SIMULTANEOUS EQUATIONS 1

Solve the following equations by substitution method.

x + y = 13

x + 3y = 27

Solution

x + y = 13 ----------- (i)

x + 3y = 27 ---------- (ii)

from equation (i)

x + y = 13

x = 13 - y ----------(iii)

substitute equation (iii) in (ii) above,

x + 3y = 27

(13 - y)  + 3y = 27

13 + 2y = 27            [since –y+3y=2y]

2y = 27 - 13

2y = 14

2      2

y = 7

From equation (iii)

x = 13 - y ----------(iii)

x = 13 - 7

x = 6

Hence x=6 and y=7.

TRY THIS………………………..

Solve the following equations by substitution method.

x + y = 10

2x + y = 13

MID POINT 1

Find the midpoint of a line from (11, 4) to (5, 10)

Solution

x1=11, x2=5, y1=4, y2=10.

Mid point = (x1 + x2, y1 + y2)

                         2             2

Mid point = (11 + 5,  4 + 10)

                        2          2

Mid point = (16, 14)

                     2    2

Hence midpoint = (8, 7)

TRY THIS……………………..

Find the midpoint of a line from (12, 8) to (2, 16)

LOGARITHMS 2

If Log_ay = Log_a6 +  Log_a3 ; find y.

solution

Log_ay = Log_a6 +  Log_a3

Log_ay = Log_a(6 x 3)         [since Log_am +  Log_an = Log_a(m x n)]  

Log_ay = Log_a18

Hence y = 18                  [since Log_a cancels out]

TRY THIS..............................

If Log_ax = Log_a5 +  Log_a60 ; find x.

DIFFERENCE OF 2 SQUARES 1

factorize completely   (x-2)² - x²  

solution

= (x-2)² - x²  

= [(x-2) - x][ (x-2) + x ]  --- since a² + b² = (a+b)(a-b) Here a is (x-2) and b is x.

= [x-2 - x][ x-2 + x ] 

= [x-x-2][ x+ x - 2  ] 

= [0-2][ 2x - 2  ] 

= [-2][ 2x - 2  ] 

Hence  (x-2)² - x²  = (-2)( 2x - 2) 

TRY THIS....................

factorize completely   (x-3)² - x²