Question

quick review
systems of equations can be solved by looking at their graphs. a system with one solution has one point of intersection. a system with infinitely many solutions has infinite points of intersection. a system with no solution has no points of intersection.
example
graph the system and determine its solution.
( y = x + 4 )
( y = - 2 x + 1 )
graph each equation in the system on the same coordinate plane.
the point of intersection is ( ( - 1, 3 ) ). this means the solution to the system is ( ( - 1, 3 ) ).
practice
graph each system and find the solution(s).

1. ( y = \frac { 1 } { 2 } x + 1 )

( - 2 x + 4 y = 4 )

1. ( y = - x - 3 )

( y + x = 2 )

1. ( 2 y = 6 x + 4 )

( y = - 2 x + 2 )

Explanation:

Step1: Analyze the first system

For the system \(y=\frac{1}{2}x + 1\) and \(-2x + 4y=4\), rewrite the second equation.

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

Since the two equations are identical, every point on the line is a solution.

Step2: Analyze the second system

For the system \(y=-x - 3\) and \(y + x=2\) (or \(y=-x + 2\)).
The slope - intercept form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the \(y\) - intercept. Here \(m=-1\) for both lines, but \(b=-3\) and \(b = 2\). Parallel lines (same slope, different \(y\) - intercepts) have no intersection.

Step3: Analyze the third system

For the system \(2y=6x + 4\) (or \(y = 3x+2\)) and \(y=-2x + 2\).
Set \(3x+2=-2x + 2\) (equating the two \(y\) values).

$$\begin{align*} 3x+2x&=2 - 2\\ 5x&=0\\ x&=0 \end{align*}$$

Substitute \(x = 0\) into \(y=-2x + 2\) to get \(y = 2\). The point \((0,2)\) is the intersection of the two lines.

Answer:

1. Solution:

• First, rewrite the second equation \(-2x + 4y=4\) in slope - intercept form (\(y=mx + b\)).
• Add \(2x\) to both sides: \(4y=2x + 4\).
• Divide by \(4\): \(y=\frac{1}{2}x + 1\).
• Since both equations \(y=\frac{1}{2}x + 1\) and \(y=\frac{1}{2}x + 1\) are the same, the system has infinitely many solutions.

1. Solution:

• Rewrite the second equation \(y + x=2\) as \(y=-x + 2\).
• The first equation is \(y=-x-3\).
• The slopes of both equations (\(m=-1\)) are the same, but the \(y\) - intercepts (\(b=-3\) and \(b = 2\)) are different.
• So, the system has no solution.

1. Solution:

• Rewrite the first equation \(2y=6x + 4\) in slope - intercept form. Divide by \(2\): \(y = 3x+2\).
• The second equation is \(y=-2x + 2\).
• Set the two equations equal to each other: \(3x+2=-2x + 2\).
• Add \(2x\) to both sides: \(3x+2x+2=-2x+2x + 2\), so \(5x+2=2\).
• Subtract \(2\) from both sides: \(5x+2 - 2=2 - 2\), so \(5x=0\).
• Divide by \(5\): \(x = 0\).
• Substitute \(x = 0\) into \(y=-2x + 2\): \(y=-2(0)+2=2\).
• The solution of the system is \((0,2)\).