$x+y=156[ x+z=183$; $y+z=139:$ what is value of $x, y, z$? Please can you help a dad teach his daughter the following problem:
$x+y=156,. 
x+z=183 ,, 
y+z=139 $
What is value of x and y and z? 
I do not know the type of equation to search "how to"!!
Apologies if this is very simple
 A: So what you want to do here is to take combinations of these equations in order to eliminate variables, like so:
If we call $x+y=156$ equation 1, $x+z =183$ equation 2 and $y+z=139$ equation 3 then subtracting equation 2 from equation 1 gives us $y-z = -27$ (we literally just subtract the LHS of equation 2 from equation 1 and the same for the RHS), call this equation 4. 
Then we have equation 3 saying $y+z =139$, and equation 4 saying $y-z = -27$. So equation 2 + equation 4 (again just add the LHS and RHS of the equations) gives us $2y = 112$, so $y=56$.
Then take $y=56$ and substitute it into equation 1 to find $x=100$, and find $z=83$ using your new found $x,y$ values and the other equations (just pick whichever is easier).
Hope this helps! 
A: After you realize $x+y+z=0.5((x+y)+(y+z)+(z+x)),$ it should be easy to find $x,y,z.$
A: Add all the equations to get, 
$$2x+2y+2z = 156 + 183 + 139 $$
$$\therefore x+y+z = 239$$
You can use this to solve for $x$, $y$, $z$ by substituting the values of $y+z$, $z+x$, $x+y$
