Question

If
the side of a square Is tripled, how
many times will its area be as
to the area of the original square?​

Answer 1

Answer:

3x + y =1

(2k-1)x+(k-1)y=2k+1

=> 3x+y-1=0

(2k-1)x+(k-1)y-2k-1 =0

We know,

a1/a2 = b1/b2 not equal to c1/c2 has no

solution

Now,we have,

a1=3 , b1=1

a2=2k-1, b2 = k-1

3/2k-1 = 1/k-1

3(k-1) = 2k -1

3k-3 = 2k -1

3k-2k = -1 +3

k = 2$\huge\red{\mid{\fbox{\tt{αηѕᴡєя}}\mid}}$

Answer 2

Answer:

3x + y =1

(2k-1)x+(k-1)y=2k+1

=> 3x+y-1=0

(2k-1)x+(k-1)y-2k-1 =0

We know,

a1/a2 = b1/b2 not equal to c1/c2 has no

solution

Now,we have,

a1=3 , b1=1

a2=2k-1, b2 = k-1

3/2k-1 = 1/k-1

3(k-1) = 2k -1

3k-3 = 2k -1

3k-2k = -1 +3

k = 2$\huge\red{\mid{\fbox{\tt{αηѕᴡєя}}\mid}}$

Answer 3

Answer:

3x + y =1

(2k-1)x+(k-1)y=2k+1

=> 3x+y-1=0

(2k-1)x+(k-1)y-2k-1 =0

We know,

a1/a2 = b1/b2 not equal to c1/c2 has no

solution

Now,we have,

a1=3 , b1=1

a2=2k-1, b2 = k-1

3/2k-1 = 1/k-1

3(k-1) = 2k -1

3k-3 = 2k -1

3k-2k = -1 +3

k = 2$\huge\red{\mid{\fbox{\tt{αηѕᴡєя}}\mid}}$