Question

worksheet
name:
per:
learning objectives:

1. solve a system of equations by graphing and check the solutions by substitution.
2. identify a system with infinite or no solutions by their graphs

for #1-6: solve each system of equations using the graphing method.

1. $\begin{cases} y = -x - 5 \\ y = x + 1 end{cases}$
2. $\begin{cases} -2x + y = 6 \\ y = -x end{cases}$
3. $\begin{cases} 4x - 2y = 8 \\ y = 2x + 3 end{cases}$
4. $\begin{cases} 2x + y = 6 \\ y = -2x + 6 end{cases}$
5. $\begin{cases} x = -3 \\ y = 2 end{cases}$
6. $\begin{cases} y = 3x - 4 \\ y = -5x + 1 end{cases}$

Explanation:

Problem 1:

Step1: Analyze the equations

We have the system \(

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

\). Both are in slope - intercept form (\(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept). The first line has a slope of \(- 1\) and a y - intercept of \(-5\). The second line has a slope of \(1\) and a y - intercept of \(1\).

Step2: Find the intersection

To find the solution, we can set the two equations equal to each other since they both equal \(y\):
\(-x - 5=x + 1\)
Add \(x\) to both sides: \(-5 = 2x+1\)
Subtract \(1\) from both sides: \(-6 = 2x\)
Divide both sides by \(2\): \(x=-3\)
Substitute \(x = - 3\) into \(y=x + 1\), we get \(y=-3 + 1=-2\)

Step3: Graphical verification (conceptual)

When we graph \(y=-x - 5\), we start at \((0,-5)\) and go down 1 unit and right 1 unit (or up 1 unit and left 1 unit) for the slope of \(-1\). For \(y=x + 1\), we start at \((0,1)\) and go up 1 unit and right 1 unit (or down 1 unit and left 1 unit) for the slope of \(1\). The two lines intersect at \((-3,-2)\)

Step1: Rewrite the first equation

The system is \(

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

\). Rewrite the first equation in slope - intercept form: \(y = 2x+6\). The slope of this line is \(2\) and the y - intercept is \(6\). The second line \(y=-x\) has a slope of \(-1\) and a y - intercept of \(0\).

Step2: Find the intersection

Set \(2x + 6=-x\)
Add \(x\) to both sides: \(3x+6 = 0\)
Subtract \(6\) from both sides: \(3x=-6\)
Divide by \(3\): \(x=-2\)
Substitute \(x = - 2\) into \(y=-x\), we get \(y = 2\)

Step3: Graphical verification (conceptual)

For \(y = 2x+6\), start at \((0,6)\) and go up 2 units and right 1 unit. For \(y=-x\), start at \((0,0)\) and go down 1 unit and right 1 unit. They intersect at \((-2,2)\)

Step1: Rewrite the first equation

The system is \(

$$\begin{cases}4x-2y = 8\\y = 2x+3\end{cases}$$

\). Rewrite the first equation:
\(-2y=-4x + 8\)
Divide both sides by \(-2\): \(y = 2x-4\)
The first line (after rewriting) has a slope of \(2\) and a y - intercept of \(-4\). The second line \(y = 2x+3\) has a slope of \(2\) and a y - intercept of \(3\)

Step2: Analyze the lines

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

Answer:

\((-3,-2)\)

Problem 2: