Should any factors be accounted for when explaining how to solve an equation?
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The system of inequalities is the following:
i) <span>y ≤ –0.75x
ii)</span><span>y ≤ 3x – 2
since </span>
, we can write the system again as
Whenever we are asked to sketch the solution of a system of linear inequalities, we:
1. Draw the lines
2. Color the regions of the inequalities.
3. The solution is the region colored twice.
A.
to draw the line
consider the points: (-4, 1) and (0, 0), or any 2 other points (x,y) for which hold.
since we have an "smaller or equal to" inequality, the line is a solid line (not dashed, or dotted).
In order to find out which region of the line to color, consider a point not on the line, for example P(1, 1), which is clearly in the upper region of the line.
For (x, y)=(1, 1) the inequality
, does not hold because
is not true,
this means that the solution is the region of the line not containing (1, 1), as shown in picture 1.
B.
similarly, to draw the solution of inequality ii) y ≤ 3x – 2,
we first draw the line y=3x-2, using the points (0, -2) and (2, 4), or any other 2 points (x,y) for which y=3x-2 holds.
after we draw the line, we can check the point P(1, 7) which clearly is above the line y=3x-2.
for (x, y) = (1, 7), the inequality y ≤ 3x – 2 does not hold
because 7 is not ≤ 3*1-2=1, so the region we color is the one not containing P(1, 7), as shown in picture 2.
The solution of the system is the region colored with both colors, the solid lines included. Check picture 3.
the lines intersect at (0.533, -0.4) because:
–0.75x=3x-2
-0.75x-3x=-2
-3.75x=-2, that is x= -2/(-3.75)=0.533
for x=0.533, y=3x-2=3(0.533)-2=-0.4
Answer: Picture 3, the half-lines included. So the graph is in the 3rd and 4th Quadrants
i) <span>y ≤ –0.75x
ii)</span><span>y ≤ 3x – 2
since </span>
Whenever we are asked to sketch the solution of a system of linear inequalities, we:
1. Draw the lines
2. Color the regions of the inequalities.
3. The solution is the region colored twice.
A.
to draw the line
consider the points: (-4, 1) and (0, 0), or any 2 other points (x,y) for which
since we have an "smaller or equal to" inequality, the line is a solid line (not dashed, or dotted).
In order to find out which region of the line to color, consider a point not on the line, for example P(1, 1), which is clearly in the upper region of the line.
For (x, y)=(1, 1) the inequality
this means that the solution is the region of the line not containing (1, 1), as shown in picture 1.
B.
similarly, to draw the solution of inequality ii) y ≤ 3x – 2,
we first draw the line y=3x-2, using the points (0, -2) and (2, 4), or any other 2 points (x,y) for which y=3x-2 holds.
after we draw the line, we can check the point P(1, 7) which clearly is above the line y=3x-2.
for (x, y) = (1, 7), the inequality y ≤ 3x – 2 does not hold
because 7 is not ≤ 3*1-2=1, so the region we color is the one not containing P(1, 7), as shown in picture 2.
The solution of the system is the region colored with both colors, the solid lines included. Check picture 3.
the lines intersect at (0.533, -0.4) because:
–0.75x=3x-2
-0.75x-3x=-2
-3.75x=-2, that is x= -2/(-3.75)=0.533
for x=0.533, y=3x-2=3(0.533)-2=-0.4
Answer: Picture 3, the half-lines included. So the graph is in the 3rd and 4th Quadrants