The problem asks to solve the system of equations:

1. $y = -2|x + 1| + 3$
2. $y - x = 1$, or equivalently, $y = x + 1$

We need to find all points $(x, y)$ that satisfy both equations.

Step 1: Set the two equations equal to each other.

From the second equation, we know $y = x + 1$. Substitute this into the first equation: $x + 1 = -2|x + 1| + 3$

Step 2: Solve the absolute value equation.

We now have two cases based on the absolute value.

Case 1: $x + 1 \geq 0$, so $|x + 1| = x + 1$

$x + 1 = -2(x + 1) + 3$ Expanding: $x + 1 = -2x - 2 + 3$ $x + 1 = -2x + 1$ Subtract 1 from both sides: $x = -2x$ Add $2x$ to both sides: $3x = 0 \quad \Rightarrow \quad x = 0$ Substitute $x = 0$ back into $y = x + 1$: $y = 0 + 1 = 1$ Thus, one solution is $(0, 1)$.

Case 2: $x + 1 < 0$, so $|x + 1| = -(x + 1)$

$x + 1 = -2(-(x + 1)) + 3$ Expanding: $x + 1 = 2(x + 1) + 3$ Simplifying: $x + 1 = 2x + 2 + 3$ $x + 1 = 2x + 5$ Subtract $x$ from both sides: $1 = x + 5$ Subtract 5 from both sides: $x = -4$ Substitute $x = -4$ back into $y = x + 1$: $y = -4 + 1 = -3$ Thus, another solution is $(-4, -3)$.

Step 3: Verify possible points from the list.

• $(-2, 1)$: Not a solution.
• $(-1, 2)$: Not a solution.
• $(0, 1)$: Is a solution.
• $(2, -3)$: Not a solution.
• $(-4, -3)$: Is a solution.
• $(-3, -4)$: Not a solution.

Final Answer:

The points that are solutions to the system are:

• $(0, 1)$
• $(-4, -3)$

Would you like further details or explanations on any step?

Here are 5 follow-up questions you could explore:

1. How do we solve systems of equations involving absolute values in more complex cases?
2. Can a system with absolute values have more than two solutions?
3. How do graphical methods compare to algebraic methods in solving systems of equations?
4. What happens when both equations are non-linear?
5. What other methods can solve this system (like substitution or elimination)?

Tip: Always consider the domain when solving absolute value equations; split cases carefully based on the absolute value definition!