Daerah Penyelesaian Sistem Pertidaksamaan Linear

Hello everyone! Today we are solving question thirty-eight. We need to find the points that define the boundary of the solution area for a system of linear inequalities as shown in the original problem image here.

To determine the solution area, we have four inequalities: one, x plus y is less than or equal to eight; two, two x minus y is greater than or equal to one; three, x is greater than or equal to one; and four, y is greater than or equal to zero. When we graph these on our axes, they form this shaded polygon.

Let's find the bottom boundary points where y equals zero. On the line x equals one, we get the vertex one, zero. On the line x plus y equals eight, setting y to zero gives us x equals eight, resulting in the vertex eight, zero.

Now for the left boundary where x equals one, we substitute this into the equation two x minus y equals one. This gives us two times one minus y equals one, which simplifies to two minus one equals y. So, y equals one, giving us the point one, one.

Finally, we find the intersection point of the two diagonal lines. We solve the system where x plus y equals eight and two x minus y equals one. By substituting eight minus y for x, we find three y equals fifteen, so y equals five and x equals three. This gives us the point three, five.

In conclusion, our boundary points are one, zero; eight, zero; three, five; and one, one. Comparing these vertices to our choices, we can see that the correct answer is option B. Thanks for watching!