Question

The pair of linear equations 2x+7y=k,kx+21y=18 has infinitely many solutions if

Answer 1

2x+7y=k
kx+21y=18

infinitely many soln. condition
a1/a2=b1/b2=c1/c2

2/k=7/21=k/18
Now, 2*21=k*7
         42=7k
         k=6←

Now, 7*18=21*k
         7*18/21=k
         6=k←
From these two +6 is common.
so k=6

Answer 2

Answer:

The value of k = 6.

Step-by-step explanation:

Given:

Let the two pairs of linear equations be

2x + 7y = k and

kx + 21y = 18

To find: the infinite solutions.

Step 1

For infinitely many solutions:

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Simplifying the pair of linear equations in the above equation, we get

$\frac{a_{1}}{a_{2}}=\frac{2}{k},$  $\frac{b_{1}}{b_{2}}=\frac{7}{21}=\frac{1}{3},$ $\frac{e_{1}}{c_{2}}=\frac{k}{18}$

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

then, k = 6

Step 2

$\\\Rightarrow \frac{k}{18}=\frac{1}{3}$

3k = 18

then, k = 6

From these two equations we get k = 6

Therefore, the value of k = 6.

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