Simultaneous equation with three unknowns

Question

I have to solve the following exercise:

Solve the following sets of equations

$$X + 2Y = 4 \\ X + 3Z = 5 \\ 2Y - Z = 1$$.

I know that I must first find two equations which have the same values. However, equations 1 and 2 both have the same $X$ values but then have a $Y$ value and a $Z$ value so you can't take away these values.

Equations 2 and 3 both have Z values but they also have a X and Y value so you can't eliminate this also.

So what would I need to do.

Answer 1

The trick is to get two equations that involve only two unknowns. See below:

Subtract #2 from #1. This gives you a new equation that I'll call #4: $$ 2Y - 3Z = -1 $$ Now we have two equations (#3 and the new #4) that only involve $Y$ and $Z$. Subtract #3 from #4. This gives $$ -2Z = -2 $$ So $Z=1$. Substitute this back into #2, and you get $X+3 = 5$, so $X=2$.

Substitute $X=2$ in #1, and we get $2+2Y=4$. So $Y=1$.

Answer 2

Subtracting eq(i) from eq(ii) gives $$ 3z-2y=1$$ Adding above equation to eq(iii) eliminates $y$ and gives $$z=1$$ Putting $z$ value in eq(iii) gives $$y=1$$ Putting $z$ value in eq(ii) gives $$x=2$$

Answer 3

From the third equation, we can substitute $2y$ with $x-4$ (because of the first equation; subtract $x$ from both sides) to get$$x+3z=5\\-x-z=-3\tag1$$ Adding the two equations together, we eliminate $x$ and arrive at$$2z=2\therefore z=1$$ Substituting $z=1$ to the first equation of $(1)$ gives $x=2$. Substituting $x=2$ into the third equation gives $y=1$. Thus, $(x,y,z)=(2,1,1)$

---

From a Larger Perspective: \begin{align*} & x+2y=4\quad\implies 2y=4-x\\ & x+3z=5\\ & 2y-z=1\end{align*} And the work:\begin{align*} & x+3z=5\\ & 4-x-z=1\\\\ & 2z+4=6\qquad\text{Adding the two equations together}\\ & z=1\\ & x=2\\ & y=1\end{align*}