Question

solve each system by graphing and sketch the graph. 1) $y = 3x + 4$ $y = -4x - 3$ 2) $y = -\frac{3}{2}x - 1$ 4) $x = -3$

Answer:

Problem 1: Solve \(

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

\) by Graphing

Step 1: Analyze the first line \( y = 3x + 4 \)

This is a linear equation in slope - intercept form (\( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept). For \( y = 3x+4 \), the slope \( m_1 = 3 \) and the y - intercept \( b_1=4 \). To graph this line, we start by plotting the y - intercept at \( (0,4) \). Then, using the slope (rise over run), since the slope is 3 (or \( \frac{3}{1} \)), we move up 3 units and right 1 unit from \( (0,4) \) to get another point, say \( (1,7) \), or down 3 units and left 1 unit to get \( (- 1,1) \).

Step 2: Analyze the second line \( y=-4x - 3 \)

For the equation \( y=-4x - 3 \), the slope \( m_2=-4 \) and the y - intercept \( b_2 = - 3 \). We plot the y - intercept at \( (0,-3) \). Using the slope (which can be thought of as \( \frac{-4}{1} \) or \( \frac{4}{-1} \)), we move down 4 units and right 1 unit from \( (0,-3) \) to get \( (1,-7) \), or up 4 units and left 1 unit to get \( (-1,1) \).

Step 3: Find the intersection point

When we graph both lines, we look for the point where they cross each other. To find the exact intersection (we can also solve the system algebraically to confirm), we set the two equations equal to each other since they both equal \( y \):
\( 3x + 4=-4x-3 \)
Add \( 4x \) to both sides: \( 3x + 4x+4=-4x + 4x-3 \), which simplifies to \( 7x+4=-3 \)
Subtract 4 from both sides: \( 7x+4 - 4=-3 - 4 \), so \( 7x=-7 \)
Divide both sides by 7: \( x=-1 \)
Now substitute \( x = - 1 \) into one of the original equations, say \( y = 3x + 4 \). Then \( y=3(-1)+4=-3 + 4 = 1 \).

So the solution to the system is the point where the two lines intersect, which is \( (-1,1) \).

Problem 2: (Incomplete in the given image, but we can explain the general process for \( y=-\frac{3}{2}x - 1 \) and a missing equation)

For a linear system with \( y =-\frac{3}{2}x-1 \) and another linear equation (let's assume the missing equation is also linear), we would:

1. Analyze the slope - intercept form of \( y =-\frac{3}{2}x-1 \): slope \( m=-\frac{3}{2} \), y - intercept \( b=-1 \). Plot the y - intercept at \( (0,-1) \), then use the slope (down 3, right 2 or up 3, left 2) to find other points.
2. Analyze the second (missing) linear equation in the same way (find slope and y - intercept, plot points).
3. Find the intersection point of the two lines (either by graphing or by solving the system algebraically by setting the two equations equal to each other if they are both solved for \( y \)).

Problem 4: (Incomplete in the given image, but for \( x=-3 \) and a likely equation \( x - y=-4 \) or \( y=-4 \) as per the partial text)

If the system is \(

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

\) (assuming the partial text \( x - y=-4 \) is a typo or miswriting and it's \( y=-4 \)):

• The line \( x=-3 \) is a vertical line passing through \( x=-3 \) (all points with \( x=-3 \) and any \( y \) - value).
• The line \( y=-4 \) is a horizontal line passing through \( y = - 4 \) (all points with \( y=-4 \) and any \( x \) - value).
• The intersection of a vertical line \( x=-3 \) and a horizontal line \( y=-4 \) is the point \( (-3,-4) \).

Final Answers (for the complete system in Problem 1)

The solution to the system \(

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

\) is the point of intersection of the two lines, which is \( \boldsymbol{(-1,1)} \).

(For the other problems, we need the complete system of equations to provide a full solution. If you can provide the missing parts of the equations for Problem 2 and Problem 4, we can solve them completely.)