# $x+y=156[ x+z=183$; $y+z=139:$ 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

• Easiest here is to subtract the second equation from the first, and then add the third after which only $\ y\$ remains as a variable. Apr 24 '20 at 12:58
• This is a system of linear equations. Apr 24 '20 at 13:00
• But maybe another method is supposed to be used. In this case, please give more context. Apr 24 '20 at 13:00
• So to add more context...Liam has these three shapes. Three shapes X, Y and Z He uses them to make different towers. He measures the height of each tower he makes. Four towers created with the shapes Tower one length x+y=156cm Tower two length x+z=183cm Tower three length y+z=139cm Liam stacks all three shapes to make one tall tower. How tall is the tower? I hope this helps
– Paul
Apr 24 '20 at 13:15
• I have watch several videos on system of linear equations and I cannot get my head around it because none of the examples they use seem to apply. Please would you be kind enough to type the working out for me so I can follow? I do not want just the answer I want yo be able to understand it so I can teach my daughter. I appreciate this is probably very simple for you guys but I am struggling to teach her with the schools closed.
– Paul
Apr 24 '20 at 13:19

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!

• Brilliant thank you very very much. Is there a name/theory for this kind of equation?
– Paul
Apr 24 '20 at 13:27
• @Paul my pleasure! Yea, if you google "system of linear equations" or "solving simultaneous equations" then you will find lots of resources on this topic. Apr 24 '20 at 14:06

After you realize $$x+y+z=0.5((x+y)+(y+z)+(z+x)),$$ it should be easy to find $$x,y,z.$$

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 $$y+z$$, $$z+x$$, $$x+y$$