+0  
 
0
100
3
avatar

I need help with this system:

x(y + z) = 39

y(x + z) = 60

z(x + y) = 63

What is x?

 Jan 19, 2021

Best Answer 

 #1
avatar+25640 
+2

I need help with this system:
\(x(y + z) = 39\\ y(x + z) = 60\\ z(x + y) = 63\)

 

What is x?

 

\(\begin{array}{|lrcll|} \hline (1) & x(y + z) &=& 39 \\ (2) & y(x + z) &=& 60 ~\text{or}~ \mathbf{yx+yz = 60} \\ (3) & z(x + y) &=& 63 ~\text{or}~ \mathbf{zx+yz = 63} \\ \hline (1)+(2)+(3): &x(y + z)+ y(x + z)+z(x + y) &=& 39+60+63 \\ & x(y + z) +yx +yz +zx+zy &=& 162 \\ & x(y + z) +x(y+z) +2yz &=& 162 \\ & 2x(y + z) +2yz &=& 162 \quad | \quad :2 \\ & x(y + z) +yz &=& 81 \quad | \quad x(y + z) = 39 \\ & 39 +yz &=& 81 \\ & yz &=& 81-39 \\ & \mathbf{yz} &=& \mathbf{42} \\ \hline & \mathbf{yx+yz} &=& \mathbf{60} \quad | \quad \mathbf{yz = 42} \\ & yx+42 &=& 60 \\ & yx &=& 60-42 \\ & \mathbf{yx} &=& \mathbf{18}~\text{or}~ \mathbf{y=\dfrac{18}{x}} \\ \hline & \mathbf{zx+yz} &=& \mathbf{63} \quad | \quad \mathbf{yz = 42} \\ & zx+42 &=& 63 \\ & zx &=& 63-42 \\ & \mathbf{zx} &=& \mathbf{21}~\text{or}~ \mathbf{z=\dfrac{21}{x}} \\ \hline & \mathbf{yz} &=& \mathbf{42} \\ & \dfrac{18}{x}*\dfrac{21}{x}&=& \mathbf{42} \\ & 42x^2 &=& 18*21 \\ & x^2 &=&\dfrac{18*21}{42} \\ & x^2 &=& \dfrac{3*6*3*7}{6*7} \\ & x^2 &=& 3*3 \\ & \mathbf{x} &=& \mathbf{3} \\ \hline & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\ & y &=& \dfrac{18}{3} \\ & \mathbf{y} &=& \mathbf{6} \\ \hline & \mathbf{z} &=& \mathbf{\dfrac{21}{x}} \\ & z &=& \dfrac{21}{3} \\ & \mathbf{z} &=& \mathbf{7} \\ \hline \end{array}\)

 

laugh

 Jan 19, 2021
edited by heureka  Jan 19, 2021
 #1
avatar+25640 
+2
Best Answer

I need help with this system:
\(x(y + z) = 39\\ y(x + z) = 60\\ z(x + y) = 63\)

 

What is x?

 

\(\begin{array}{|lrcll|} \hline (1) & x(y + z) &=& 39 \\ (2) & y(x + z) &=& 60 ~\text{or}~ \mathbf{yx+yz = 60} \\ (3) & z(x + y) &=& 63 ~\text{or}~ \mathbf{zx+yz = 63} \\ \hline (1)+(2)+(3): &x(y + z)+ y(x + z)+z(x + y) &=& 39+60+63 \\ & x(y + z) +yx +yz +zx+zy &=& 162 \\ & x(y + z) +x(y+z) +2yz &=& 162 \\ & 2x(y + z) +2yz &=& 162 \quad | \quad :2 \\ & x(y + z) +yz &=& 81 \quad | \quad x(y + z) = 39 \\ & 39 +yz &=& 81 \\ & yz &=& 81-39 \\ & \mathbf{yz} &=& \mathbf{42} \\ \hline & \mathbf{yx+yz} &=& \mathbf{60} \quad | \quad \mathbf{yz = 42} \\ & yx+42 &=& 60 \\ & yx &=& 60-42 \\ & \mathbf{yx} &=& \mathbf{18}~\text{or}~ \mathbf{y=\dfrac{18}{x}} \\ \hline & \mathbf{zx+yz} &=& \mathbf{63} \quad | \quad \mathbf{yz = 42} \\ & zx+42 &=& 63 \\ & zx &=& 63-42 \\ & \mathbf{zx} &=& \mathbf{21}~\text{or}~ \mathbf{z=\dfrac{21}{x}} \\ \hline & \mathbf{yz} &=& \mathbf{42} \\ & \dfrac{18}{x}*\dfrac{21}{x}&=& \mathbf{42} \\ & 42x^2 &=& 18*21 \\ & x^2 &=&\dfrac{18*21}{42} \\ & x^2 &=& \dfrac{3*6*3*7}{6*7} \\ & x^2 &=& 3*3 \\ & \mathbf{x} &=& \mathbf{3} \\ \hline & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\ & y &=& \dfrac{18}{3} \\ & \mathbf{y} &=& \mathbf{6} \\ \hline & \mathbf{z} &=& \mathbf{\dfrac{21}{x}} \\ & z &=& \dfrac{21}{3} \\ & \mathbf{z} &=& \mathbf{7} \\ \hline \end{array}\)

 

laugh

heureka Jan 19, 2021
edited by heureka  Jan 19, 2021
 #2
avatar+116126 
+1

Nice method, heureka    !!!!

 

 

cool cool cool

CPhill  Jan 19, 2021
 #3
avatar+25640 
+2

Thank you, CPhill !

 

laugh

heureka  Jan 20, 2021

83 Online Users

avatar
avatar
avatar
avatar