

Like x2+x3+x mean we can change x3+x2+x

Step 2:

Now  we can  divide the first term of the dividend

by the1st  term of the divisor, it mean 4x2/4=x2 .It gives the first terms of quotient.

Step 3:

Now we got the first term of  quotient and then subtract the Multiplication of first terms quotient and dividend 4x2-4=x2

Step 4:

Again we can  divide the second term term of the dividend   by the first term of the divisor 4, it mean   4x/4 =1 .It should provide the second terms  after then multiply the first term with quotient after  then subtract

Step 5:

Same procedure for the constant term Now we got the final answer  x2+x+1

Quadratic Equations Activity

An equation with one variable, in which  the highest power of the variable is two is called a quadratic equation.

For example, ( i ) 3x2 + 5 x - 8 = 0

(ii) 2y2 - 48 = 0

(iii) 6x2 + 5x = 0

(iv ) y2 = 4 are all quadratic equations. Let us do some activity or problems using quadratic equations.

Activities of Quadratic Equations:

Activity 1 :

Solve the quadratic equation 2x2 - 7x = 39

2x2 - 7x - 39 - 0

`=>`              2x2  - 13x + 6x - 39 = 0  ( factorising the left hand side )

`=>`              x ( 2x - 13 ) + 3 (2 x - 13 ) = 0

`=>`             ( 2x - 13 ) ( x + 3 ) = 0

2x - 13 = 0 or x + 3 = 0

`rArr`              x  =  `13/2`   or  x  =  -3

Hence the quadratic equation is solved by factorisation method.

Activity 2 :

Find the quadratic equation whose solution set is { -2, 3 }

Since solution set is { -2, 3}

we have x = -2 or x = 3

x + 2 = 0 or x - 3 = 0

`rArr`           ( x + 2 ) ( x - 3 ) = 0

`rArr`           x2 - 3x + 2x - 6 = 0

`rArr`           x2 - x - 6 = 0 is the required quadratic equation.

Activity of Quadratic Equations(continued):

Activity 3 :

Solve the quadratic equation 5x2 - 2x - 3 = 0 using the formula.

The roots of the standard quadratic equation ax2 + bx + c = 0 where a`!=` 0, are given by the formula

x   =   `( -b stackrel(+)(-) sqrt ( b^2 - 4ac )) / ( 2a)`

Comparing 5x2 - 2x - 3 = 0 with ax2 + bx + c = 0 we get a = 5, b = -2 and c = -3.

so, x = `(2 stackrel(+)(-) sqrt((-2)^2 - 4. 5. (-3))/(2.5))`

=  `(2 stackrel( +)(-) sqrt ( 64)) / ( 10)`

= `(2 stackrel(+)(-) 8)/10`

= `(2-8)/10` = 1 and `-3/5`

Hence 1 and `(-3)/5` are the roots of the given quadratic equation.

Activity 4 :

Solve the equation 2x4 - 5x2 + 3 + 0 which is reducible to quadratic equation.

Let x2 = y

Then, 2x4 - 5x2 + 3 = 0   `rArr`   2y2 - 5y + 3 + 0

`rArr`   ( y - 1 ) ( 2y - 3 ) = 0

`rArr`   y = 1 or   y   = `3/2`

When y = 1, x2 = 1 `rArr` x = 1 or -1

When y =  `3/2`   x2 = `3/2` `rArr` x = `sqrt(3/2)`   or `-sqrt(3/2)`

Hence the fourth degree equation is solved using the quadratic equation technique.

Slope Ratio Calculator

The slope is defined as ratio of change of x axis to change of y axis. The slope intercept form is y = mx + b. Where m is slope and b is y intercept.

Slope formula is (m) =`"vertical" /"horizontal"`.

The slope form (m) is = `(y_2 - y_1)/(x_2 - x_1)` = `"rise"/"run"` .

We will learn about the slope ratio calculator example problems and practice problems are given below.

Example Problems for Slope Ratio Calculator:

Slope ratio calculator:





Example problem 1:

Find the slope ratio of a line, which contains two points A (0, 1), B (10, 2).

Solution:



The slope of a line which contains two points (x1, y1) and (x2, y2) is given by,

Here, x1 = 0, x2 = 10, y1 = 1, y2 = 2.

Slope of the line, m = `(y_2 - y_1)/(x_2 - x_1)`

= `(2 - 1)/(10 - 0)`

After simplify this, we get

= `1/(10)`

Slope of the line (m) = `1/(10)`

So, the slope ratio of a line, which contains two points A (0, 1), B (10, 2) is `1/(10)` = 0.1



Example problem 2:

Find the slope ratio of a line, which contains the two points A (-13, 10), B (2, -20).



Solution:

The slope of a line which contains two points (x1, y1) and (x2, y2) is given by,

Here, x1 = -13, x2 = 2, y1 = 10, y2 = -20.

Slope of the line, m = ` (y_2 - y_1)/(x_2 - x_1)`

=  `(-20 - 10)/(2 + 13)`

After simplify this, we get

= ` (-30)/(15)`

Slope of the line (m) = -2

So, the slope ratio of a line, which contains the two points A (-13, 10), B (2, -20) is -2

These examples problem are helpful to study of slope ratio calculator.

Practice Problems for Slope Ratio Calculator:

Practice problem 1:

Find the slope of a line, which contains two points A (9, 0), B (0, 3).

Answer: Slope (m) = -0.3333

Practice problem 2:

Find the slope of a line, which contains two points A (10, 200), B (300, 20).

Answer: Slope (m) = -0.620

Algebra Probability Help

Mathematical numbers are studied by algebra. Algebra is also used to learn the polynomials and the equations etc. Probability is a method of state knowledge or principle that an occurrence will happen. In mathematics the idea has been given a correct sense in probability theory, that is used widely in such areas of learn as mathematics, finance, statistics etc. Here we will see the examples and solved with the help of algebra probability.

Example Problems for Algebra

1) What is the multiplication of following two numbers with the help of algebra? 115*118

Solution

115*118 = (100+15)*(100+18)

= (100)2+(100*18)+(15*100)+(15*18)

=10000+1800+1500+270

=13570.

2) 18x+12y+12x+8a. Simplify the given equation in algebra.

Solution

The given equation is 18x+12y+12x+8a

There are two related groups are available. So connect the groups.

The new equation is,

(18x+12x)+12y+8a

Add the numbers inside the bracket. We get 30x+12y+8a.

Assemble the numbers and we get the correct format.

=8a+30x+12y.

Example for Probability

There are 45 things are available in a shop. In those things, 16 are the books, 12 are the bags and 17 are the caps. What is the probability for the following outcomes?

i) Select the books

ii) Select the bags

iii) Select the caps.

Solution:

Total number of things n(S) =45

Number of books n (A) =16

Number of bags n (B) =12

Number of caps n(C) =17

i) Assume P(A) is the probability for select the books.

P(A)=`(n(A))/(n(S))`

=`(16)/(45)` .

ii) Assume P(B) is the probability for select the bags.

P(B) =`(n(B))/(n(S))`

= `(12)/(45)`

=`(4)/(15)` .

iii) Assume P(C) is the probability for select the caps.

P(C)=`(n(C))/(n(S))`

=`(17)/(45)` .

Practice Problems

Practice problem 1

What is the product of the following numbers in algebra ? 111*128

Answer:

101*108=14208.

Practice problem 2

8x+14y+12a+12y. Simplify the terms with the help of algebra.

Answer:

12a+8x+26y.

Practice problem 3

Jenifer has the 6 papers, Stephen has 7 papers. What is the probability for select the Jenifer’s papers.

Answer:

Probability for select the Jenifer’s papers=6/13.

These algebra and probability problems are helping to study these concepts

Trigonometry Sine Function

In trigonometry one of the ratios is the sine function. It is defined by sinx = (Opposite side/Hypotenuse) = (Perpendicular/Hypotenuse)

This trigonometry sine function has applications on solving some practical problems in finding the height of the wall, length of a ladder leaning on the wall and the distance between the wall and the foot of the ladder.

Let us learn some sine values for standard angles.





The above table will help us in solving problems involving sine functions.  Now let us solve few problems on the topic trigonometry sine function.

Example Problems on Trigonometry Sine Function

Ex 1: From the below diagram, find the value of x using sine function.

Sol: Sin30 = `x/12`

This is implies, x = 12 `xx` sin30

                            = 12 `xx` `(1/2)` [table value for sin30 = `1/2` ]

                            = 6 cm.

Therefore, the value of x = 6cm.

Ex 2: From the below diagram, find the value of x using sine function.

Sol: Sin30 = `3/x`

               x = `3/sin30`

                  = `3/(1/2)`

                  = 3 `xx` 2 = 6cm.

Therefore, the value of x = 6cm.

Ex 3: If 2 sinA = 1, what is the value of A?

Sol: Given: 2sinA = 1

                     SinA = `1/2`

This implies that the value of A = `30^0` .     [Table value Sin30 = `1/2` ]

Ex 4: Simplify: sin60 + sin30.

Sol: sin60 + sin30 = `sqrt(3)/2` + `(1/2)`

                              = `(((sqrt(3)) + 1)/2)` .            [Table value Sin60 = `sqrt(3)/2` ]

Ex 5: In the triangle, find the value of x^0.

Sol: We know that Sin`x^0` = `sqrt (3)/2`

This implies `x^0` = `60^0` [Table value, sin60 = `sqrt (3)/2` ]

Ex 6: If 4sin²x – 3 = 0, x is an acute angle, find (i) sinx (ii) x.

Sol: Given: 4sin²x – 3 = 0

                     Sin²x = `(3/4)`

                     Sinx = `+-` `sqrt(3)/2`

(ii) Since, sinx = `+-` `sqrt(3)/2` ,

As per the table value, x =  `60^0` .

Therefore, x = `+-` `60^0` .

Ex 7: Solve for x:

Sin(x + 10) = ½

Sol: Since Sin(x + 10) = `(1/2)`

                        X + 10 = `30^0`

Therefore, x = 30 – 10

                     = `20^0` .

Therefore, the value of x = `20^0` .

Practice Problems on Trigonometry Sine Function

1. Solve for x: Sin² x + sin²30 = 1.

[Answer: x = `+-` `60^0` ]

2. Find the acute angle A and B, if Sin (A+B) = 1 and Sin(A – B) = `(1/2)`

[Answer: A = `60^0` , B =` 30^0` ].

Areas of Combinations of Plane Figures

Areas of Combinations of plane figure:

Areas of Combinations of plane figure is the process of calculating the areas of different combinations of figures. these types of figures  We come across in our daily life and also in the form of various interesting designs.

Flower beds, drain covers, window designs, We come across, designs   on th etable covers, are some of such examples.We illustrate the  process of calculating  areas of these  figures through some examples.

The following   examples are combined with some plane figures.

Areas of Combinations of Plane Figure Problems:

Example:

Two circular flower beds have been shown on two sides of  a square lawn ABCD  of side  50m.If the center of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.





Solution:

Area of  the square lawn ABCD = 50 x 50 m2  --------------------- (1)

Let   OA = OB = x metres

So            x2 + x2 = 502

Or             2x2 = 50 x  50

X2  = 25 x 50                                        -----------------------(2)

Now ,

Area of sector OAB  = `(90)/(360)` *`Pi` *x2

=`(1)/(4)` * `Pi` i*x2

= `(1)/(4)`  x `(22)/(7)` x 25 x 50m2     [from (2)] ------(3)

Also, area of  `Delta` OAD = `(1)/(4)` * 50 * 50 m2  (<AOB=90)------(4)

So, area of flower bed                 AB = (`(1)/(4)`* `(22)/(7)`*25*50 – `(1)/(4)`*50*50)m2   [from (3) and (4)]


= `(1)/(4)`*25*50(`(22)/(7)` -2)m2

=`(1)/(4)`*25*50* `(8)/(7)` m2   ----------------(5)

Similarly area of the other flower bed

= `(1)/(4)` * 25 * 50 * `(8)/(7)` m2    -------------------------(6)

Therefore,

Total area  =(50*50 + `(1)/(4)` *25*50* `(8)/(7)` +`(1)/(4)`*25*50*`(8)/(7)`)m2    [from (1),(5) and (6)]

=25*50(2+`(2)/(7)`+`(2)/(7)`)m2

=25*50* `(18)/(7)` m2

=  3214. 29 m2

Areas of Combinations of Plane Figure Example 2:

Example 2:

Find the area of the shaded region in the following figure , where ABCD is a square of a side  10 cm.





Solution:

Area of square ABCD

= 10 * 10 cm2

= 100 cm2

Diameter of each circle    = `(10)/(2)`cm=5cm

So, radius of each circle      =(5)/(2) cm

<br>

=`(22)/(7)` *`(5)/(2)` * `(5)/(2)`cm2

=`(550)/(28)` cm2

Therefore area of the four circles = 4* `(550)/(28)`cm2 =  78.57Cm2

Hence area of the shaded region = (100-78.57)cm2= 21.43Cm2