Answer:
(2,3) is the answer
Step-by-step explanation:
[tex]5x - 4y = - 2.....equation1 \\ 2x + 4y = 16 \\ 4y = 16 - 2x......equation2 \\ now \: substituting \: the \: value \: of \: 4y \: in \: eq1 \\ 5x - (16 - 2x) = - 2 \\ 5x - 16 + 2x = - 2 \\ 7x = - 2 +16 \\ x = \frac{14}{7} \\ x = 2 \\ putting \: the \: value \: of \: x \: in \: eq2 \\ 4y = 16 - 2x \\ y = \frac{16 - 2 \times 2}{4} \\ y = \frac{12}{4} \\ y = 3[/tex]
