Question

use a graphing calculator to graph each system of equations. determine which quadrant the solution lies in.
i
ii
iii
iv
3x + 2y = 10
x - y = -3
y = 2x - 9
3x - 2y = -19

Answer:

To solve this, we analyze each system of equations:

System 1: \( 3x + 2y = 10 \) and let's assume the second equation (maybe a typo, but let's solve \( 3x + 2y = 10 \))

Let's solve for \( y \): \( 2y = -3x + 10 \Rightarrow y = -\frac{3}{2}x + 5 \)
This line has a negative slope and y - intercept 5.

System 2: \( x - y = -3 \)

Rewrite as \( y = x + 3 \). This has a positive slope and y - intercept 3.

System 3: \( y = 2x - 9 \)

This is a line with positive slope 2 and y - intercept - 9.

System 4: \( 3x - 2y = - 19 \)

Rewrite as \( 2y=3x + 19\Rightarrow y=\frac{3}{2}x+\frac{19}{2} \). Positive slope and positive y - intercept.

Now, let's find the intersection (solution) for each pair (assuming we pair them correctly, maybe the original problem has pairs of equations for each system). But since we need to find the quadrant of the solution:

• For a system with equations like \( y = x + 3 \) and \( y=2x - 9 \) (just an example of pairing), set \( x + 3=2x - 9\Rightarrow x = 12,y=15 \) (Quadrant I, since \( x>0,y>0 \))
• For a system with \( 3x + 2y = 10 \) and \( 3x - 2y=-19 \), add the two equations: \( 6x=-9\Rightarrow x =-\frac{3}{2},y=\frac{3x + 19}{2}=\frac{3\times(-\frac{3}{2})+19}{2}=\frac{- \frac{9}{2}+19}{2}=\frac{\frac{-9 + 38}{2}}{2}=\frac{29}{4}\) (Quadrant II, \( x<0,y>0 \))
• For a system with \( y = 2x-9 \) and \( x - y=-3 \) ( \( y=x + 3 \) ), set \( 2x-9=x + 3\Rightarrow x = 12,y = 15 \) (Quadrant I)
• For a system with \( 3x + 2y = 10 \) and \( y=x + 3 \), set \( x + 3=-\frac{3}{2}x+5\Rightarrow x+\frac{3}{2}x=5 - 3\Rightarrow\frac{5}{2}x = 2\Rightarrow x=\frac{4}{5},y=\frac{4}{5}+3=\frac{19}{5}\) (Quadrant I)

But since the problem is about determining the quadrant of the solution for each system:

• If we consider the system \( 3x + 2y=10 \) and \( 3x - 2y = - 19 \), the solution is \( x=-\frac{3}{2},y=\frac{29}{4} \) (Quadrant II)
• For \( x - y=-3 \) and \( y = 2x-9 \), solution is \( x = 12,y = 15 \) (Quadrant I)
• For \( 3x+2y = 10 \) and \( y=x + 3 \), solution is \( x=\frac{4}{5},y=\frac{19}{5} \) (Quadrant I)
• For \( x - y=-3 \) and \( 3x - 2y=-19 \), let's solve: from \( y=x + 3 \), substitute into \( 3x-2(x + 3)=-19\Rightarrow3x-2x-6=-19\Rightarrow x=-13,y=-10 \) (Quadrant III, \( x<0,y<0 \))

Since the problem is a bit unclear due to formatting, but if we assume the systems are:

1. \( 3x + 2y = 10 \) and \( x - y=-3 \)

• Solve \( 3x + 2y = 10 \) and \( y=x + 3 \)
• Substitute \( y=x + 3 \) into \( 3x + 2(x + 3)=10\Rightarrow3x+2x + 6 = 10\Rightarrow5x=4\Rightarrow x=\frac{4}{5},y=\frac{19}{5} \) (Quadrant I)

1. \( x - y=-3 \) and \( y = 2x-9 \)

• \( x + 3=2x-9\Rightarrow x = 12,y = 15 \) (Quadrant I)

1. \( y = 2x-9 \) and \( 3x - 2y=-19 \)

• Substitute \( y = 2x-9 \) into \( 3x-2(2x - 9)=-19\Rightarrow3x-4x + 18=-19\Rightarrow - x=-37\Rightarrow x = 37,y=2\times37-9 = 65 \) (Quadrant I)

1. \( 3x + 2y = 10 \) and \( 3x - 2y=-19 \)

• Add the equations: \( 6x=-9\Rightarrow x=-\frac{3}{2},y=\frac{3x + 19}{2}=\frac{3\times(-\frac{3}{2})+19}{2}=\frac{- \frac{9}{2}+19}{2}=\frac{\frac{29}{2}}{2}=\frac{29}{4} \) (Quadrant II)

If we assume the four systems (I, II, III, IV) have solutions:

• System I: Solution in Quadrant I ( \( x>0,y>0 \))
• System II: Solution in Quadrant I ( \( x>0,y>0 \))
• System III: Solution in Quadrant I ( \( x>0,y>0 \))
• System IV: Solution in Quadrant II ( \( x<0,y>0 \))

But since the problem is not fully clear in terms of equation pairing, a more accurate way is to solve each system properly. However, if we take the system \( 3x + 2y = 10 \) and \( 3x - 2y=-19 \):

1. Add the two equations:

• \( (3x + 2y)+(3x - 2y)=10+(-19) \)
• \( 6x=-9\Rightarrow x =-\frac{3}{2}=-1.5 \)

1. Substitute \( x = - 1.5 \) into \( 3x + 2y = 10 \)

• \( 3\times(-1.5)+2y = 10\Rightarrow - 4.5+2y = 10\Rightarrow2y=14.5\Rightarrow y = 7.25 \)

Since \( x=-1.5<0 \) and \( y = 7.25>0 \), the solution is in Quadrant II

For the system \( x - y=-3 \) ( \( y=x + 3 \)) and \( y = 2x-9 \):

1. Set \( x + 3=2x-9 \)

• \( x=12 \)

1. Substitute \( x = 12 \) into \( y=x + 3 \)

• \( y = 15 \)

Since \( x = 12>0 \) and \( y = 15>0 \), the solution is in Quadrant I

If we consider the system \( 3x + 2y = 10 \) and \( y=x + 3 \):

1. Substitute \( y=x + 3 \) into \( 3x + 2y = 10 \)

• \( 3x+2(x + 3)=10\Rightarrow5x + 6 = 10\Rightarrow5x = 4\Rightarrow x = 0.8 \)

1. \( y=0.8 + 3=3.8 \)

Solution in Quadrant I

For the system \( y = 2x-9 \) and \( 3x - 2y=-19 \):

1. Substitute \( y = 2x-9 \) into \( 3x - 2y=-19 \)

• \( 3x-2(2x - 9)=-19\Rightarrow3x-4x + 18=-19\Rightarrow - x=-37\Rightarrow x = 37 \)

1. \( y=2\times37-9 = 65 \)

Solution in Quadrant I

So, depending on the system, the quadrants are:

• For \( 3x + 2y = 10 \) and \( 3x - 2y=-19 \): Quadrant II
• For the other systems (with positive x and y values): Quadrant I

If we assume the four systems (I, II, III, IV) have solutions:

• System I: Quadrant I
• System II: Quadrant I
• System III: Quadrant I
• System IV: Quadrant II

(Note: The exact pairing of equations is a bit unclear from the given image, but the above is a way to determine the quadrant of the solution for a system of linear equations by finding the intersection point \((x,y)\) and checking the signs of \(x\) and \(y\) to determine the quadrant:

• Quadrant I: \( x>0,y>0 \)
• Quadrant II: \( x<0,y>0 \)
• Quadrant III: \( x<0,y<0 \)
• Quadrant IV: \( x>0,y<0 \))