Question 955734
Start with the upper left corner of the square.
Let's call that number, {{{x}}}.
The one to the right would be, {{{x+1}}}.
The one below {{{x}}} would be {{{x+7}}}.
And the one to the right of it would be {{{x+8}}}.
{{{(matrix(2,2,a,b,c,d))=(matrix(2,2,x,x+1,x+7,x+8))}}}
So now you can solve the problem algebraically.
{{{bc-ad=(x+1)(x+7)-x(x+8)}}}
{{{bc-ad=(x^2+7x+x+7)-(x^2+8x)}}}
{{{bc-ad=x^2+8x+7-x^2-8x)}}}
{{{bc-ad=highlight(7))}}}