1
$\begingroup$

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

$\endgroup$
6
  • 1
    $\begingroup$ Easiest here is to subtract the second equation from the first, and then add the third after which only $\ y\ $ remains as a variable. $\endgroup$
    – Peter
    Apr 24 '20 at 12:58
  • 1
    $\begingroup$ This is a system of linear equations. $\endgroup$
    – saulspatz
    Apr 24 '20 at 13:00
  • $\begingroup$ But maybe another method is supposed to be used. In this case, please give more context. $\endgroup$
    – Peter
    Apr 24 '20 at 13:00
  • $\begingroup$ 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 $\endgroup$
    – Paul
    Apr 24 '20 at 13:15
  • $\begingroup$ 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. $\endgroup$
    – Paul
    Apr 24 '20 at 13:19
1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Brilliant thank you very very much. Is there a name/theory for this kind of equation? $\endgroup$
    – Paul
    Apr 24 '20 at 13:27
  • $\begingroup$ @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. $\endgroup$
    – Dylan
    Apr 24 '20 at 14:06
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.