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 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

To solve the system by graphing, we find the point where the two lines intersect. Another way is to set the two equations equal to each other since at the intersection point, the \( y \) - values (and \( x \) - values) are equal. So we set \( -x + 5=x + 1 \).
Add \( x \) to both sides: \( 5=2x + 1 \).
Subtract 1 from both sides: \( 4 = 2x \).
Divide both sides by 2: \( x = 2 \).
Now substitute \( x = 2 \) into one of the equations, say \( y=x + 1 \). Then \( y=2 + 1=3 \). So the intersection point is \( (2,3) \). When we graph the two lines, we will see that they cross at the point \( (2,3) \).

Answer:

The solution to the system of linear equations \(

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

\) is \( x = 2 \), \( y = 3 \) or the ordered pair \( (2,3) \).