Question

What is the value of k if the system of equations 2x + 3y = 5 & 4x +ky =10 has infinitely many solutions?




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Answer 1

GiveN :

• Two equations are : 2x + 3y = 5 and 4x + ky = 10

To FinD :

• Value of k

SolutioN :

We've formula for infinite solutions :

$\implies \boxed{\sf{\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}}}$

Where,

• $\sf{a_1\ =\ 2 \: \: \: \: and \: \: \: \: a_2\ =\ 4}$
• $\sf{b_1 \: = \: 3 \: \: and \: b_2 \: = \: k}$
• $\sf{c_1 \: = \: 5 \: and \: c_2 \: = \: 10}$

$\implies \sf{\dfrac{4}{2} \: = \: \dfrac{3}{k}}$

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$\implies \sf{2 = \dfrac{3}{k}}$

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$\implies \sf{3 = 2k}$

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$\implies \sf{k = \dfrac{3}{2}}$

Answer 2

GiveN :-

We have two equations which system has infinitely many solutions →

• $2x + 3y = 5$
• $4x + ky = 10$

To FinD :-

• What is the value of k.

Solution :-

$i) \: 2x + 3y = 5 \\ \\ \implies \: 2x + 3y - 5 = 0$

$ii) \: 4x + ky = 10 \\ \\ \implies4x + ky - 10 = 0$

Now, comparing both equations with ax + by + c = 0

$a _1 = 2 \: \: \: \: a_2 = 4 \\ \\ b_1 = 3 \: \: \: \: b_2 = k \\ \\ c_1 = - 5 \: \: \: \: c_2 = - 10$

We know the formula of infinity system

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

So,

$\implies \: \frac{2}{4} = \frac{3}{k} = \frac{ - 5}{ - 10} \\ \\ \implies \frac{1}{2} = \frac{3}{k} = \frac{1}{2}$

So,

$\implies \frac{3}{k} = \frac{1}{2} \\ \\ \implies \: k = 3 \times 2 \\ \\ \implies \: \red{\boxed{k = 6}}$