Question

find the values of k for which the system of equations x+2y-3=0 and ky+5x+7=0 has a unique solution.

Please do solve this according to given equations with explanation.​

Answer 1

Step-by-step explanation:

a1=1 b1 =2 c1= -3

a2 =5 b2 = k c2 = 7

a1/a2 =1/5

b1/b2 = 2/k

1/5× 2/k

1×k= 5×2

k= 10

k =10 ans

Answer 2

➣ Answer

• K can have any value but it should not be equal to 10.

➣ Given

• The pair of linear equation in two variables i.e. x + 2y - 3 = 0 and ky + 5x + 7 = 0 has a unique solution.

➣ To Find

• The value of k.

➣ Step By Step Explanation

Given that the pair of linear equation in two variables i.e. x + 2y - 3 = 0 and ky + 5x + 7 = 0 has a unique solution.

We need to find the value of k.

So let's do it !!

➵ Formula Used

We will use this $\underline{\boxed{\frak{\red{\cfrac{a_{1}}{a_{2}}\neq\cfrac{b_{1}}{b_{2}}}}}}$ because it has a unique solution.

➵ Assembling

Before we substitute the values we need to assemble them in the form of $\tt a_{1} + b_{1} + c_{1} = 0\:and\:a_{2} + b_{2}+ c_{2} = 0$

x + 2y - 3 = 0

ky + 5x + 7 = 0 => 5x + ky + 7 = 0

➵ By substituting the values

$\longmapsto\tt\cfrac{a_{1}}{a_{2}}\neq\cfrac{b_{1}}{b_{2}} \\ \\\longmapsto\tt\cfrac{1}{5} \neq \cfrac{2}{k} \\ \\\longmapsto \tt k \neq5 \times 2 \\ \\\longmapsto \tt k \neq10$

Therefore, k can have any value but should not be equal to 10.

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