I'm studying High School Algebra and it had this question:
Solve the system by equations:
\begin{align*} x + y - z &= \,0 \\ 2x + 4y - 2z &= 6 \\ 3x + 6y - 3z &= \,9 \end{align*}
The solution was:
infinitely many solutions (x, 3, z) where x = z − 3; y = 3; z is any real number
I've spent hours on the problem. The textbook just gave a vague explanation and I can't seem to get how it works. Can someone please intuitively explain how this is?
On
If $x,y,x\in\mathbb R$ are such that $2x+4y-2z=6$, then $\frac32(2x+4y-2z)=\frac32\times6$; in other words, $3x+6y-3z=9$, which is the thir equation. Therefore, solving that system is the same thing as solving the system$$\left\{\begin{array}{l}x+y-z=0\\2x+4y-2z=6,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}x+y=z\\2x+4y=2z+6.\end{array}\right.$$Solving this system, you will get the solutions that you mentioned.
Hint: Dividing the second equation by $2$ and the third by $3$ we get $$x+y-z=0$$ $$x+2y-z=3$$ $$x+2y-z=3$$ the second and the third equation are the same. Multiplying the first equation by $-1$ and adding to the second we get $y=3$. Plgging this into the first and second equation we get $$x-z=-3$$ $$x-z=-3$$ so we obtain the solutions $$x,3,x+3$$