Online Course Discussion Forum

Math Challenge IC Handout 7

 
 
PengChristina的头像
Math Challenge IC Handout 7
PengChristina - 2018年01月25日 Thursday 22:04
 
Just one question this time! :D


What does 7.22 mean? I'm confused. Do I pick one of the two equations on the bottom question as my expression? Do I try to calculate x,y,z according to the equations? I did that, but I don't think those values are what the question is looking for. 


Thank you so much!

 
ReynosoDavid的头像
Re: Math Challenge IC Handout 7
ReynosoDavid - 2018年01月26日 Friday 10:46
 

The problem is first showing you the same example as in 7.2. The equations you want to use are $x+y = 2$ and $y-z = 1$. Try using one of the equations to solve for $x$. Then use the other equation to write $x$ in terms of $z$ as well.

PengChristina的头像
Re: Math Challenge IC Handout 7
PengChristina - 2018年01月26日 Friday 23:18
 

Thanks! Um... How exactly do you solve for x? I tried coming up with different variations of the parent equations and substituting... It just keeps going back to the parent equations.

ReynosoDavid的头像
Re: Math Challenge IC Handout 7
ReynosoDavid - 2018年01月29日 Monday 10:18
 

In the example that is mentioned in the problem, solving for $y$ in the equation $x - y = 0$ yields, $y = x$, so you can turn the equation $x+y+z=1$ into an equation that has only $x$ and $z$: $x+x+z = 1$, so $\displaystyle x = \frac{1-z}{2}$. Since we already knew that $x=y$, we have the line is determined by the equation $\displaystyle x = y = \frac{z-1}{2}$. Note we have $$\text{something in terms of only x} = \text{something in terms of only y} = \text{something in terms of only z}.$$