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

Multiply by 6xy to clear the fractions.

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


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


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





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


Answer: (44/19, 33/20)
We know
in the graph two points
point 1 (2,1) and point 2 (-1,-4)
the equation in point-slope form is ----------------> (y-y1)=m(x-x1)
step 1
find the value of m
m=(y2-y1)/(x2-x1)----------> (-4-1)/(-1-2)--------> m=5/3
step 2
with the point (-1,-4) and m=5/3
find the equation in point-slope form
(y-(-4))=(5/3)*(x-(-1))-------------> (y+4)=(5/3)(x+1)
the answer is the option d) ( y+ 4 ) =+ 5/3 ( x + 1 )
Answer:
a = 1
b = -1
c = -2
Step-by-step explanation:
We reorder the equation in such way that let us see the usual
ax² + bx + c = 0
Then the original quation is:
-2 = - x + x² - 4 ⇒ 0 = 2 - x + x² - 4 ⇒ x² - x - 2 = 0
Now we are able by simply inspection to identify a, b . c comparing our equation with the general equation so :
a = 1
b = -1
c = -2