E.400. Ştiind că 2a+3b−c=112a + 3b − c = 112a+3b−c=11 şi 3a+b+2c=27,3a + b + 2c = 27,3a+b+2c=27, să se calculeze 4a−b+5c.4a − b + 5c.4a−b+5c.
43.43.43.
4a−b+5c=2(3a+b+2c)−(2a+3b−c=11)=2⋅27−11=43.4a − b + 5c = 2(3a + b + 2c) - (2a + 3b − c = 11) = 2 \cdot 27 - 11 = 43.4a−b+5c=2(3a+b+2c)−(2a+3b−c=11)=2⋅27−11=43.