+0

# help

+1
50
3
+103

Solve for real numbers x, y, z:

x + yz = 6,

y + xz = 6,

z + xy = 6.

Dec 8, 2019

#1
+2

doesn't $$x = y = z = 2$$?

i didn't do it algebraically, but all of these equations are asking for "something" + "square of something" = 6, which for $$x = y = z = 2$$ that results in

$$2 + 2^2 = 6$$.

the only other way i could think  to solve it logically is to think of the combinations that you can use to make 6, i.e. 1 + 1 x 6, 2 + 2 x 2. but the only values that satisfy all 3 equations are $$x = y = z = 2$$ :) hope this helps

Dec 8, 2019
#2
+106519
+1

WolframAlpha  shows several possibilities  :

https://www.wolframalpha.com/input/?i=x+%2B+yz+%3D+6%2C+++y+%2B+xz+%3D+6%2C++z+%2B+xy+%3D+6

Dec 9, 2019
#3
+23812
+1

Solve for real numbers x, y, z:
x + yz = 6,
y + xz = 6,
z + xy = 6.

$$\begin{array}{|lrcll|} \hline (1) & \mathbf{x + yz} &=& \mathbf{6} \\ & yz &=& 6-x \\ & \mathbf{z} &=& \mathbf{ \dfrac{6-x}{y} } \quad | \quad \boxed{ y\neq 0 !} \\ \hline (2) & \mathbf{y + xz} &=& \mathbf{6} \\ & xz &=& 6-y \\ & x\left(\dfrac{6-x}{y}\right) &=& 6-y \\ & x(6-x) &=& y(6-y) \\ & 6x-x^2 &=& 6y-y^2 \\ & \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ \hline (3) & \mathbf{z + xy} &=& \mathbf{6} \qquad \text{ or } \qquad \mathbf{z = 6-xy} \\ & xy &=& 6-z \\ & xy &=& 6-\left(\dfrac{6-x}{y}\right) \\ & xy &=& \dfrac{6y-(6-x)}{y} \\ & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x^2-6x+6y-y^2} &=& \mathbf{0} \\ x &=& \dfrac{ 6\pm \sqrt{36-4(6y-y^2)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4y^2-24y+36} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y^2-6y+9)} } {2} \\ x &=& \dfrac{ 6\pm \sqrt{4(y-3)^2} } {2} \\ x &=& \dfrac{ 6\pm 2(y-3) } {2} \\ x &=& 3 \pm (y-3) \\ \\ \mathbf{ x = y } \quad &\text{or}& \quad \mathbf{x=6-y} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{xy^2} &=& \mathbf{6y-6+x} \\ \hline x_1 = y: & y^3 &=& 6y-6+y \\ & y^3-7y+6 &=& 0 \\ & y_1 &=& -3 \qquad \Rightarrow x_1 = y_1 = -3 \\ & y_2 &=& 1 \qquad \Rightarrow x_2 = y_2 = 1 \\ & y_3 &=& 2 \qquad \Rightarrow x_3 = y_3 = 2 \\ \hline x_2 = 6-y: & (6-y)y^2 &=& 6y-6+(6-y) \\ & 6y^2-y^3 &=& 5y \\ & y^3- 6y^2 +5y &=& 0 \\ & \underbrace{y}_{y\neq 0} \underbrace{(y^2- 6y +5)}_{=0} &=& 0 \\\\ & y^2- 6y +5 &=& 0 \\ & (y-5)(y-1) &=& 0 \\ & y_4 &=& 5 \qquad \Rightarrow x_4 = 6-y_4 = 1 \\ & y_5 &=& 1 \qquad \Rightarrow x_5 = 6-y_5 = 5 \\ \hline \end{array}$$

$$\begin{array}{|c|r|r|r|} \hline & x & y & z= 6-xy \\ \hline 1. & -3 & -3 & -3 \\ 2. & 1 & 1 & 5 \\ 3. & 2 & 2 & 2 \\ 4. & 1 & 5 & 1 \\ 5. & 5 & 1 & 1 \\ \hline \end{array}$$

Dec 9, 2019