Question

giai hpt

a.\(\left\{{}\begin{matrix}x-2\left(y-1\right)=3x\\3x-2\left(y+1\right)=3\left(x-1\right)\end{matrix}\right.\)

b.\(\left\{{}\begin{matrix}3\left(x+1\right)-2y=5-y\\4x-2\left(y+1\right)=-3\end{matrix}\right.\)

Answer 1

a.Vì x-2(y-1) = 3x <=> -2(y-1) = -2x <=> y-1=x

Thay vào, ta có (y-1)-2(y-1) = 3(y-1) <=> -(y-1) = 3(y-1)

<=> y-1 = 0 <=> y = 1 => x = 0

b.Ta có 3(x+1)−2y = 5−y <=> 3x+3-2y = 5-y

<=> 3x-2y = 2-y <=> -2y = 2-y-3x(1)

Lại có 4x−2(y+1) = −3 <=> 4x-2y-2 = -3

<=> 4x-2y = -1 <=> -2y = -1-4x(2)
Từ (1) và (2), ta có 2-y-3x = -1-4x <=> -1-x = 2-y

<=> -x+y = 3 <=> x-y = -3

Lại có 4x−2(y+1) = −3 => 4x-2(y+1) = x-y

<=> 4x-2y-2 = x-y <=> 3x-y = 2

Mà x-y = -3 => (3x-y)-(x-y) = -5

=> 2x = -5 <=> x = -5/2 => y = 1/2

Vậy...