A word problem on algebra linear equations

Equations play a crucial role in modern mathematics and form the basis for mathematical modeling of numerous phenomena and processes in science and engineering.

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Let's see a word problem from 8th grade word problems worksheets.

Question:-

Jack is thinking two positive numbers.one of them is 4 more than the other and the sum of their squares is 208.what are the two numbers?

Answer:-

Let one number be x.

the other number is 4 more ,x+4

given sum of the squares =208

x²+(x+4)²=208

x²+x²+8x+16=208

expanding the 2nd term

2x²+8x+16=208

dividing by 2 as it is common for all terms

x²+4x+8=104

subtract 104 on both sides

x²+4x-96=0

x²+12x-8x-96=0

x(x+12)-8(x+12)=0

(x+12)(x-8)=0

x=-12 or x=8

here we want only positive number ,so x=8

and the numbers are 8 and 12

Factors of a Number:

Topic:-Factors of a Number:

Factoring is like taking a number apart. It means to express a number as the product of its factors. Factors are either composite numbers or prime numbers (except that 0 and 1 are neither prime nor composite).
This algebra help gives us few examples as well.

•A factor of a number N is a number which divides N exactly.

•Example: the factors of 12 are 1,2,3,4,6,and 12

•Note that every number has itself and 1 as its factors.

•When a number is greater than 1 and has just itself and 1 as factors, the number is prime.
Multiples of a Number:

•A number which is the product of a given number and another factor.

•Example: 12 is a multiple of each of the numbers 1,2,3,4, and 6
Example:
5 x 6 = 30
The numbers that are multiplied 5 and 6 are called factors.
The product formed by multiplying the factors that is 30 is called multiple.
30 is a multiple of 6 and 5
Example: Which number is a factor of 16?
_ 8
_ 5
_ 9
Answer: 8

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problem on time and speed

Topic :- Time and Speed


Time problems and speed problems are there in both math and Physics .Here is a simple math problem which help you to understand to how to find time when speed value is given . This problem solved
by using distance formula ,So first we find the distance based on which we get the time .

Question:-
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
Answer:-
Let the time taken by the second cyclist to catch up the first be 'x' hours.

Then the time taken by the first to reach than the point wil be (x+3) hours.

Speed of the second cyclist=10 miles/hour.

So the distance covered =speed*time
= 10 * x = 10x miles

Speed of the first cyclist = 6 miles/hour

So distance covered = 6(x+3)
=(6x+18) miles

Now distnce covered by both are same .

So

10x = 6x+18
-6x   -6x
------------------
4x  =  18

4x     18
---- = ----
4      4

   18     9        1
x= ---- = ---- = 4----
   4     2         2

So the time taken by the second
cyclist = 4 hours 30 minutes

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