Question

for what value of k does the pair of linear equations 3(k-1)x+4y= 24 and 15x+20y= 8(k+3) have infinitely many solutions​

Answer 1

Answer:

Step-by-step explanation: using formula

A1/A2=B1/B2=C1/C2

3(k-1)/15=4/20

3k-3/15=4/20

3k-3=3

3k=6

k=2

Answer 2

Answer: k = 2 or 12

Step-by-step explanation:

Given :  3(k - 1)x + 4y = 24

             15x + 20y = 8(k + 3)

To find : value of k for infinite solutions

Solution :

⇒ 3(k - 1)x + 4y = 24   ...(i)

⇒ 15x + 20y = 8(k + 3)   ...(ii)

We will have infinite number of solutions when :

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

⇒ $\frac{3(k - 1)}{15} = \frac{4}{20} = \frac{24}{8(k + 3)}$

⇒ $\frac{k- 1}{5} = \frac{1}{5} = \frac{3}{k + 3}$

⇒ k = 2 or k = 12

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