Question

4x^2-8y^2=19
2x+3y=1

Answer 1

What are we doing?
I will assume you are finding the intersection of the straight line
and the hyperbola (solving the two equations)

form 2x + 3y = 1 ----> x = (1-3y)/2
sub into the hyperbola
4(1-3y)^2 / 4 - 8y^2 = 19
1 - 6y + 9y^2 - 8y^1 = 19
y^2 - 6y - 18 = 0
y = (6 ± √108)/2 = 3 ± 3√3 = appr 8.2

sub into the first:
2x+3y = 1
2x + 3(3+3√3) = 1
2x = 1 - 9 - 9√3
2x = -8-9√3
x = -4 - (9/2)√3 = appr -11.8

if y = 3 - 3√3 = appr -2.2
2x + 3(3-3√3) = 1
2x = 1 - 9 + 9√3
x = -4 + (9/2)√3 = appr 3.8

They intersect at appr (-11.8 , 8.2) and (3.8 , -2.2)

You can check these are correct by using the following to graph it

www.desmos.com/calculator