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?

Guest Jan 25, 2022

#1**0 **

this question was already answered on a web2.0calc post. enter your question on google and you will find it

XxmathguyxX Jan 26, 2022

#2**+4 **

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 = **\(1499\over9\)

proyaop Jan 26, 2022