{y=-1/2x-2 {y=-|x-2| solve algebraically and graphically

To solve the system of equations algebraically and graphically, we'll go through each equation step by step:

### 1. The equations are:
\[
y = -\frac{1}{2}x - 2 \tag{1}
\]
\[
y = -|x - 2| \tag{2}
\]

### Algebraic Solution:
We are looking for the intersection points between these two equations.

#### Step 1: Set the equations equal to each other
Since both equations are equal to \(y\), we set them equal to each other:
\[
-\frac{1}{2}x - 2 = -|x - 2|
\]

#### Step 2: Solve for \(x\)
Now, we'll solve this equation. First, recall that the absolute value function behaves differently depending on whether \(x - 2\) is positive or negative.

1. **Case 1: \(x - 2 \geq 0\) (i.e., \(x \geq 2\))**

   In this case, \(|x - 2| = x - 2\). So the equation becomes:
   \[
   -\frac{1}{2}x - 2 = -(x - 2)
   \]
   Simplifying:
   \[
   -\frac{1}{2}x - 2 = -x + 2
   \]
   Add \(x\) to both sides:
   \[
   \frac{1}{2}x - 2 = 2
   \]
   Add 2 to both sides:
   \[
   \frac{1}{2}x = 4
   \]
   Multiply both sides by 2:
   \[
   x = 8
   \]
   Now substitute \(x = 8\) into either of the original equations to find \(y\). Using \(y = -\frac{1}{2}x - 2\):
   \[
   y = -\frac{1}{2}(8) - 2 = -4 - 2 = -6
   \]
   So, one solution is \((x, y) = (8, -6)\).

2. **Case 2: \(x - 2 < 0\) (i.e., \(x < 2\))**

   In this case, \(|x - 2| = -(x - 2)\). So the equation becomes:
   \[
   -\frac{1}{2}x - 2 = -( -(x - 2)) = x - 2
   \]
   Simplifying:
   \[
   -\frac{1}{2}x - 2 = x - 2
   \]
   Add \(2\) to both sides:
   \[
   -\frac{1}{2}x = x
   \]
   Add \(\frac{1}{2}x\) to both sides:
   \[
   0 = \frac{3}{2}x
   \]
   Thus, \(x = 0\). Now substitute \(x = 0\) into either of the original equations to find \(y\). Using \(y = -\frac{1}{2}x - 2\):
   \[
   y = -\frac{1}{2}(0) - 2 = -2
   \]
   So, another solution is \((x, y) = (0, -2)\).

### Summary of Algebraic Solutions:
The two solutions are:
\[
(8, -6) \quad \text{and} \quad (0, -2)
\]

### Graphical Solution:

To graph these equations:

1. **Equation 1: \(y = -\frac{1}{2}x - 2\)**  
   This is a linear equation with a slope of \(-\frac{1}{2}\) and a y-intercept of \(-2\). The line will slope downwards.

2. **Equation 2: \(y = -|x - 2|\)**  
   This is a V-shaped graph that opens downwards, with the vertex at \((2, 0)\). The absolute value function reflects points below the x-axis.

The intersection points between the two graphs will be the same as those found algebraically: \((8, -6)\) and \((0, -2)\).

Would you like me to plot these graphs for you?

y = -1/2x - 2 and y = -|x - 2|, solve algebraically and graphically

Solve and graph the system of equations y = -1/2x - 2 and y = -|x - 2|. Learn how to solve algebraically by finding intersections and graphically by plotting the linear and V-shaped graphs. Suitable for high school algebra students, Grades 9-11.

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