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Elan Coil [88]
1 year ago
14

To solve a system of inequalities so you can graph it how do you change these two equations into something like the two that are

on number one?Problem #2.

Mathematics
1 answer:
Zigmanuir [339]1 year ago
8 0
Explanation

Problem #2

We must find the solution to the following system of inequalities:

\begin{gathered} 3x-2y\leq4, \\ x+3y\leq6. \end{gathered}

(1) We solve for y the first inequality:

-2y\leq4-3x.

Now, we multiply both sides of the inequality by (-1), this changes the signs on both sides and inverts the inequality symbol:

\begin{gathered} 2y\ge-4+3x, \\ y\ge\frac{3}{2}x-2. \end{gathered}

The solution to this inequality is the set of all the points (x, y) over the line:

y=\frac{3}{2}x-2.

This line has:

• slope m = 3/2,

,

• y-intercept b = -2.

(2) We solve for y the second inequality:

\begin{gathered} x+3y\leq6, \\ 3y\leq6-x, \\ y\leq-\frac{1}{3}x+2. \end{gathered}

The solution to this inequality is the set of all the points (x, y) below the line:

y=-\frac{1}{3}x+2.

This line has:

• slope m = -1/3,

,

• y-intercept b = 2.

(3) Plotting the lines of points (1) and (2), and painting the region:

• over the line from point (1),

,

• and below the line from point (2),

we get the following graph:

Answer

The points that satisfy both inequalities are given by the intersection of the blue and red regions:

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Please help! These mathematics are very confusing. need help right away.
Ostrovityanka [42]
<span>
Exercise #1:
Point H = (–2, 2)
Point J = (–2, –3)
Point K = (3, –3)

It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

</span><span>Notice that points J and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.

Notice that point-J is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-J is one corner of the square.

H is another corner of the square.  It's 5 units above J.

K is another corner of the square.  It's 5 units to the right of J.

The fourth corner is (2, 3) ... 5 to the right of H,
                                       and 5 above K.
____________________________________

Exercise #2:
</span><span>Point H = (6, 2)
Point J = (–2, –4)
Point K = (-2, y) .

</span><span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

</span><span>Notice that points J and K have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

We need K to connect to point-H in such a way that it's on
the same horizontal line as H.  Then the vertical and horizontal
lines that meet at K will be perpendicular, and we'll have the
right angle that we need there to make the right triangle.
So K and H need to have the same y-coordinate.
H is the point (6, 2).  So K has to be up at  (2, 2) .
____________________________________________

Exercise #3:
</span>
<span>Point H = (-6, 2)
Point J = (–6, –1)
Point K = (4, 2) .
</span>
<span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it.  Then
you'd immediately see what's going on.

This exercise is exactly the same as #1, except that it's a
rectangle instead of a square.  It's still make of horizontal
and vertical lines, and that's all we need to know in order
to solve it.</span><span>

Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.

</span><span>Notice that points H and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.

Notice that point-H is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-H is one corner of the rectangle.

J is another corner of the rectangle.  It's 3 units below H.

K is another corner of the square.  It's 4 units to the right of H.

The fourth corner is (2, -1) ... 4 to the right of J,
                                       and  3 below K.

</span>
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