Basic Algebra Problems: Mathematics Problem Solving!

QNO. 01: What is the answer to (a-b) x (a-b) = ?

Solution:

To put it simply, we should use the “rainbow expansion method”

(a-b) x (a-b)

Step 1: a x a = a2

Step 2: a x (-b) = -ab

Step 3: -b x a = -ab

Step 4: -b x -b = b2

Thus (a-b) x (a-b) = a2 -ab -ab + b2 = a2 -2ab + b2

Note that I bolded a and b in the first bracket to clearly mark my steps of expansion, but one should note that this clearly doesn’t matter.

Afterwards you should try it out with (a+b)(a-b) and (a+b)(a+b)/(a+b)2 too!

The results are always as follow: (FORMULA MEMORISATION METHOD)

(a+b) x (a+b) = (a+b)2 = a2 + 2ab + b2

(a+b) x (a-b) = a2 – b2

QNO. 02: If x÷4 = 4÷x, what is the value of x?

Solution:

Write the expression in fraction form

 X/4 = 4/X

Cross multiply the expression:

X^2 = 16

Taking square root on both sides:

X = +4 

X = -4

QNO. 03: What is the X value needed to fill the numbers 1, 3, 5, 6, 9, 12, 13, X?

Solution:

This series is related to three different parts:

First part: 1,3,5

Which have the difference of 2

3-1 = 2

5-3 = 2

Second part: 6,9,12

Which have the difference of 3, and 3 comes after 2.

9-6 = 3

12-9 = 3

The third part is: 13, X

It will have a difference of 4,

Given value X will be

X - 13 = 4

X = 4 + 13

X= 17

QNO. 04: How do you solve the system algebraically, 2x + y - 10 = 0 x - y - 4 = 0? What is the value of y, 1/3 2/3 14/3?

Solution:

Give the number to the equations:

2X + Y - 10 = 0--------------------- equ 1

X - Y - 4 = 0 ---------------------- equ 2

By adding both equations, Get the value of X:

3X - 14 = 0

X = 14/3

For Y, Put the value of X in equation 2:

14/3 - Y - 4 = 0

Y = 14/3 - 4

Y = (14 - 12) / 3

Y = 2/3

QNO. 05: What is the center of a circle whose equation is x2 + y2 – 12x – 2y + 12 = 0?

Solution:

Given equation:

X^2 + Y^2 - 12X - 2Y + 12 = 0 ------------------------equ 1

As we know that the equation of a circle is:

X^2 + Y^2 + 2gX + 2fY + C = 0

By comparing given equation to equation of a circle:

-12X = 2gX                 OR     2fY = -2Y

 g = -6                                    Y = -1

As we know that the center of the circle is:

( -g , -f) = ( 6 , 1 )

Algebra and Calculus problem 1.2