Question

Pair of linear equation 2x-5y+4 and 2x+y-8 has

Answer 1

Step-by-step explanation:

$2x - 5y + 4 = 0 - (i) \\ 2x + y - 4 = 0 - (ii) \\ \blue{\mathfrak{\underline{\large{Subtract \: {eq}^{n} \: (ii) \: in \: (i)}}}:} \\ 2x - 5y + 4 = 0 \\ 2x \: + \: y \: - \: 4 = 0 \\ - \: \: \: \: \: - \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - - - \\ - 6y + 8 = 0 \\ \boxed{y = \frac{4}{3} } \\ \blue{\mathfrak{\underline{\large{Put \:the \: value \: of \: y \: in \: {eq}^{n} \: (ii) }}}:} \\ 2x + \frac{4}{3} - 4 = 0 \\ \frac{6x + 4 - 12}{3} = 0 \\ 6x + 4 - 12 = 0 \\ 6x = 8 \\ \boxed{x = \frac{4}{3} } \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

Answer:

$here \: \frac{a1}{a2} ≠ \frac{b1}{b2}$

$∴it \: has \: an \: unique \: solution$