Question

# 3) y> 4x -3 y2- 2x +3 ...

Solve the following systems of inequalities graphically then give three ordered pairs satisfying the inequalities. Show that the ordered pairs satisfy the inequalities.

Transcribed: 3) y> 4x -3 y2- 2x +3

Given that,

$y>4x-3\phantom{\rule{0ex}{0ex}}y\ge -2x+3$

Write the given inequality in equation form,

Equation (1),

When x = 0, y=-3,

When y = 0, $x=\frac{3}{4}=0.75$

Equation (1) passes through (0, -3) and (0.75, 0).

Equation (2),

When x = 0, y = 3

When y = 0, $x=\frac{3}{2}=1.5$

Equation (2) passes through (0, 3) and (1.5, 0).

Subtracting (1) from (2),

$y-y=-2x+3-4x+3\phantom{\rule{0ex}{0ex}}-6x+6=0\phantom{\rule{0ex}{0ex}}6x=6\phantom{\rule{0ex}{0ex}}x=1$

Substitute x = 1 in (1),

$y=4-3=1$

So intersection point of (1) and (2) is (1, 1).

Now take three points (1, 6), (0, 4) and (2, 7).

Substitute these three points in given inequality,

At (1, 6),

$6>4-3⇒6>1\phantom{\rule{0ex}{0ex}}6\ge -2+3⇒6\ge 1$

At (0, 4),

$4>0-3⇒4>-3\phantom{\rule{0ex}{0ex}}4\ge 0+3⇒4\ge 3$

At (2, 7),

$7>8-3⇒7>5\phantom{\rule{0ex}{0ex}}7\ge -4+3⇒7\ge -1$

Hence these three points (1, 6), (0, 4) and (2, 7) satisfy the given inequality.

Graph:

[Shaded region represents the solution set of given equality included (1, 6), (0, 4) and (2, 7)]

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