Question 256779
you have 2 equations.


they are


x + y = 5
x^2 + 3xy + 2y^2 = 40


solve for y in the first equation to get y = 5-x


substitute for y in the second equation to get:


x^2 + 3x(5-x) + 2(5-x)^2 = 40


simplify to get:


x^2 + 15x - 3x^2 + 2 * (25 - 10x + x^2) = 40


simplify further to get:


x^2 + 15x - 3x^2 + 50 - 20x + 2x^2 = 40


combine like terms to get:


-5x + 50 = 40 


subtract 50 from both sides of eqution to get:


-5x = 40 - 50 = -10


multiply both sides by -1 to get:


5x = 10


divide both sides by 5 to get:


x = 2


substitute in first equation to get:


x + y = 5 becomes 2 + y = 5 becomes y = 3


you have x = 2 and y = 3


2x + 4y = 2*2 + 4*3 = 4 + 12 = 16


your answer is (2x + 4y) = 16.