# Solve the system of inequalities graphically y < x 4x^2-16x-3y^2-12y < 8

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### 1 Answer

We are asked to solve the system of inequalitites graphically:

`y<x`

`4x^2-16x-3y^2-12y<8`

First graph the inequality `y<x` : this is the line `y=x` , drawn dotted, and shaded "beneath" (the side that includes the point (1,-1)).

Now we consider the other inequality. We know that it is a conic section, perhaps degenerate. Use completing the square to write in standard form:

`4x^2-16x-3y^2-12y<8`

`4(x^2-4x)-3(y^2+4y)<8`

`4(x^2-4x+4)-3(y^2+4y+4)<8+16-12`

`4(x-2)^2-3(y+2)^2<12`

`((x-2)^2)/3-((y+2)^2)/4<1`

This is a hyperbola, centered at (2,-2) whose branches open left/right. The area to shade is between the branches.

To draw the hyperbola lightly draw a rectangle whose center is at (2,-2) and whose vertices are `(2+sqrt(3),0),(2-sqrt(3),0),(2+sqrt(3),-4),(2-sqrt(3),-4)`

Then draw the diagonals -- the diagonals are the asymptotes for the branches.

The vertices of the branches are at `(2-sqrt(3),-2),(2+sqrt(3),-2)`

**The solution set will include points between the branches that are beneath the line y=x.**

** The shaded region is bounded at the top, and unbounded at the bottom.**

(There is an artifact of the grapher -- there should not be lines connecting the branches of the hyperbola)

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