Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
 now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
 
        
             
        
        
        
Hello What is it that need to be answer ow I see
 
        
        
        
Answer:
see explanation
Step-by-step explanation:
The equation of a line in point- slope form is
y - b = m(x - a)
where m is the slope and (a, b) a point on the line
Here m =  and (a, b) = (3, - 2) , thus
 and (a, b) = (3, - 2) , thus
y - (- 2) =  (x - 3) , that is
 (x - 3) , that is
y + 2 =  ( x - 3) ← in point- slope form
( x - 3) ← in point- slope form
(b)
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m =  , thus
 , thus
y =  x + c ← is the partial equation
 x + c ← is the partial equation
To find c substitute (3, - 2) into the partial equation
- 2 =  + c ⇒ c = - 2 -
 + c ⇒ c = - 2 -  = -
 = - 
y =  x -
 x -  ← in slope- intercept form
 ← in slope- intercept form