Question

y = x + 1 et
y = -3x - 15

déterminez si le couple (2, -1) est une solution de chacun des systèmy = 2x - 1 et y = -x + 2
b) y = 3,8x - 8,6

Explanation:

Step1: Check first equation

Substitute \( x = 2 \), \( y=-1 \) into \( y = 2x - 1 \).
Right - hand side: \( 2(2)-1=4 - 1 = 3 \).
Left - hand side: \( y=-1 \).
Since \( -1
eq3 \), wait, maybe a miscalculation. Wait, let's re - check. Wait, the first equation is \( y = 2x-1 \), substituting \( x = 2 \): \( y=2\times2 - 1=4 - 1 = 3 \). But the given \( y=-1 \), no, wait maybe the second part. Wait, the system is \( y = 2x-1 \) and \( y=-x + 2 \). Let's check the second equation.

Step2: Check second equation

Substitute \( x = 2 \), \( y = - 1 \) into \( y=-x + 2 \).
Right - hand side: \( -2 + 2=0 \).
Left - hand side: \( y=-1 \).
Wait, no, maybe I misread the couple. Wait, the couple is \( (2,-1) \)? Wait, let's re - do. Wait, for \( y = 2x-1 \), when \( x = 2 \), \( y=2(2)-1 = 3 \). For \( y=-x + 2 \), when \( x = 2 \), \( y=-2 + 2=0 \). The couple \( (2,-1) \): when \( x = 2 \), \( y=-1 \) does not satisfy \( y = 2x-1 \) (since \( 2x-1=3
eq - 1 \)) and does not satisfy \( y=-x + 2 \) (since \( -x + 2 = 0
eq - 1 \)). Wait, maybe there is a typo, but assuming we follow the steps:

Wait, maybe the couple is \( (1,1) \)? No, the problem says \( (2,-1) \). Wait, let's check again.

Wait, perhaps I made a mistake. Let's solve the system \( y = 2x-1 \) and \( y=-x + 2 \) to find the solution. Set \( 2x-1=-x + 2 \), then \( 2x+x=2 + 1 \), \( 3x=3 \), \( x = 1 \), then \( y=2(1)-1 = 1 \). So the solution is \( (1,1) \), so \( (2,-1) \) is not a solution. But let's check the substitution again for \( (2,-1) \):

For \( y = 2x-1 \): \( 2(2)-1 = 3
eq - 1 \)

For \( y=-x + 2 \): \( -2 + 2=0
eq - 1 \)

So the couple \( (2,-1) \) is not a solution of the system \( y = 2x-1 \) and \( y=-x + 2 \).

Answer:

The couple \( (2,-1) \) is not a solution of the system \(

$$\begin{cases}y = 2x-1\\y=-x + 2\end{cases}$$

\)