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

6 0

3 years ago