Question

linear systems unit – ls #2 solving linear systems graphically (pencil and paper) a system of equations consists of two or more equations that are considered together. in our course we will be working with a system of 2 equations. a solution to a system of equations is the ordered pair of the point where the two lines meet on the graph (if they meet). ex: solve by graphing. a) ( y = 1 - 2x ) ( y = x - 5 ) these lines are in slope y-intercept form. this means, to graph them, start by plotting the y-intercept, then, use the slope to find other points on the line. b) ( x + y = 3 ) ( 2x + 2y = 6 ) these lines are in x and y intercept form. this means, to graph them, make your x and y intercept chart, then plot your x and y intercept and draw a line through them. c) ( y = 2x + 3 ) ( y = 2x - 4 )

Explanation:

Part (a)

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

This is in slope - intercept form \( y=mx + b \), where \( m=-2 \) (slope) and \( b = 1 \) (y - intercept). To graph, start at \( (0,1) \), then use the slope: from \( (0,1) \), go down 2 units and right 1 unit (or up 2 units and left 1 unit) to get another point, e.g., \( (1,-1) \).

Step 2: Analyze \( y=x - 5 \)

This is in slope - intercept form with \( m = 1 \) (slope) and \( b=-5 \) (y - intercept). Start at \( (0,-5) \), then use the slope: from \( (0,-5) \), go up 1 unit and right 1 unit (or down 1 unit and left 1 unit) to get another point, e.g., \( (1,-4) \).

Step 3: Find the intersection

Graph both lines. The point where they intersect is the solution. By graphing, we can see that the two lines intersect at \( x = 2 \), \( y=-3 \). We can also solve algebraically: set \( 1-2x=x - 5 \), then \( 1 + 5=3x \), \( 6 = 3x \), \( x = 2 \). Substitute \( x = 2 \) into \( y=x - 5 \), \( y=2-5=-3 \).

Step 1: Rewrite \( x + y=3 \) in slope - intercept form

Solve for \( y \): \( y=-x + 3 \). Here, \( m=-1 \) (slope) and \( b = 3 \) (y - intercept).

Step 2: Rewrite \( 2x+2y = 6 \) in slope - intercept form

Divide both sides by 2: \( x + y=3 \), which is the same as \( y=-x + 3 \). So the two equations represent the same line.

Step 3: Determine the solution

Since the two equations are equivalent, they have infinitely many solutions (all points on the line \( y=-x + 3 \)).

Step 1: Analyze \( y = 2x+3 \)

This is in slope - intercept form with \( m = 2 \) (slope) and \( b = 3 \) (y - intercept). Start at \( (0,3) \), then use the slope: from \( (0,3) \), go up 2 units and right 1 unit (or down 2 units and left 1 unit) to get another point, e.g., \( (1,5) \).

Step 2: Analyze \( y = 2x-4 \)

This is in slope - intercept form with \( m = 2 \) (slope) and \( b=-4 \) (y - intercept). Start at \( (0,-4) \), then use the slope: from \( (0,-4) \), go up 2 units and right 1 unit (or down 2 units and left 1 unit) to get another point, e.g., \( (1,-2) \).

Step 3: Determine the relationship

Since the two lines have the same slope (\( m = 2 \)) but different y - intercepts (\( 3
eq - 4 \)), they are parallel lines. Parallel lines never intersect.

Answer:

The solution is \( (2,-3) \)

Part (b)