Question

By comparing the ratios a1/a2, b1/b2 and c1/c2, find out for what value (s) of α, the lines
representing the following equations have a unique solution, no solution or infinitely many
solution:
αx + 3y = α - 3
12x + αy = α

Answer 1

Answer:

α = 6,  the lines representing the following equations have a many solutions

Step-by-step explanation:

Two lines dx+ey+g=0  and  mx+ny+p=0  

have no solution if  $\frac{d}{m}$ = $\frac{e}{n}$ ≠ $\frac{g}{p}$  

infinite solution if $\frac{d}{m}$ = $\frac{e}{n}$ = $\frac{g}{p}$

Unique solution if $\frac{d}{m}$ ≠ $\frac{e}{n}$

Now,

Given

ax+3y−a+3 = 012

x+ay−a = 0

For Unique solution

$\frac{d}{m}$ ≠ $\frac{e}{n}$

$\frac{a}{12}$≠ $\frac{3}{a}$

$a^{2}$ = 36

a = ±6

Therefore for all values of a expect 6 and −6 the system has unique solution

For no solution

$\frac{d}{m}$ = $\frac{e}{n}$ ≠ $\frac{g}{p}$

$\frac{a}{12}$ =  $\frac{3}{a}$ ≠ $\frac{-(a-3)}{-a}$

$\frac{3}{a}$ ≠ $\frac{-(a-3)}{-a}$

$\frac{3}{a}$ ≠ $\frac{(a-3)}{a}$

a ≠ 6

Now

$\frac{a}{12}$ ≠ $\frac{3}{a}$

$a^{2}$ = 36

a = ±6

a = −6        (as we have already found that a≠6)

Therefore for a = −6  the system has no solution

Many solution

$\frac{d}{m}$ = $\frac{e}{n}$ = $\frac{g}{p}$

$\frac{a}{12}$ = $\frac{3}{a}$ = $\frac{-(a-3)}{-a}$

$\frac{a}{12}$ = $\frac{3}{a}$

$a^{2}$ = 36

a = ±6

now

$\frac{3}{a}$ = $\frac{-(a-3)}{-a}$

3 = a−3

a = 6

Therefore for a = 6 the system has many solution.

#SPJ3

Answer 2

Step-by-step explanation:

The given linear equation are

⇒5x−4y+8=0....eq1

⇒a

1

=5,b

1

=−4,c

1

=8

⇒7x+6y−9=0...eq2

⇒a

2

=7,b

2

=6,c

2

=−9

⇒

a

2

a

1

=

7

5

⇒

b

2

b

1

=

6

−4

⇒

c

2

c

1

=

9

8

comparing

⇒

a

2

a

1

,

b

2

b

1

,

c

2

c

1

⇒

a

2

a

1



=

b

2

b

1

Hence,the line represented by eq1 and eq2 intersect at a point