Question

resolva no método de adição:

a) { x + y = 20
{ 2x - 3y = 4

b) { x + y = 18
{ 4x + 2y = 56

c) { x + y = 45
{ 4x + 2y = 130

d) { x + y = 27
{ 2x + 3y = 69​

Answer 1

Resposta:

$a:\ x=\frac{64}{5}\ e\ y=\frac{36}{5}\\\\b:\ :\ x=10\ e\ y=8\\\\c:\ x=20\ e\ y=25\\\\d:\ x=15\ e\ y=12$

Explicação passo a passo:

$a)\ \left\{\begin{array}{ccc}x+y=20\ (i)\\2x-3y=4\ (ii)\end{array}\right\\\\\\Somando\ (i)\ e\ (ii):\\\\x+y+2x-3y=20+4\\\\3x-2y=24\\\\3x=24+2y\\\\x=\frac{24+2y}{3}\ (iii)\\\\\\Substituindo\ (iii)\ em\ (i):\\\\\frac{24+2y}{3}+y=20\\\\3\bullet(\frac{24+2y}{3}+y)=3\bullet20\\\\24+2y+3y=60\\\\2y+3y=60-24\\\\5y=36\\\\y=\frac{36}{5}\\\\\\Substituindo\ o\ valor\ de\ y\ em\ (iii):\\\\x=\frac{24+2(\frac{36}{5} )}{3}\\\\x=\frac{24+\frac{72}{5} }{3}$

$x=\frac{\frac{120+72}{5} }{3}\\\\x=\frac{\frac{192}{5} }{3} \\\\x=\frac{192}{5}\bullet\frac{1}{3}\\\\x=\frac{192}{15}\\\\x=\frac{64}{5}$

$b)\ \left\{\begin{array}{ccc}x+y=18\ (i)\\4x+2y=56\ (ii)\end{array}\right\\\\\\Somando\ (i)\ e\ (ii):\\\\x+y+4x+2y=18+56\\\\5x+3y=74\\\\5x=148-3y\\\\x=\frac{74-3y}{5}\ (iii)\\\\\\Substituindo\ (iii)\ em\ (i):\\\\\frac{74-3y}{5}+y=18\\\\5\bullet(\frac{74-3y}{5}+y)=5\bullet18\\\\74-3y+5y=90\\\\-3y+5y=90-74\\\\2y=16\\\\y=\frac{16}{2}\\\\y=8\\\\\\Substituindo\ o\ valor\ de\ y\ em\ (iii):\\\\x=\frac{74-3\bullet(8)}{5}\\\\x=\frac{74-24}{5}\\\\x=\frac{50}{5}\\\\x=10$

$c)\ \left\{\begin{array}{ccc}x+y=45\ (i)\\4x+2y=130\ (ii)\end{array}\right\\\\\\Somando\ (i)\ e\ (ii):\\\\x+y+4x+2y=45+130\\\\5x+3y=175\\\\5x=175-3y\\\\x=\frac{175-3y}{5}\ (iii)\\\\\\Substituindo\ (iii)\ em\ (i):\\\\\frac{175-3y}{5}+y=45\\\\5\bullet(\frac{175-3y}{5}+y)=5\bullet45\\\\175-3y+5y=225\\\\-3y+5y=225-175\\\\2y=50\\\\y=25\\\\\\Substituindo\ y\ em\ (iii):\\\\x=\frac{175-3\bullet(25)}{5}\\\\x=\frac{175-75}{5}\\\\x=\frac{100}{5}\\\\x=20$

$d)\ \left\{\begin{array}{ccc}x+y=27\ (i)\\2x+3y=69\ (ii)\end{array}\right\\\\\\Somando\ (i)\ e\ (ii):\\\\x+y+2x+3y=27+69\\\\3x+4y=96\\\\3x=96-4y\\\\x=\frac{96-4y}{3}\ (iii)\\\\\\Substiutindo\ (iii)\ em\ (i):\\\\\frac{96-4y}{3}+y=27\\\\3\bullet(\frac{96-4y}{3}+y)=3\bullet27\\\\96-4y+3y=\\\\-4y+3y=81-96\\\\-y=-15\\\\y=15\\\\\\Substituindo\ y\ em\ (iii):\\\\x=\frac{96-4\bullet(15)}{3}\\\\x=\frac{96-60}{3}\\\\x=\frac{36}{3}\\\\x=12$