Maths Teachers in Bhola Nath nagar, Delhi

Solving in the Substution Method we get:

5x+2y+13=0

∴5x+2y=-13

and 7x-5y+26=0

∴7x-5y=-26 Suppose: Consider; 5x+2y=-13→(1) and 7x-5y=-26→(2) Taking equation (1), we have 5x+2y=-13 ⇒5x=-2y-13 ⇒x=-2y-135→(3) Putting x=-2y-135 in equation (2), we get  7x-5y=-26 7(-2y-135)-5y=-26 ⇒-14y-91-25y=-130 ⇒-39y-91=-130 ⇒-39y=-130+91 ⇒-39y=-39 ⇒y=1→(4) Now, Putting y=1 in equation (3), we get x=-2y-13 x=-2(1)-135 ⇒x=-2-135 ⇒x=-155 ⇒x=-3 ∴x=-3 and y=1.

Solving in the Elimination Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y=-13→(1) 7x-5y=-26→(2) equation(1)×5⇒25x+10y=-65 equation(2)×2⇒14x-10y=-52  Adding ⇒39x=-117 ⇒x=-11739 ⇒x=-3 Putting x=-3 in equation (1), we have 5(-3)+2y=-13 ⇒2y=-13+15 ⇒2y=2 ⇒y=1 ∴x=-3 and y=1.

Solving in the Cross Multiplication Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y+13=0→(1) 7x-5y+26=0→(2) Here, a1=5,b1=2,c1=13 a2=7,b2=-5,c2=26  x=(b1⋅c2-b2⋅c1)/(a1⋅b2-a2⋅b1) =(2)(26)-(-5)(13)(5)(-5)-(7)(2) =(52)-(-65)(-25)-(14) =117-39 x=-3 y=(c1⋅a2-c2⋅a1)/(a1⋅b2-a2⋅b1) =(13)(7)-(26)(5)(5)(-5)-(7)(2) =(91)-(130)(-25)-(14) =-39-39 y=1 ∴x=-3 and y=1.

10 •••••••••• + 2 •• 12

It has two answers either add x-2 or x²+3x-5

We put value of given polynomial = 0 and value of x = -4. After solving equation we'll get value of k=9

X=1, y=2

K=1

3(a+b)+3(b+c)+3(c+a) = 3a+3b+3b+3c+3c+3a =6a+6b+6c =6(a+b+c)

Since, one root is negative of the other, here the sum of those two roots = 0

As per the given polynomial, the sum of the two roots = (-b)/a = [6(1-k)]/2 = 0 . 

Or, 3.(1 - k) = 0 .   => 1 - k = 0 .

Hence, k = 1   .   (Ans.)

Given, unpainted area of the wall= 60%. Then, wall area painted= 100%-60%= 40%. Then, area painted by Anil and Umar combined= 40%. Again, Anil alone has painted 25% of the wall area (given). Then, area painted by Umar alone= 40%-25%=15%. According to the question, Umar has painted 30 sq.m. i.e. 15% of wall area=30 sq.m. => 30/100 of area= 20 sq.m. So, total area of the wall= 100%= (30×100)/15= 200 sq.m. Thus, area of the wall painted by Anil alone= 25% of the wall area= (25×200)/100= 50 sq.m. answer.

Here, a = 2, b = -6(k - 1) 

ATP, sum of the zeros = (-b)/a = 0    [since, one zero is -ve of the other]

=> 3(k - 1) = 0       => k - 1 = 0 

Thus, k = 1     ... Ans.