Question

How would I solve the system of inequalities?Is the procedure correct?

Answer 1

To plot the line of the inequality , we begin by inserting the values x = 0 and y = 0 one after the other.

`\begin{gathered} 4x+5y>20 \\ \text{When x = 0, then} \\ 4(0)+5y=20 \\ 5y=20 \\ y=4 \\ \text{That gives us the point (0, 4)} \\ \text{Similarly when y = 0, then} \\ 4x+5(0)=20 \\ 4x=20 \\ x=5 \\ \text{That gives us the point (5, 0)} \\ \text{With these two points we can now plot a line by joining both points with a ruler} \\ \text{Note also that the inequality can now be expressed as;} \\ 4x+5y>20 \\ 5y>20-4x \\ y>(20-4x)/(5) \\ y>(20)/(5)-(4x)/(5) \\ y>4-(4)/(5)x \\ y>-(4)/(5)x+4 \end{gathered}`

Letus now derive two points for the second inequality given just like we did for the first one above;

`\begin{gathered} -3x+7y\ge7 \\ \text{When x = 0, then } \\ -3(0)+7y=7 \\ 7y=7 \\ y=1 \\ \text{This gives us the point (0, 1)} \\ \text{Also, when y = 0, then} \\ -3x+7(0)=7 \\ -3x=7 \\ x=-(7)/(3)\text{ (OR x=-2}(1)/(3)) \\ \text{That gives us the point (-2}(1)/(3),0) \\ \text{The equation can now be expressed as} \\ -3x+7y\ge7 \\ 7y\ge7+3x \\ y\ge(7+3x)/(7) \\ y\ge(7)/(7)+(3x)/(7) \\ y\ge1+(3)/(7)x \\ y\ge(3)/(7)x+1 \end{gathered}`

The graph of both inequalities is now shown below;

Observe carefully that the solution lies in the region shaded BLUE and RED.

One point in this region is (4, 4)

The blue and red graphs are represented by;

`\begin{gathered} y>-(4)/(5)x+4\text{ (RED graph)} \\ y\ge1+(3)/(7)x\text{ (BLUE graph)} \end{gathered}`