184k views
5 votes
Let x, y and z be real numbers such that

\begin{align*} x+y-z &= -8, \\ x-y &= 18,\text{ and} \\ y+z &= 30. \\ \end{align*}
Compute xyz.

User Nicoowr
by
8.4k points

1 Answer

4 votes

Answer:


xyz=(19952)/(27)

Explanation:

Given system of equations:


\begin{aligned} x+y-z &= -8 \\ x-y &= 18 \\ y+z &= 30\end{aligned}

Before we can compute xyz, we first need to find the values of x, y and z. To do this, we must solve the given system of equations.

Rearrange the second and third equations to isolate x and z, respectively:


\begin{aligned}x-y&=18\\x-y+y&=18+y\\x&=18+y\end{aligned}
\begin{aligned}y+z&=30\\y+z-y&=30-y\\z&=30-y\end{aligned}

Substitute these into the first equation and solve for y:


\begin{aligned}x+y-z&=-8\\(18+y)+y-(30-y)&=-8\\18+y+y-30+y&=-8\\3y-12&=-8\\3y&=4\\y&=(4)/(3)\end{aligned}

Now, substitute the found value of y into the two rearranged equations to find the values of x and z:


\begin{aligned}x&=18+(4)/(3)\\\\x&=(54)/(3)+(4)/(3)\\\\x&=(58)/(3)\end{aligned}
\begin{aligned}z&=30-(4)/(3)\\\\z&=(90)/(3)-(4)/(3)\\\\z&=(86)/(3)\end{aligned}

Now we have found the values of x, y and z, we can multiply them to compute xyz:


\begin{aligned}xyz&=(58)/(3) \cdot (4)/(3)\cdot (86)/(3)\\\\xyz&=(58\cdot4\cdot86)/(3\cdot3\cdot3)\\\\xyz&=(19952)/(27)\end{aligned}

Therefore, the computation of xyz is:


\large\boxed{(19952)/(27)}

User Abhishek Menon
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories