Question

por favor me ajudem a resolver

Answer 1

a) $\begin{cases} x+y=3 \\ x-y=9\end{cases}$

Somando: $(x+y)+(x-y)=3+9~~\Rightarrow~~2x=12~~\Rightarrow~~\boxed{x=6}$

$x+y=3~~\Rightarrow~~6+y=3~~\Rightaerrow~~\boxed{y=-3}$.


b) $\begin{cases} 2x-2y=12 \\ x+2y=11\end{cases}$

Somando: $3x=23~~\Rightarrow~~\boxed{x=\dfrac{23}{3}}$

$2x-2y=12~~\Rightarrow~~\dfrac{46}{3}-2y=12~~\Rightarrow~~46-6y=36$

$6y=10~~\Rightarrow~~y=\dfrac{10}{6}~~\Rightarrow~~\boxed{y=\dfrac{5}{3}}$.


c) $\begin{cases} x+y=7 \\ x+2y=11\end{cases}~~\Rightarrow~~\begin{cases} -x-y=-7 \\ x+2y=11\end{cases}$

Somando: $(-x-y)+(x+2y)=-7+11~~\Rightarrow~~\boxed{y=4}$

$x+y=7~~\Rightarrow~~x+4=7~~\Rightarrow~~\boxed{x=3}$.


d) $\begin{cases} -3x-2y=8 \\ x-5y=3\end{cases}$

Da segunda equação, $x=5y+3$. Substituindo na primeira:

$-3(5y+3)-2y=8~~~\Rightarrow~~-15y-9-2y=8~~\Rightarrow~~-17y=17~~\Rightarrow~~\boxed{y=-1}$

$x=5y+3~~\Rightarrow~~x=-5+3~~\Rightarrow~~\boxed{x=-2}$


e) $\begin{cases} x+y=12 \\ x-y=8\end{cases}$

Somando: $(x+y)+(x-y)=12+8~~\Rightarrow~~2x=20~~\Rightarrow~~\boxed{x=10}$

$x+y=12~~\Rightarrow~~10+y=12~~\Rightarrow~~\boxed{y=2}$.


f) $\begin{cases} 3x-2y=-1 \\ 2x+5y=12\end{cases}~~\Rightarrow~~\begin{cases} -6x+4y=2 \\ 6x+15y=36\end{cases}$

Somando: $(-6x+4y)+(6x+15y)=2+36~~\Rightarrow~~19y=38~~\Rightarrow~~\boxed{y=2}$

$3x-2y=-1~~\Rightarrow~~3x-4=-1~~\Rightarrow~~3x=3~~\Rightarrow~~\boxed{x=1}$.



g) $\dfrac{x}{3}+\dfrac{y}{4}=-\dfrac{1}{3}$

$\dfrac{4x+3y}{12}=\dfrac{-4}{12}$

$\boxed{4x+3y=-4}$

$x-\dfrac{y}{8}=\dfrac{5}{2}$

$\dfrac{8x-y}{8}=\dfrac{20}{8}$

$\boxed{8x-y=20}$

$\begin{cases} 4x+3y=-4 \\ 8x-y=20\end{cases}$

Da segunda equação, $y=8x-20$. Substituindo na primeira:

$4x+3(8x-20)=-4~~\Rightarrow~~4x+24x-60=-4~~\Rightarrow~~28x=56~~\Rightarrow~~\boxed{x=2}$

$y=8x-20~~\Rightarrow~~y=16-20~~\Rightarrow~~\boxed{y=-4}$



h) $2x-\dfrac{4y}{5}=2-\dfrac{x}{3}$

$\dfrac{30x-12y}{15}=\dfrac{30-5x}{15}$

$30x-12y=30-5x~~\Rightarrow~~\boxed{35x-12y=30}$


$y-5x=-\dfrac{3y}{5}-6$

$5y-25x=-3y-30$

$25x-3y-5y=30~~\Rightarrow~~\boxed{25x-8y=30}$.

$\begin{cases} 35x-12y=30 \\ 25x-8y=30\end{cases}$

Igualando, $35x-12y=25x-8y~~\Rightarrow~~10x=4y~~\Rightarrow~~x=\dfrac{4y}{10}=\dfrac{2y}{5}$.

Substituindo em $25x-8y=30$:

$25\cdot\dfrac{2y}{5}-8y=30~~\Rightarrow~~10y-8y=30~~\Rightarrow~~2y=30~~\Rightarrow~~\boxed{y=15}$.


$x=\dfrac{2y}{5}~~\Rightarrow~~x=\dfrac{2\cdot15}{5}~~\Rightarrow~~x=\dfrac{30}{5}$

$\boxed{x=6}$