Yesoka

From the two equations you have given in the system there is no k and m to get infinite solutions (a,b) could you double check your equations?

For this question we know the shortest altitude would come from the longest side which in this case can be either BC or AC because they both equal 20. Now we need to find the area of triangle ABC. For this question we should use a formula known as Heron's formula. Which basically says the area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$ s = the semiperimeter or (a+b+c)/2 and a,b,c are the three sides of a triangle. Using heron's formula we get the semiperimeter = 55/2. $\sqrt{55/2(55/2-20)(55/2-20)(55/2-15)}$ this simplifies to $\sqrt{55/2(15/2)(15/2)(25/2)} = \sqrt{(15^2*55*25)/2^4}$ the denominator when square rooted equals 4. 15^2 goes out to 15 and 25 goes out to 5 leaving 55. Giving us the area of ABC as $75\sqrt{55}/4$. Assigning the altitude as x. We get (x*20)/2 = $75\sqrt{55}/4$. Which gets us our final answer of:

$x = 75\sqrt{55}/40$

Could you give us the diagram?

Could you give a image of how the grid looks like labeled with the letters?

Since AC is the diagonal of square ABCD we know that it is a right triangle.

There are 2 methods.

1st Method(45-45-90 Triangle)

Since AC is a diagonal of square ABCD. We know that triangle ABC is a 45-45-90 triangle.
Using our knowledge that it is a 45-45-90 triangle and the ratios will go 1:1:$x = \sqrt{2}$. We have x:x:2 = 1:1:$\sqrt{2}$ 
Dividing 2 by $\sqrt{2}$ we get $2/\sqrt{2}$ which rationalizing the denominator equals $x = \sqrt{2}$.

$x = \sqrt{2}$

Also rationalizing the demoninator involves multiplying a fraction with an irrational denominator with itself/itself. For example rationalizing $2/\sqrt{2}$ in this case would involve multiplying the fraction by $\sqrt{2}/\sqrt{2} = 1$

Another method is easier for beginners(Pythagorean Theorem)

Knowing that the figure above is a square we know BC is also equal to x. Using the pythagorean theorem we get $x^2 + x^2 = 2^2 or 2x^2 = 4$ . Dividing by 2 on both sides we get $x^2 = 2$. Squarerooting on both sides we get $x = +- \sqrt{2}$ However, since a length cannot have negative length.

$x = \sqrt{2}$

Assigning the original as the variable n.

Turning the words into an equation we get 5n/(n+7) = 0.

The only way for 5n/(n+7) to equal 0 is when 5n = 0.
So I believe the original number is 0.