Question

4x-5y+16=0 and 2x+y-6=0

Answer 1

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}$

$\red{\bold{\underline{\underline{❥Question᎓}}}}$4x-5y+16=0 and 2x+y-6=0

Answer:

⟹$4x - 5y + 16 = 0$

⟹ $x = \frac{16 + 5y}{4}$

$\bold{put\: value\: of\: x\: in \:another \:equation}$

⟹ $2( \frac{16 + 5y}{4} ) + y - 6 = 0$

⟹ $\frac{32 + 10y}{4} + y - 6 = 0$

⟹ $32 + 10y + 4y - 24 = 0$

⟹ $14y + 8 = 0$

⟹ $14y = - 8$

⟹ $y = - \frac{8}{14} = - \frac{4}{7} [/</p><p>tex]</p><p>[tex]\bold{put\: y \:value\: in }$

⟹ $x = \frac{32 }+ 10 \times - \frac{4}</p><p>{7} {4} = 8 + 10 \times - \frac{4}{7}$

⟹ $18 \times - \frac{4}{7} = - \frac{72}{7}$

Hope it helps you..!!!

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Answer 2

According to the question

$⟹4x - 5y + 16 = 0 \\ \\ </p><p></p><p>⟹ x = \frac{16 + 5y}{4} \\ \\ </p><p></p><p>\bold{put\: value\: of\: x\: in \:another \:equation} \\ \\ </p><p></p><p>⟹ 2( \frac{16 + 5y}{4} ) + y - 6 = 0 \\ \\ </p><p></p><p>⟹ \frac{32 + 10y}{4} + y - 6 = 0 \\ \\ </p><p></p><p>⟹ 32 + 10y + 4y - 24 = 0 \\ \\ </p><p></p><p>⟹ 14y + 8 = 0 \\ \\ </p><p></p><p>⟹ 14y = - 8 \\ \\ </p><p></p><p>⟹ y = - \frac{8}{14} = - \frac{4}{7}$ $\bold{put\: y \:value\: in } \\ \\ </p><p>⟹ x = \frac{32 }+ 10 \times - \frac{4} {7} {4} = 8 + 10 \times - \frac{4}{7} \\ \\ </p><p></p><p>⟹ 18 \times - \frac{4}{7} = - \frac{72}{7}$