Question

make solution set of this using cross multiplication method:9x+11y+15=0
7x-13y-25=0​

Answer 1

$\huge\boxed{\underline{\underline{\green{\tt Solution}}}} </p><p>$

The given equations are

$9x + 11y + 15 = 0 \: \: ........(1)$

$7x - 13y - 25 = 0 \: \: .........(2)$

On Cross Multiplication we get

$\frac{x}{11 \times ( - 25) - ( - 13) \times 15} = \frac{y}{7 \times 15 - 9 \times ( - 25)} = \frac{1}{9 \times ( - 13) - 7 \times 11}$

$\frac{x}{( - 275 + 195)} = \frac{y}{(105 + 225)} = \frac{1}{( - 117 - 77)}$

$\frac{x}{ - 80} = \frac{y}{330} = \frac{1}{ - 194}$

Above gives

$\frac{x}{ - 80} = \frac{1}{ - 194}$

$x = \frac{80}{194} = \frac{40}{97}$

$\frac{y}{330} = \frac{1}{ - 194}$

$y = - \frac{330}{194} = - \frac{165}{97}$

So the required solution set is

$x = \frac{40}{97}$ & $y = - \frac{165}{97}$

$</p><p>\displaystyle\textcolor{red}{Please \: Mark \: it \: Brainliest}$

Answer 2

✿solution✿☘️

the given equation written as ;

$9x + 11y + 15 = 0 \\ 7x - 13y - 25 = 0$

this is customarily written as :

$\frac{x}{b1c2 - b2c1} = \frac{y}{c1a2 - c2a1} = \frac{1}{a1b2 - a2b1}$

$\frac{x}{11 \times( - 25) - ( - 13) \times 25} = \frac{y}{7 \times 15 - 9 \times ( - 25)} = \frac{1}{9 \times ( - 13) - 7 \times 11}$

$\frac{x}{ - 275 + 195} = \frac{y}{105 + 225} = \frac{1}{ - 117 - 77}$

$\frac{x}{ - 80 } = \frac{y}{330} = \frac{1}{194}$

$x = \frac{ - 80}{ - 194} = \frac{40}{ 97} \\ y = \frac{330}{ - 194} = \frac{165}{ - 97}$

hence ,

$x = \frac{40}{97} and \: y = \frac{165}{ - 97}$