Question

find the value of k for which system of equation have infinite many solutions 2x +3y = 7 and 8 x + ( k +4 ) y -28= 0​

Answer 1

Answer:

k=8

Step-by-step explanation:

2x+3y=7.....(1)

8x+(k+4)y=28....(2)

let equation (1)×4 ;

8x+12y=28

8x+(k+4)y=28

-. -. -

_______________

12y-(k+4)y=0

12y=(k+4)y

then k=8

Answer 2

Answer:

8 is the required value of  k .

Step-by-step explanation:

Explanation:

Given , two equations which are 2x + 3y = 7 and 8 x+ (k+ 4)y = 28

Let 2x + 3y - 7  = 0 .........(i) and

Let 8x + (k+ 4)y - 28 = 0   ........(ii)

As we know that , if two line $a_1x+ b_1y + c_1 = 0 and \ a_2x + b_2y +c_2 = 0$ having infinite many solution than ,

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Step 1:

Therefore , from the question we have ,

$a_1 = 2 , a_2 = 8 , b_1 = 3 ,b_2= (k + 4) , c_1 = -7\ and \ c_2= -28$

Now ,  according to the condition  given in the question ,  equation have infinite solutions

⇒ $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

⇒$\frac{2}{8} =\frac{3}{(k+4)} =\frac{-7}{-28}$

∴$\frac{2}{8} = \frac{3}{(k+4)}$ and $\frac{3}{(k +4)} = \frac{-7}{-28}$

On Applying cross multiplication method we get ,

2(k+ 4) =  3 × 8      and       3 (-28) = -7 (k+ 4)

⇒2k + 8 = 24         and       84 = 7k + 28         [minus sign are cancel out ]

⇒2k  = 16               and 7k =  56

⇒ k = $\frac{16}{2}$ = 8            and  k = $\frac{56}{7}$ = 8

Final answer:

Hence ,  the value of k is 8  .

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