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The graph of the ineequation 2x + 3y ge ...

The graph of the ineequation `2x + 3y ge 6` does not lie in the first quadrant.(True/False)

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To determine whether the graph of the inequality \(2x + 3y \geq 6\) lies in the first quadrant, we will follow these steps: ### Step 1: Rewrite the Inequality We start with the inequality: \[ 2x + 3y \geq 6 \] ### Step 2: Find the Boundary Line To analyze the inequality, we first consider the corresponding equation: \[ 2x + 3y = 6 \] This equation represents the boundary line of the inequality. ### Step 3: Find Intercepts To plot the line, we find the x-intercept and y-intercept. - **X-intercept**: Set \(y = 0\): \[ 2x + 3(0) = 6 \implies 2x = 6 \implies x = 3 \] So, the x-intercept is \( (3, 0) \). - **Y-intercept**: Set \(x = 0\): \[ 2(0) + 3y = 6 \implies 3y = 6 \implies y = 2 \] So, the y-intercept is \( (0, 2) \). ### Step 4: Plot the Line Now we plot the points \( (3, 0) \) and \( (0, 2) \) on the graph. Draw a straight line through these points. Since the inequality is \( \geq \), we will use a solid line to indicate that points on the line are included in the solution set. ### Step 5: Determine the Region To determine which side of the line satisfies the inequality \(2x + 3y \geq 6\), we can test a point not on the line. A common choice is the origin \( (0, 0) \): \[ 2(0) + 3(0) = 0 \not\geq 6 \] Since the origin does not satisfy the inequality, the solution set lies on the opposite side of the line. ### Step 6: Analyze the Quadrants The first quadrant is where both \(x\) and \(y\) are positive. The line divides the plane, and since the region satisfying the inequality does not include the origin, we need to check if there are points in the first quadrant that satisfy the inequality. ### Step 7: Conclusion The region defined by \(2x + 3y \geq 6\) does not include any points in the first quadrant, as we can see from the graph. Therefore, the statement "The graph of the inequality \(2x + 3y \geq 6\) does not lie in the first quadrant" is **True**.
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Knowledge Check

  • The solution set formed by the inequations x + y ge3, x + y le 4, x le 2 in the first quadrant represents a

    A
    triangle
    B
    parallelogram
    C
    rectangle
    D
    rhombus
  • The graph of 2x-3 ge 0 is

    A
    B
    C
    D
  • The area of the region bounded by x^(2)=y-2, y=4, y=6 and the Y-axis in the first quadrant is

    A
    `2/3`
    B
    `2/3(8-sqrt(2))`
    C
    `2/3(8-2sqrt(2))`
    D
    `3/2(8-sqrt(2))`
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