Q. 401 - Intermediate Algebra

Question

Solve a System of Linear Inequalities by Graphing
In the following exercises, solve each system by graphing.

$\left\{\begin{array}{l}2x-3y<6\\ 3x+4y\ge 12\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of inequality $\left\{\begin{array}{l}2x-3y<6\\ 3x+4y\ge 12\end{array}\right.$ is the overlapped region that contains the point

Step 1. Given

The system of inequality is $\left\{\begin{array}{l}2x-3y<6\\ 3x+4y\ge 12\end{array}\right.$

To find the solution of inequality by graphing 

Step 2. Graph the first inequality

Graph the line $2x-3y=6$

It is a dashed line since it contains the inequality $<$.

And test the point $(0,0)$

It is a solution to the given inequality, so shade the region that contains the point $(0,0)$

Step 3. Graph the second inequality

Graph the line $3x+4y=12$

It is a solid line since it contains the inequality $\ge$.

And test the point $(0,0)$.

It is not a solution to the given inequality, so shade the region that does not contain the point $(0,0)$

Step 4. Solution of the inequality

The point where the boundary line intersect is not a solution since it is not a solution to the inequality $2x-3y<6$.

The solution is all the points in the area shaded twice which appears as the darkest shaded region.

Step 5. Check the solution by choosing the point

Choose $(0,4)$ as a test point

Test for first inequality: 

      $2x-3y<6$

$2\left(0\right)-3\left(4\right)<6$

          $-12<6$ is true.

Test for second inequality: 

     $3x+4y\ge 12$

$3\left(0\right)+4\left(4\right)\ge 16$

             $16\ge 16$ is true.

The region containing $(0,4)$ is the solution to the system.