Question

3
$y > 2x - 4$
$6x + 3y < 9$
(grid coordinate system)
.a. $(0, 0)$
.b. $(-3, -7)$

Explanation:

Step1: Analyze the first inequality \( y > 2x - 4 \)

We can test the points in this inequality. For a point \((x,y)\), substitute into the inequality to see if it holds.

Step2: Analyze the second inequality \( 6x + 3y < 9 \)

Simplify this inequality first. Divide all terms by 3: \( 2x + y < 3 \), or \( y < -2x + 3 \).

Step3: Test point a: \((0,0)\)

• For \( y > 2x - 4 \): Substitute \( x = 0, y = 0 \). We get \( 0 > 2(0) - 4 \) which is \( 0 > -4 \). This is true.
• For \( y < -2x + 3 \): Substitute \( x = 0, y = 0 \). We get \( 0 < -2(0) + 3 \) which is \( 0 < 3 \). This is true. So \((0,0)\) satisfies both inequalities.

Step4: Test point b: \((-3,-7)\)

• For \( y > 2x - 4 \): Substitute \( x = -3, y = -7 \). We get \( -7 > 2(-3) - 4 \) which is \( -7 > -6 - 4 \) or \( -7 > -10 \). This is true.
• For \( y < -2x + 3 \): Substitute \( x = -3, y = -7 \). We get \( -7 < -2(-3) + 3 \) which is \( -7 < 6 + 3 \) or \( -7 < 9 \). This is also true? Wait, maybe there was a miscalculation. Wait, let's re - check the second inequality for \((-3,-7)\):

Wait, the original second inequality is \( 6x + 3y < 9 \). Substitute \( x=-3,y = -7 \):

\( 6(-3)+3(-7)=-18 - 21=-39 \). And \( -39 < 9 \), which is true. Wait, but maybe the question is about graphing or identifying the solution region. But from the given, for point a \((0,0)\), as we saw, it satisfies both inequalities. For point b \((-3,-7)\), let's check the first inequality \( y>2x - 4 \): \( -7>2(-3)-4=-6 - 4=-10 \), which is true. And \( 6(-3)+3(-7)=-18 - 21=-39 < 9 \), which is true. But maybe the user's hand - written notes have some context. But from the testing, \((0,0)\) satisfies both, and \((-3,-7)\) also satisfies? Wait, maybe the question is to determine which point is in the solution set. But based on the calculations, \((0,0)\) is in the solution set (as both inequalities are satisfied) and \((-3,-7)\) is also in? Wait, maybe I made a mistake. Wait, let's re - check the second inequality simplification. The original inequality is \( 6x + 3y < 9 \). Divide by 3: \( 2x + y < 3 \), so \( y < - 2x+3 \). For \((-3,-7)\), \( y=-7 \), \( -2x + 3=-2(-3)+3 = 6 + 3=9 \). And \( -7 < 9 \), which is true. And for the first inequality, \( y=-7 \), \( 2x - 4=2(-3)-4=-10 \), and \( -7 > -10 \), which is true. But maybe the graph (not fully visible) has a region where \((0,0)\) is in and \((-3,-7)\) is not? Wait, maybe the initial analysis of the graph is needed. But from the algebraic test, \((0,0)\) satisfies both inequalities.

Answer:

For point a \((0,0)\), it satisfies both \( y > 2x - 4 \) and \( 6x + 3y < 9 \). For point b \((-3,-7)\), it also satisfies both inequalities algebraically. But if we consider the graph (partially shown), assuming the standard graphing of these inequalities:

The first inequality \( y>2x - 4 \) is a dashed line with slope 2 and y - intercept - 4, shading above. The second inequality \( 6x + 3y < 9 \) (or \( y < - 2x+3 \)) is a dashed line with slope - 2 and y - intercept 3, shading below. The intersection region is where we shade above \( y = 2x - 4 \) and below \( y=-2x + 3 \).

Testing \((0,0)\): In the intersection region (above \( y = 2x-4 \) and below \( y=-2x + 3 \)).

Testing \((-3,-7)\): Let's see the y - value for \( y = 2x-4 \) at \( x=-3 \) is \( -10 \), and \( y=-7>-10 \) (good for first inequality). For \( y=-2x + 3 \) at \( x = - 3 \) is \( 9 \), and \( y=-7 < 9 \) (good for second inequality). But maybe the graph has a different scale. However, from the algebraic test, \((0,0)\) is a valid solution point.