Answer :

[tex]\left\{\begin{array}{ccc}4x+13y=40\\4x+3y=-40&/\cdot(-1)\end{array}\right\\+\left\{\begin{array}{ccc}4x+13y=40\\-4x-3y=40\end{array}\right\\-----------\\.\ \ \ \ \ \ 10y=80\ \ \ \ /:10\\.\ \ \ \ \ \ y=8\\\\4x+3\cdot8=-40\\4x+24=-40\\4x=-40-24\\4x=-64\ \ \ \ /:4\\x=-16\\\\Solution:x=-16\ and\ y=8.[/tex]
doss370
4x + 13y = 40
4x + 3y = -40
using elimination method
10y = 80
y = 8
substitute y in eq1
4x + 13(8) = 40
4x + 104 = 40
4x = 40 - 104
4x = - 64
x = -16

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