Question

name:
algebra ii
solving linear systems by graphing worksheet #2
solve the system by graphing.

1. $y = -x + 5$

$y = x + 1$
2.

Explanation:

Step1: Analyze the first equation \( y = -x + 5 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where the slope \( m=- 1 \) and the y - intercept \( b = 5 \). To graph this line, we can start by plotting the y - intercept at the point \( (0,5) \). Then, using the slope (rise over run), since the slope is - 1 (which can be written as \( \frac{-1}{1} \)), from the point \( (0,5) \), we move down 1 unit and right 1 unit (or up 1 unit and left 1 unit) to get other points on the line. For example, when \( x = 1 \), \( y=-1 + 5=4 \), so the point \( (1,4) \) is on the line. When \( x = 5 \), \( y=-5 + 5 = 0 \), so the point \( (5,0) \) is on the line.

Step2: Analyze the second equation \( y=x + 1 \)

This is also in slope - intercept form with slope \( m = 1 \) and y - intercept \( b=1 \). We plot the y - intercept at \( (0,1) \). Using the slope of 1 ( \( \frac{1}{1} \) ), from the point \( (0,1) \), we move up 1 unit and right 1 unit (or down 1 unit and left 1 unit) to get other points. For example, when \( x = 1 \), \( y=1 + 1=2 \), so the point \( (1,2) \) is on the line. When \( x=-1 \), \( y=-1 + 1 = 0 \), so the point \( (-1,0) \) is on the line.

Step3: Find the intersection point

The solution to the system of linear equations is the point where the two lines intersect. By graphing both lines (either by plotting the points we found or using the slope - intercept method), we can see that the two lines \( y=-x + 5 \) and \( y=x + 1 \) intersect at the point where \( -x + 5=x + 1 \). Solving for \( x \):
\[

$$\begin{align*} -x+5&=x + 1\\ 5-1&=x+x\\ 4&=2x\\ x&=2 \end{align*}$$

\]
Substitute \( x = 2 \) into \( y=x + 1 \), we get \( y=2 + 1=3 \). So the intersection point is \( (2,3) \). We can also see this from the graph: when we plot both lines, they cross at \( (2,3) \).

Answer:

The solution to the system \(

$$\begin{cases}y=-x + 5\\y=x + 1\end{cases}$$

\) is \( (2,3) \)