Do the equations 4x + 3y - 1 = 5 and 12x + 9y | vedclass

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(B) Given equations are:$4x + 3y - 1 = 5 \implies 4x + 3y = 6$$12x + 9y = 15$Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_

Vedclass · Accepted Answer

No

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(B) Given equations are:$4x + 3y - 1 = 5 \implies 4x + 3y = 6$$12x + 9y = 15$Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:$a_1 = 4, b_1 = 3, c_1 = 6$$a_2 = 12, b_2 = 9, c_2 = 15$Calculating the ratios:$\frac{a_1}{a_2} = \frac{4}{12} = \frac{1}{3}$$\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$$\frac{c_1}{c_2} = \frac{6}{15} = \frac{2}{5}$Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$,the lines are parallel and not coincident. Therefore,the answer is No.

Do the equations $4x + 3y - 1 = 5$ and $12x + 9y = 15$ represent a pair of coincident lines? Justify your answer.

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