For all real values of c, the pair of equatio | vedclass

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(B) The given pair of linear equations is:$x - 2y = 8$$5x - 10y = c$Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y +

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False

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(B) The given pair of linear equations is:$x - 2y = 8$$5x - 10y = c$Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = -2, c_1 = -8$$a_2 = 5, b_2 = -10, c_2 = -c$For a system of linear equations to have a unique solution,the condition is $\frac{a_1}{a_2} 
eq \frac{b_1}{b_2}$.Here,$\frac{a_1}{a_2} = \frac{1}{5}$ and $\frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5}$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2}$,the system can never have a unique solution regardless of the value of $c$.If $c = 40$,then $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{5}$,which means the system has infinitely many solutions.If $c 
eq 40$,the system has no solution.Therefore,the statement is False.

For all real values of $c,$ the pair of equations $x-2y=8$ and $5x-10y=c$ have a unique solution. Justify whether it is true or false.

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