Substitution, elimination and a graphical check

Label $2x+y=3$ as (1) and $4x-3y=0$ as (2). From (1), $y=3-2x$. Substitute into (2): $4x-3(3-2x)=0$, so $10x=9$ and $x=0.9$. Then $y=3-2(0.9)=1.2$. Exact fractions are $x=9/10$, $y=6/5$.

Multiply (1) by 3 to obtain $6x+3y=9$. Add (2): $10x=9$, hence $x=0.9$ and, from (1), $y=1.2$. Check $2(0.9)+1.2=3$ and $4(0.9)-3(1.2)=0$. The source also plots $Y_1=3-2X$ and $Y_2=\frac43X$: the unique intersection $(0.9,1.2)$ is the simultaneous solution. The diagram below redraws the mathematical content of its entry and intersection screens.

Label $x+y+z=6$, $x+2y+z=8$, $2x+y+z=7$ as (1), (2), (3). From (1), $x=6-y-z$. Substitution into (2) gives $(6-y-z)+2y+z=8$, hence $y=2$. Substitution into (3) gives $2(6-y-z)+y+z=7$, hence $y+z=5$. Therefore $z=3$ and (1) gives $x=1$.

Subtract (1) from (2): $y=2$. Substitute into (1) and (3): $x+z=4$ and $2x+z=5$. Subtract these equations: $x=1$, then $z=3$. Verify all three originals: $1+2+3=6$, $1+2(2)+3=8$, $2(1)+2+3=7$. Thus $(x,y,z)=(1,2,3)$.