Question

$\frak{\underline{Question}}:-$
3) The pair of linear equations 3x - 5y - 4 = 0 and 2x + 7y - 12 = 0 have

$\\ \sf(a) \: \frac{1}{7} \\ \\ \sf(b) \: \frac{2}{7} \\ \\ \sf(c) \: \frac{3}{7} \\ \\ \sf(d) \: \frac{5}{7}$
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Answer 1

Step-by-step explanation:

I don't know what your question means

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hope it helps

Answer 2

Step-by-step explanation:

$3x - 5y - 4 = 0 \\ = > 3x - 5y = 4 \: \: \: - (1) \\ = > 2x + 7y = 12 \: \: \: - (2)$

multiplying (1) by 7 and (2) by 5 , we get ,

$21x - 35y = 28 \\ = > 10x + 35y = 60 \\ = > 31x = 88 \\ = > x = \frac{88}{31} \\ = > x = 2 \frac{26}{31}$