The slope/y-int form would be y=9/2x+2 you have to move the x value and whole number(y-int) to the other side which would make them negative and then divide by the -2 and there you go.

4 0

2 years ago

Can anyone plzzzz help me with this question of linear equation in two varaiable plzzzzzz... Its very important... Plz solve it

kotegsom [21]

That's not a linear system, but you have an awesome school system for giving you this problem.

$\dfrac 5 y - \dfrac 2 x = \dfrac{13}6$

Multiply by 6xy to clear the fractions.

$30x - 12y = 13 xy$

That's a second degree equation, also known as a conic. That one happens to be a hyperbola.

$30x = y(13 x +12)$

$y = \dfrac{30x}{13 x + 12}$

Let's clear the fractions from the second equation, multiplying out common denominator xy:

$\dfrac {36} x - \dfrac {24} y = 1$

$36y - 24 x = xy$

We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.

Substituting,

$36 \left( \dfrac{30x}{13 x + 12} \right) - 24 x = x \left( \dfrac{30x}{13 x + 12} \right)$

$36(30x) -24 x(13x + 12) = 30x^2$

$1080 x - 312 x^2- 288 x = 30x^2$

$x(792 - 342 x)= 0$

$x = 0 \textrm{ or } x=792/342 = \dfrac{44}{19}$

We have to rule out x=0 because it's in the denominator.

$y = \dfrac{30x}{13 x + 12} = \dfrac{30(44/19)}{13(44/19)+12}$

$y = \dfrac{33}{20}$

Answer: (44/19, 33/20)

8 0

2 years ago

{(−1, 5), (2, 8), (5, 3), (13, −4)} what are the range

makkiz [27]

$(x;\ y)\\\\(-1, \fbox5), (2, \fbox8), (5, \fbox3), (13, \fbox{-4})\\\\\boxed{The\ range:y\in\{-4;\ 3;\ 5;\ 8\}}$

7 0

3 years ago

13. Factor 12 + 8. *<br>O 3(2+4)<br>4(3+2)<br>2(4+3)<br>4+3(2)

e-lub [12.9K]

2752 is this because this is hard ending as

4 0

2 years ago