Solve the inequality 3(x + 1) + 2(x + 2) > 0.
1 answer:
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Answer:
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
To solve this problem, we must imagine the triangles and parallel lines which are formed. It is best to draw the triangle described in the problem so that you can clearly understand what I will be talking about.
The first step we have to do is to make an equality equation in triangle ABC.
In triangle ABC, we are given that lines XY and BC are two parallel lines (XY || BC). Therefore this means that:
AX / XB = AY / YC ---> 1
The next step is to make an equality equation in triangle AXC.
We are given that lines ZY and XC are two parallel lines (ZY || XC). Therefore this also means that:
AZ / ZX = AY / YC ---> 2
Combining 1 and 2 since they have both AY / YC in common:
AX / XB = AZ / ZX
we are given that:
AZ = 8, ZX = 4 therefore AX = AZ + ZX = 12, hence
12 / XB = 8 / 4
XB = 6
