If w, x, y, and z are real numbers satisfying:
w+x+y = -2
w+x+z = -4
w+y+z = 19
x+y+z = 12,
what is wx + yz?
+364 this question was already answered on a web2.0calc post. enter your question on google and you will find it
+1632 We have four equations, four unknowns. Fortunately, in each equation the variable has a coefficient of 1, so here is what we can do.
First we add up all 4 equations getting:
\(3w + 3x + 3y + 3z = 25\)
Then we divide both sides of the equation by 3, getting:
\(w + x + y + z = {25\over3}\), let's call this equation5, and the other equations we will list from equation1 - equation4.
Now we take equation5 and subtract it from equation1. We get:
\(w + x + y + z - w - x - y = 25/3 + 6/3\)
This simplifies to \(z = {31\over3}\).
Equation5 - Equation2:
\(y = {37\over3}\)
Equation5 - Equation3:
\(x = {-32\over3}\)
Equation5 - Equation4:
\(w = {-11\over3}\)
Now to multiply:
wx + yz = 352/9 + 1147/9 = \(1499\over9\)
