Question

look at the system of inequalities.
$y \geq 3x$
$y \leq x + 6$
$y \leq 7x$
the solution set is the triangular region where all the inequalities are true.
what are the vertices of that triangular region?
$(\square, \square)$
$(\square, \square)$
$(\square, \square)$

Explanation:

Step1: Encontrar la intersección de \( y \geq 3x \) y \( y \leq x + 6 \)

Igualamos las dos ecuaciones: \( 3x = x + 6 \)
Resolviendo para \( x \): \( 3x - x = 6 \) → \( 2x = 6 \) → \( x = 3 \)
Sustituimos \( x = 3 \) en \( y = 3x \): \( y = 3(3) = 9 \)? No, espera, \( y = 3x \) y \( y = x + 6 \), entonces \( 3x = x + 6 \) → \( x = 3 \), \( y = 3(3) = 9 \)? Wait, no, \( y = x + 6 = 3 + 6 = 9 \). So the point is \( (3, 9) \)? Wait, no, wait, let's check again. Wait, \( y \geq 3x \) and \( y \leq x + 6 \). So when we solve \( 3x = x + 6 \), we get \( x = 3 \), \( y = 9 \). But wait, maybe I made a mistake. Wait, let's check the other intersections.

Step2: Encontrar la intersección de \( y \geq 3x \) y \( y \leq 7x \)

Igualamos \( 3x = 7x \) → \( 4x = 0 \) → \( x = 0 \), entonces \( y = 0 \). So the point is \( (0, 0) \).

Step3: Encontrar la intersección de \( y \leq x + 6 \) y \( y \leq 7x \)

Igualamos \( x + 6 = 7x \) → \( 6 = 6x \) → \( x = 1 \)
Sustituimos \( x = 1 \) en \( y = 7x \): \( y = 7(1) = 7 \). Wait, or \( y = x + 6 = 1 + 6 = 7 \). So the point is \( (1, 7) \)? Wait, no, let's do it correctly.

Wait, let's re-examine:

1. Intersection of \( y = 3x \) and \( y = 7x \): \( 3x = 7x \) → \( 4x = 0 \) → \( x = 0 \), \( y = 0 \). So (0, 0).

2. Intersection of \( y = 3x \) and \( y = x + 6 \): \( 3x = x + 6 \) → \( 2x = 6 \) → \( x = 3 \), \( y = 3*3 = 9 \). Wait, but \( y = x + 6 = 3 + 6 = 9 \). So (3, 9)? Wait, no, that can't be, because \( y \leq 7x \), so when \( x = 3 \), \( 7x = 21 \), so \( y = 9 \) is less than 21, so that's okay. But wait, maybe the third intersection is between \( y = x + 6 \) and \( y = 7x \). Let's solve \( x + 6 = 7x \) → \( 6 = 6x \) → \( x = 1 \), then \( y = 7(1) = 7 \), and \( y = 1 + 6 = 7 \), so that's (1, 7).

Wait, but let's check the regions. The inequalities are \( y \geq 3x \) (above the line \( y = 3x \)), \( y \leq x + 6 \) (below the line \( y = x + 6 \)), and \( y \leq 7x \) (below the line \( y = 7x \)). So the vertices are where two of these lines intersect, and are part of the solution region.

So first, intersection of \( y = 3x \) and \( y = 7x \): (0, 0).

Second, intersection of \( y = 3x \) and \( y = x + 6 \): solve \( 3x = x + 6 \) → \( x = 3 \), \( y = 9 \). So (3, 9). Wait, but does \( y = 9 \) satisfy \( y \leq 7x \)? \( 7x = 21 \), so 9 ≤ 21, yes.

Third, intersection of \( y = x + 6 \) and \( y = 7x \): solve \( x + 6 = 7x \) → \( x = 1 \), \( y = 7 \). So (1, 7).

Wait, but let's verify with the inequalities. For (0, 0): \( 0 \geq 0 \) (yes, \( y \geq 3x \)), \( 0 \leq 0 + 6 \) (yes), \( 0 \leq 0 \) (yes, \( y \leq 7x \)). Good.

For (1, 7): \( 7 \geq 3(1) = 3 \) (yes), \( 7 \leq 1 + 6 = 7 \) (yes), \( 7 \leq 7(1) = 7 \) (yes). Good.

For (3, 9): \( 9 \geq 3(3) = 9 \) (yes, equality), \( 9 \leq 3 + 6 = 9 \) (yes, equality), \( 9 \leq 7(3) = 21 \) (yes). Good.

Wait, but maybe I made a mistake in the second intersection. Wait, let's check again. Let's graph the lines:

- \( y = 3x \): passes through (0,0) with slope 3.

- \( y = x + 6 \): passes through (0,6) with slope 1.

- \( y = 7x \): passes through (0,0) with slope 7.

So the region is bounded by these three lines. The vertices are where two lines intersect and are part of the feasible region.

So:

1. Intersection of \( y = 3x \) and \( y = 7x \): (0, 0).

2. Intersection of \( y = 3x \) and \( y = x + 6 \): (3, 9).

3. Intersection of \( y = x + 6 \) and \( y = 7x \): (1, 7).

Wait, but let's check if (1, 7) is indeed a vertex. Let's see the region: above \( y = 3x \), below \( y = x + 6 \), and below \( y = 7x \). So between \( x = 0 \) and \( x = 1 \), the upper boundary is \( y = 7x \) (since \( 7x \) is above \( 3x \) when \( x > 0 \)), and between \( x = 1 \) and \( x = 3 \), the upper boundary is \( y = x + 6 \) (since \( x + 6 \) is above \( 7x \) when \( x < 1 \)? Wait, no, when \( x = 1 \), \( 7x = 7 \) and \( x + 6 = 7 \), so they meet at (1,7). For \( x < 1 \), say \( x = 0.5 \), \( 7x = 3.5 \), \( x + 6 = 6.5 \), so \( 7x < x + 6 \), so the upper boundary is \( x + 6 \)? Wait, no, the inequality is \( y \leq 7x \) and \( y \leq x + 6 \), so the lower of the two. So for \( x < 1 \), \( 7x < x + 6 \), so \( y \leq 7x \). For \( x > 1 \), \( x + 6 < 7x \), so \( y \leq x + 6 \). And \( y \geq 3x \) always. So the region is bounded by \( y = 3x \) (lower), \( y = 7x \) (upper for \( x \in [0,1] \)), and \( y = x + 6 \) (upper for \( x \in [1,3] \)). So the vertices are:

- (0, 0): intersection of \( y = 3x \) and \( y = 7x \).

- (1, 7): intersection of \( y = 7x \) and \( y = x + 6 \).

- (3, 9): intersection of \( y = x + 6 \) and \( y = 3x \).

Yes, that makes sense. So the three vertices are (0, 0), (1, 7), and (3, 9).

Wait, but let's confirm with another approach. Let's list all pairs of lines:

1. \( y = 3x \) and \( y = x + 6 \): solve \( 3x = x + 6 \) → \( x = 3 \), \( y = 9 \) → (3, 9).

2. \( y = 3x \) and \( y = 7x \): solve \( 3x = 7x \) → \( x = 0 \), \( y = 0 \) → (0, 0).

3. \( y = x + 6 \) and \( y = 7x \): solve \( x + 6 = 7x \) → \( x = 1 \), \( y = 7 \) → (1, 7).

All three points are in the solution region:

- (0,0): \( 0 \geq 0 \), \( 0 \leq 0 + 6 \), \( 0 \leq 0 \) → yes.

- (1,7): \( 7 \geq 3(1) = 3 \), \( 7 \leq 1 + 6 = 7 \), \( 7 \leq 7(1) = 7 \) → yes.

- (3,9): \( 9 \geq 3(3) = 9 \), \( 9 \leq 3 + 6 = 9 \), \( 9 \leq 7(3) = 21 \) → yes.

So those are the vertices.

Answer:

(0, 0)
(1, 7)
(3, 9)