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Given $3 x - 4 y = 7$ and $x + cy = 13$ , for what value of $c$ will the two equations not have a solution?

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\frac{3}{4}

\frac{4}{3}

- 4

\frac{- 4}{3}

answer is D.

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Detailed Solution

Given equations are,

3 x - 4 y = 7

x + cy = 13

\Rightarrow 3 x - 4 y - 7 = 0

x + cy - 13 = 0

Comparing with the general equations

a_{1} x + b_{1} y + c_{1} = 0; a_{2} x + b_{2} y + c_{2} = 0

\Rightarrow a_{1} = 3, b_{1} = - 4, c_{1} = - 7, a_{2} = 1, b_{2} = c, c_{2} = - 13

As we know the system of linear equation has no solution if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

∴ \frac{3}{1} = \frac{- 4}{c} \neq \frac{- 7}{- 13}

c = \frac{- 4}{3}

Therefore, option (4) is correct.

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