Question

algebra 1
worksheet 7.1
solve each system by graphing.

1. $x = 4$

$y = -\frac{1}{2}x + 3$

1. $y = x - 2$

$y = 6x + 3$

1. $2x - y = -1$

$2x - y = -4$

1. $x - 2y = 6$

$5x - 2y = -2$

Explanation:

Problem 1: Solve \(

$$\begin{cases} x = 4 \\ y = -\frac{1}{2}x + 3 \end{cases}$$

\) by graphing

Step 1: Analyze the first equation

The equation \( x = 4 \) is a vertical line passing through \( x = 4 \) on the x - axis. All points on this line have an x - coordinate of 4.

Step 2: Analyze the second equation

The equation \( y=-\frac{1}{2}x + 3 \) is in slope - intercept form \( y = mx + b \), where the slope \( m=-\frac{1}{2} \) and the y - intercept \( b = 3 \). To graph this line, we can start at the y - intercept \( (0,3) \) and then use the slope. The slope \( -\frac{1}{2} \) means we go down 1 unit and right 2 units (or up 1 unit and left 2 units) from the y - intercept to get other points on the line.

Step 3: Find the intersection point

To find the solution of the system, we need to find the point that lies on both lines. Since \( x = 4 \), we substitute \( x = 4 \) into the second equation \( y=-\frac{1}{2}x + 3 \).
Substitute \( x = 4 \) into \( y=-\frac{1}{2}x + 3 \):
\( y=-\frac{1}{2}(4)+3=-2 + 3=1 \)
So the lines \( x = 4 \) and \( y=-\frac{1}{2}x + 3 \) intersect at the point \( (4,1) \).

Step 1: Analyze the first equation \( y=x - 2 \)

This is in slope - intercept form \( y=mx + b \), with slope \( m = 1 \) and y - intercept \( b=-2 \). We start at the point \( (0,-2) \) and use the slope (up 1, right 1 or down 1, left 1) to plot other points on the line.

Step 2: Analyze the second equation \( y = 6x+3 \)

This is also in slope - intercept form, with slope \( m = 6 \) and y - intercept \( b = 3 \). We start at the point \( (0,3) \) and use the slope (up 6, right 1 or down 6, left 1) to plot other points on the line.

Step 3: Find the intersection point

To find the intersection, we can set the two equations equal to each other since they both equal \( y \):
\( x-2=6x + 3 \)
Subtract \( x \) from both sides: \( - 2=5x+3 \)
Subtract 3 from both sides: \( -5 = 5x \)
Divide both sides by 5: \( x=-1 \)
Now substitute \( x = - 1 \) into \( y=x - 2 \): \( y=-1-2=-3 \)
So the lines \( y=x - 2 \) and \( y = 6x+3 \) intersect at the point \( (-1,-3) \).

Step 1: Rewrite the equations in slope - intercept form

For the first equation \( 2x-y=-1 \), we can rewrite it as \( y=2x + 1 \) (add \( y \) to both sides and add 1 to both sides). The slope \( m = 2 \) and the y - intercept \( b = 1 \).
For the second equation \( 2x - y=-4 \), we can rewrite it as \( y=2x + 4 \). The slope \( m = 2 \) and the y - intercept \( b = 4 \).

Step 2: Analyze the lines

Since both lines have the same slope (\( m = 2 \)) but different y - intercepts (\( 1\) and \( 4\)), the two lines are parallel. Parallel lines never intersect.

Answer:

The solution of the system is \( (4,1) \)

Problem 2: Solve \(

$$\begin{cases} y=x - 2 \\ y = 6x+3 \end{cases}$$

\) by graphing