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For what values of $‘k’$ will the pair of equations $3x + 4y =12$ and $kx + 12y = 30$ $NOT$ have a unique solution ?

- $9$
- $12$
- $3$
- $7.5$

+1 vote

**Option A. 9**

To not have a unique solution, lines should be parallel.

Two parallel lines have equal slopes.

Slope of the line 3x+4y = 12, = -3/4

Slope of the line kx+12y = 30, = -k/12

3/4 = k/12 or** k=9**

0 votes

A system of linear equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ will have a unique solution if the two lines represented by the equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ intersect at a point.

i.e., if the two lines are neither parallel nor coincident. Essentially, the slopes of the two lines should be different.

- Condition for unique solution$:\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$
- Condition for no solution$:\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$
- Condition for infinitely many solutions$:\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

Now, question asked about not unique solution ,the the condition will be $\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$

Given the equations, $3x + 4y =12$ and $kx + 12y = 30$

Here, $a_{1} = 3,b_{1} = 4,a_{2} = k,b_{2} = 12$

Now, if above equation does not have unique solution , then $\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$

$\implies \dfrac{3}{k} = \dfrac{4}{12}$

$\implies k = 9.$

So, the correct answer is $(A).$

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