Question

Q. Find the value of k for which the given system of equations has infinite many solutions: kx+3y=2k+1 2(k+1)x+9y=7k+1

Answer 1

Answer: k = 2

Step-by-step explanation:

In these equations ,

a = k , b = 3 , c = (2k+1)

A = 2(k+1) , B = 9 , C = (7k+1)

These equations has infinity solutions . so ,

a/A = b/B = c/C

k/2k+2 = 3/9 = 2k+1/7k+1

(1) k/2k+2 =3/9

9k = 6k +6

3k = 6

k = 2

So , the value of k is 2

Answer 2

Value of k is 2.

Given:

• Pair of linear equations.
• $kx + 3y = 2k + 1$ and
• $2(k + 1)x + 9y = 7k + 1$

To find:

• Find the value of k, if equations has infinite many solutions.

Solution:

Formula/Concept to be used:

If $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ are the standard linear equations, then these have infinite many solutions if $\bf \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Step 1:

Find the coefficients of both equations.

On comparison with standard equation, it is clear that

$a_1 = k$

$b_1 = 3$

$c_1 = - (2k + 1)$

$a_2 = 2(k + 1)$

$b_2 = 9$

$c_2 = - (7k + 1)$

Step 2:

Put the values of coefficients in the condition.

$\frac{k}{2k + 2} = \frac{3}{9} = \frac{ - (2k + 1)}{ - (7k + 1)}$

or

$\frac{k}{2k + 2} = \frac{1}{3} = \frac{ 2k + 1}{ 7k + 1}$

Take first two fractions:

$\frac{k}{2k + 2} = \frac{1}{3}$

or

$3k = 2k + 2$

or

$\bf k = 2$

Take last two fractions:

$\frac{1}{3} = \frac{ 2k + 1}{ 7k + 1}$

or

$7k + 1 = 6k + 3$

or

$7k - 6k = 3 - 1$

or

$\bf k = 2$

Thus,

Value of k is 2.

Learn more:

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