Question

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Answer 1

Question:

Find 'c' if the system of the equations cx + 3y + (3 - c) = 0 ; and 12x + cy + c = 0 has infinity many solutions.

SOLUTION:

The two equations are given,

cx + 3y + (3 - c) = 0 ........(i)

12x + cy + c = 0  ............(ii)

Now,

a₁ = c ; b₁ = 3 ; c₁ = (3 - c)

a₂ = 12 ; b₂ = c ; c₂ = ( -c)

For infinity may solution we need,

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

By putting the values respectively, we get,

$\implies \dfrac{c}{12}=\dfrac{3}{c}=\dfrac{(3-c)}{(-c)}$

Here,

Taking the first two fractions,

$\implies \frac{c}{12}=\frac{3}{c}$

By cross multiplication we get,

⇒ 12 * (3) = c * c

⇒ 36 = c²

⇒ √36 = c

⇒ 6 = c

Hence,

The value of 'c' should be (6) for which the two equations will have infinity many solutions.

-

Verification,

$\implies \dfrac{c}{12}=\dfrac{3}{c}=\dfrac{(3-c)}{(-c)}$

By putting the value of c,

$\implies \dfrac{6}{12}=\dfrac{3}{6}=\dfrac{(3-6)}{(-6)}$

By simplifying the fraction we get

$\implies \dfrac{1}{2}=\dfrac{1}{2}=\dfrac{(-3)}{(-6)}$

$\implies \dfrac{1}{2}=\dfrac{1}{2}=\dfrac{1}{2}$

Hence verified.

Answer 2

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