Question

Find the values of k for which the pair of equations kx +2y =3, 3x+6y= 10 have no
solution

Answer 1

Answer:

for no solution the condition is

$\large{a1/a2=b1/b2≠c1/c2}$

given that

$\large{</strong><strong>kx</strong><strong>+</strong><strong>2</strong><strong>y</strong><strong>=</strong><strong>3</strong><strong>}$

$\large{</strong><strong>3</strong><strong>x</strong><strong>+</strong><strong>6</strong><strong>y</strong><strong>=</strong><strong>1</strong><strong>0</strong><strong>}$

we have

$\large{</strong><strong>k</strong><strong>/</strong><strong>3</strong><strong>=</strong><strong>2</strong><strong>/</strong><strong>6</strong><strong>≠</strong><strong>3</strong><strong>/</strong><strong>1</strong><strong>0</strong><strong>}$

$\large{</strong><strong>k</strong><strong>/</strong><strong>3</strong><strong>=</strong><strong>2</strong><strong>/</strong><strong>6</strong><strong>}$

$\large{</strong><strong>6</strong><strong>=</strong><strong>6</strong><strong>k</strong><strong>}$

$\large{</strong><strong>k</strong><strong>=</strong><strong>1</strong><strong>}$

Answer 2

Answer:

k=1

Step-by-step explanation:

kx+2y=3

3x+6y=10

They Have No Solution.

∴$\frac{k}{3}$=$\frac{2}{6} \neq \frac{3}{10}$

$\frac{k}{3}$=$\frac{2}{6}$

6k=6

∴k=1

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