Question

o produto das soluções do sistema { 2x - 3y = -1 { - 3x + 4y = 10

Answer 1

{ 2x - 3y = -1 ===> *(4)
{ -3x + 4y = 10 ==> *(3) ↓

{ 8x - 12y = -4
{ -9x +12y = 30
-------------------------- (+)
-x = 26
x = -26 ✓

{ -3x + 4y = 10
-3(-26) +4y = 10
78 +4y = 10
4y = -68
y = -17 ✓

=> x . y = (-26) .(-17 )= 442 ✓

Answer 2

Resposta: x = -26; y = -17.

Explicação passo-a-passo:

Pode-se usar o método de substituição.

$\begin{cases} \mathtt{Eq_1:}&\mathtt{+2x-3y=-1}\\ \mathtt{Eq_2:}&\mathtt{-3x+4y=10} \end{cases}\\\\ \mathtt{Eq_1:~2x-3y=-1}\\\\ \mathtt{Eq_1:~2x=-1+3y}\\\\ \mathtt{Eq_1:~x=\dfrac{-1+3y}{2}}$

Substituindo na segunda equação.

$\mathtt{Eq_2:~-3x+4y=10}\\\\ \mathtt{Eq_2:~-3\left(\dfrac{-1+3y}{2}\right)+4y=10}\\\\\\ \mathtt{Eq_2:~\dfrac{-3(-1+3y)}{2}+4y=10}\\\\\\ \mathtt{Eq_2:~\dfrac{3-9y}{2}+4y=10}\\\\\\ \mathtt{Eq_2:~\dfrac{3-9y}{2}+\dfrac{8y}{2}=10}\\\\\\ \mathtt{Eq_2:~\dfrac{3-9y+8y}{2}=10}\\\\\\ \mathtt{Eq_2:~3-y=10\times2}\\\\ \mathtt{Eq_2:~-y=20-3}\\\\ \mathtt{Eq_2:~-y=17\times(-1)}\\\\ \mathtt{Eq_2:~y=-17}$

Substituindo o valor de y na primeira equação.

$\mathtt{2x-3y=-1}\\\\ \mathtt{2x-3(-17)=-1}\\\\ \mathtt{2x+51=-1}\\\\ \mathtt{2x=-1-51}\\\\ \mathtt{2x=-52}\\\\ \mathtt{x=\dfrac{-52}{2}}\\\\ \mathtt{x=-26}$