We have these system of inequalities:
[tex]\left \{ {{3x+y>-3} \atop {x+2y<4}} \right.[/tex]
In this way, we have two straight lines that can be solved as follows:
[tex](1) \ 3x+y=-3 \therefore y=-3x-3 \\ \\ (2) \ x+2y=4 \therefore y=-\frac{x}{2}+2[/tex]
This two lines has been plotted in the Figure below and the region of the inequality is indicated in gray color. Therefore, each point in this region satisfies the system of inequalities. To illustrate this, let's take a point within the gray region, say (0, 0):
[tex]\left \{ {{3x+y>-3} \atop {x+2y<4}} \right. \\ \\ \left \{ {{3(0)+(0)>-3} \atop {(0)+2(0)<4}} \right. \\ \\ \left \{ {{0>-3} \atop {0<4}} \right. \ True![/tex]
There are infinite points in the solution set of system of inequalities [tex]3x+y>-3[/tex] and [tex]x+2y<4[/tex].
Further explanation:
Given:
The system of inequalities is given as follows:
[tex]\begin{cases}3x+y>-3&\\x+2y<4\end{cases}[/tex]
Calculation:
The set of inequalities are,
[tex]3x+y>-3[/tex] …… (1)
[tex]x+2y<4[/tex] …… (2)
Both the inequalities are strict inequalities; therefore they are graphed with dashed lines.
The corresponding equation of inequality (1) can be expressed as follows:
[tex]\boxed{3x+y=-3}[/tex] …… (3)
The corresponding equation of inequality (2) can be expressed as follows:
[tex]\boxed{x+2y=4}[/tex] …… (4)
Multiply equation (3) by [tex]2[/tex] as shown below:
[tex]\begin{aligned}2\cdot (3x+y)&=-3\cdot 2\\6x+2y&=-6\end{aligned}[/tex]
Subtract equation (5) from equation (4) as follows:
[tex]\begin{aligned}(6x+2y)-(x+2y)&=-6-4\\(6x-x)+(2y-2y)&=-10\\5x&=-10\end{aligned}[/tex]
Simplify equation [tex]5x=-10[/tex] to find the value of [tex]x[/tex] as follows:
[tex]\begin{aligned}5x&=-10\\x&=-\dfrac{10}{5}\\x&=-2\end{aligned}[/tex]
Substitute [tex]-2[/tex] for [tex]x[/tex] in equation (4) to obtain the value of [tex]y[/tex] as follows:
[tex]-2+2y=4[/tex]
Add [tex]2[/tex] on both sides of above equation as follows:
[tex]\begin{aligned}-2+2y+2&=4+2\\2y&=6\end{aligned}[/tex]
Now, divide both sides by 2 as follows:
[tex]\begin{aligned}\dfrac{2y}{2}&=\dfrac{6}{2}\\y&=3\end{aligned}[/tex]
The intersection of the lines [tex]3x+y=-3[/tex] and [tex]x+2y=4[/tex] is the point [tex](-2,3)[/tex].
Consider test point as [tex](1,1)[/tex].
Substitute [tex]1[/tex] for [tex]x[/tex] and [tex]1[/tex] for [tex]y[/tex] in inequality (1) as follows:
[tex]\begin{aligned}(3\cdot 1)+1&\ ^{?}_{>}}-3\\3+1&\ ^{?}_{>}-3\\4&>-3\end{aligned}[/tex]
The point [tex](1,1)[/tex] satisfies the inequality (1) and this point [tex](1,1)[/tex] lies above the line [tex]3x+y=-3[/tex].
Substitute [tex]1[/tex] for [tex]x[/tex] and [tex]1[/tex] for [tex]y[/tex] in inequality (2) as follows:
[tex]\begin{aligned}1+(2\cdot 1)&\ ^{?}_{<}\ 4\\1+2&\ ^{?}_{<}\ 4\\3&<4\end{aligned}[/tex]
The point [tex](1,1)[/tex] satisfies the inequality (2) and this point [tex](1,1)[/tex] lies below the line [tex]x+2y=4[/tex].
The graph of the solution set is the shaded region as shown in Figure 1 (attached in the end).
The solution set is the common region for both the regions of inequalities [tex]3x+2y>-3[/tex] and [tex]x+2y<4[/tex].
There can be infinite number of points in the solution set of both the inequalities.
Therefore, there are infinite points in the solution set of system of the given inequalities.
Learn more:
1. Learn more about equations https://brainly.com/question/1473992
2. Learn about solving equation https://brainly.com/question/5723059
Answer details:
Grade: High school
Subject: Mathematics
Chapter: Linear Inequalities
Keywords: Point, solution set, system, inequalities, 3x+y>-3, x+2y<4, (-2, 3), graph, linear, strict, infinite, intersection, dashed, infinte points, common region, shded region, inequalities, equation.
