Find two consecutive even numbers that sum to 270. Also, describe the region R defined by x >= -1, 3y - x >= -5, and 3x + 5y < 15.

Let the first even number be $x$.
Since the numbers are consecutive even numbers, the next even number is $x+2$.

The sum of the two numbers is 270:

The first number is $x = 134$.
The second number is $x+2 = 134+2 = 136$.

The two consecutive even numbers are $\boxed{\text{134 and 136}}$.

:
To draw and label the region R, we first need to draw the boundary lines for each inequality.

The boundary line is $x = -1$. This is a solid vertical line passing through $x=-1$.
The region $x \ge -1$ is to the right of this line.

The boundary line is $3y - x = -5$. This is a solid line.
To draw it, find two points:

• If $x=0$, $3y = -5 \implies y = -\frac{5}{3} \approx -1.67$. Point: $(0, -\frac{5}{3})$.
• If $y=0$, $-x = -5 \implies x = 5$. Point: $(5, 0)$. Plot these points and draw a solid line through them. To determine the region, test a point, e.g., $(0,0)$: $3(0) - 0 \ge -5 \implies 0 \ge -5$, which is true. So, the region $3y - x \ge -5$ is above the line (containing the origin).

The boundary line is $3x + 5y = 15$. This is a dashed line because of the strict inequality $(<)$.
To draw it, find two points:

• If $x=0$, $5y = 15 \implies y = 3$. Point: $(0, 3)$.
• If $y=0$, $3x = 15 \implies x = 5$. Point: $(5, 0)$. Plot these points and draw a dashed line through them. To determine the region, test a point, e.g., $(0,0)$: $3(0) + 5(0) < 15 \implies 0 < 15$, which is true. So, the region $3x + 5y < 15$ is below the line (containing the origin).

Region R is the area where all three shaded regions overlap. It will be a triangular region bounded by the lines $x=-1$, $3y-x=-5$, and $3x+5y=15$.

• The region is to the right of $x=-1$.
• The region is above $3y-x=-5$.
• The region is below $3x+5y=15$. Shade this common region and label it R.

:

Let the original number of males be $M$ and females be $F$.
The ratio of males to females is 2:3.

This means $M = 2k$ and $F = 3k$ for some constant $k$.

On Tuesday:

• Ten males were absent, so the number of males present is $M - 10 = 2k - 10$.
• Six new female members joined, so the number of females present is $F + 6 = 3k + 6$.

On Tuesday, the ratio of males to females was 1:3.

Cross-multiply:

Subtract $3k$ from both sides:

Add 30 to both sides:

Divide by 3:

Original number of males $M = 2k = 2(12) = 24$.
Original number of females $F = 3k = 3(12) = 36$.
The total number of regular members is $M + F = 24 + 36 = 60$.

The group has $\boxed{\text{60 regular members}}$.