Question

4xsquare-9ysquare=19,
2x+3y=1

Answer 1

4x^2 - 9y^2 = 19
(2x-3y)(2x+3y) = 19
2x-3y = 19

Not sure where you want to go with that.

Answer 2

2 x + 3 y = 1

Subtract 2 x to both sides

3 y = 1 - 2 x

Divide both sides by 3

y = ( 1 - 2 x ) / 3

Plug this into: 4 x² - 9 y² = 19

4 x² - 9 [ ( 1 - 2 x ) / 3 ]² = 19

4 x² - 9 ( 1 - 2 x )² / 3² = 19

4 x² - 9 [ 1² - 2 ∙ 1 ∙ 2 x + ( 2 x )² ] / 9 = 19

4 x² - ( 1 - 4 x + 4 x² ) = 19

4 x² - 1 + 4 x - 4 x² = 19

- 1 + 4 x = 19

Add 1 to both sides

4 x = 20

Divide both sides by 4

x = 5

Plug this into: y = ( 1 - 2 x ) / 3

y = ( 1 - 2 ∙ 5 ) / 3

y = ( 1 - 10 ) / 3

y = - 9 / 3

y = - 3

Final solution:

x = 5 , y = - 3

Proof:

4 x² - 9 y² = 19

4 ( 5 )² - 9 ( - 3 )² = 19

4 ∙ 25 - 9 ∙ 9 = 19

100 - 81 = 19

19 = 19

2 x + 3 y = 1

2 ∙ 5 + 3 ∙ ( - 3 ) = 1

10 - 9 = 1

1 = 1

Answer 3

4x^2 - 9y^2 = 19 and 2x+3y = 1

the first one factors to
(2x+3y)(2x-3y) = 19
(1)(2x-3y) = 19
2x-3y = 19 with 2x+3y = 1
add them
4x=20
x=5
then 10 + 3y = 1
y = -3

https://www.wolframalpha.com/input/?i=solve+4x%5E2+-+9y%5E2+%3D+19+%2C+2x%2B3y+%3D+1