 Let us solve the problem today. Draw the class of the equations x-y plus 1 is equal to 0 and 3x plus 2y minus 12 is equal to 0. Redamine the coordinates of the vertices of the triangle formed by these lines and the x-axis and share the triangular region. Now, given equations to us are minus y plus 1 is equal to 0 and 3x plus 2y minus 12 is equal to 0. Now, let us find the corresponding values for x and y to plot the graph. We will write the corresponding values for x is equal to 0, y is equal to 1 and for x is equal to 4, y is equal to 5 and for this equation we have for x is equal to 0, y is equal to 2, y is equal to 3. Now, let us plot the point 0, 1, 4, 5, 0, 6 and 2, 3 on a separate graph paper. Now, this we have as our graph paper. So, plotting 0, 1 first taking 0 on x-axis and 1 on y-axis we get this point as our 0, 1 naming it as a we get our required point. Now, plotting 4, 5 taking 4 on x-axis and 5 on y-axis we get this point as our 0.4, 5 naming it as b we get 4, 5. Now, joining a and b, point 0, 6 taking 0 on x-axis and 6 on y-axis we get this point as our 0, 6 and naming it as c and only d plotting d as 2, 3 taking 2 on x-axis and 3 on y-axis we get this point as our join CD and we have got this line. Now, this line is our equation x-y is equal to minus 1 and this line is 3x plus 2y is equal to 12. Here we can see that a triangle is formed with vertices 2, 3 and here minus 1, 0 this vertex and this vertex that is 4, 0 therefore required vertices are this one is f so comma 0 d is our solution d as 2, 3 so we write 2, 3 comma 0 comma 0 therefore this is our required answer. This CD portion is our required triangle. We answer the problem by and have a nice day.