# Solve the system of inequalities graphically y < x 4x^2-16x-3y^2-12y < 8Use step by step solution to explain the answer.

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### 1 Answer

First we want to do a procedure called completing the square.

From the `x` and `x^2` terms, you factor out the number that is next to the `x^2`, so:

`4x^2-16x = 4(x^2-4x)`

Do the same with the ys:

`-3y^2-12y = -3(y^2+4y)`

Now take the number that is next to the x, divide it by 2, and square it:

`-4 / 2 = -2` , `(-2)^2 = 4`

This is the number you want to add and subtract from the `x^2 and `x` terms:

`4(x^2-4x)=4(x^2-4x+4-4)`

Do the same for the ys:

`-3(y^2+4y) = -3(y^2+4y+4-4)`

Distribute the last number (the one being subtracted):

`4(x^2-4x+4-4)=4(x^2-4x+4)-16`

`-3(y^2+4y+4-4)=-3(y^2+4y+4)+12`

What is left in parentheses can be factored into a perfect square:

`4(x^2-4x+4)-16=4(x-2)^2-16`

`-3(y^2+4y+4)+12=-3(y+2)^2+12`

So: `4x^2-16x-3y^2-12y = 4(x-2)^2-16 -3(y+2)^2+12 =4(x-2)^2-3(y+2)^2-4`

Thus: `4x^2-16x-3y^2-12y < 8` means 4(x-2)^2-3(y+2)^2 <12` , or:

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

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

This is a hyperbola:

It's center is at (2,-2), because the `(x-2)` shifts the graph 2 units to the right, and the `(y+2)` shifts the graph 2 units down.

It's asymptotes are the lines that go through (2,-2) and have slope `+- (2)/(sqrt(3))`

It is a left-right hyperbola instead of an up-down hyperbola, because the form of the equation is:

`x^2 - y^2` instead of `y^2 - x^2`

Thus our picture is:

(note: this is if we had an equality, not an inequality

Now if we want the inequality, pretend the middle stripe of the graph were shaded. (A quick method to tell which region to shade is to take some sample points, plug them into the equation, and see if the inequality is true. For us, an easy point to pick is (2,-2). If you plug that in, then it is true that:

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

Thus we want the shaded region to be the one containing (2,-2).

Now, we also want to satisfy the inequality y<x

The line `y=x` looks like:

We want `y<x` so we want the region below the line.

So, to satisfy both inequalities, we want the region that is shaded in both:

(I can't shade in the graph, but it is the region in the middle of the curved lines, and below the straight line. Also, I don't know why the left curved line is missing more than the right curved line, but they should both be connected to their lower counterparts.)

One final note: both these are strict inequalities, which means you don't want to include the actual lines in the allowable area. That is, the boundary of the region isn't actually part of the allowable region.