How do you solve $3u+z=15$ and $u+2z=10$?

Hint: There are two unknowns and two equations to solve. We solve the equations equating the coefficients of one variable and omitting the variable. The other variable remains with the constants. Using the binary operation, we find the value of the other variable.

Complete step by step answer:
The given equations $3u+z=15$ and $u+2z=10$ are linear equations of two variables.
We know that the number of equations has to be equal to the number of unknowns to solve them.
We take the equations as $3u+z=15.....(i)$ and $u+2z=10......(ii)$.
We multiply 2 to the both sides of the first equation and get
$\begin{align}
  & 2\times \left( 3u+z \right)=2\times 15 \\
 & \Rightarrow 6u+2z=30 \\
\end{align}$
We take the equation as $6u+2z=30.....(iii)$.
Now we subtract the equation (ii) from equation (iii) and get
$\left( 6u+2z \right)-\left( u+2z \right)=30-10$.
We take the variables together and the constants on the other side.
Simplifying the equation, we get
$\begin{align}
  & \left( 6u+2z \right)-\left( u+2z \right)=30-10 \\
 & \Rightarrow 5u=20 \\
 & \Rightarrow u=\dfrac{20}{5}=4 \\
\end{align}$
The value of u is 4. Now putting the value in the equation $3u+z=15.....(i)$, we get
$\begin{align}
  & 3\times 4+z=15 \\
 & \Rightarrow z=15-12=3 \\
\end{align}$.

Therefore, the values are $u=4,z=3$.

Note: We can also find the value of one variable z with respect to u based on the equation
$3u+z=15$ where $z=15-3u$. We replace the value of z in the second equation of
$u+2z=10$ and get
\[\begin{align}
  & u+2z=10 \\
 & \Rightarrow u+2\left( 15-3u \right)=10 \\
 & \Rightarrow u+30-6u=10 \\
\end{align}\]
We get the equation of u and solve
$\begin{align}
  & 30-5u=10 \\
 & \Rightarrow 5u=30-10=20 \\
 & \Rightarrow u=\dfrac{20}{5}=4 \\
\end{align}$
Putting the value of u we get \[z=15-3u=z=15-3\times 4=15-12=3\].
Therefore, the values are $u=4,z=3$.