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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?

 Jan 25, 2022
 #1
avatar+360 
0

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

 Jan 26, 2022
 #3
avatar+514 
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The question being asked here stated that w + x + z = -4

The question being asked when you google this question stated that w + x + z = 4.

 

I answered this one where w + x + z = -4, I am not sure if he is asking the same question, or a very similiar question. :)

proyaop  Jan 26, 2022
 #2
avatar+514 
+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 = \(1499\over9\)

 Jan 26, 2022

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