Question

the pair of equations 2x-5y+4=0 and 2x+y-8=0 has how many solutions​

Answer 1

Answer:

let both equation be 1st and 2nd

now, subtract 2nd from 1st equation, we have

Step-by-step explanation:

2x-5y+4=0

2x+y - 8 =0

- - +

--------------------

0 - 6y+12=0

now ,

6y=12

y=2

hence y =2.

put the vaule of y =2 in 1st eqn. then

2x=6

x=3

Hence y=2and x=3 is answer

mention this answer is brainliest

Answer 2

$\Huge{\textbf{\textsf{{\purple{Ans}}{\pink{wer}}{\color{pink}{:}}}}}$

$</p><p>If\\</p><p>a_1x + b_1y + c_1 = 0 \:\:\:and \\</p><p>a_2x + b_2y + c_2 = 0 \:\:are \: two \: equations\:\:then \\</p><p>they \:have \: unique \: \: solution \: \: if \\ \\ \frac{a_1}{a_2} \neq \: \frac{b_1}{b_2} </p><p> \\ = \frac{2}{2} \neq \: \frac{ - 5}{1} \\ = 1 \neq \: - 5$

so they have unique solution.