First let's find the solution of this system of equations:
[tex]\begin{gathered} \mleft\{\begin{aligned}4x+6y=24 \\ 2x+y=8\end{aligned}\mright. \\ \text{From the second equation:} \\ y=8-2x \\ \text{ Using this value of y in the first equation:} \\ 4x+6\cdot(8-2x)=24 \\ 4x+48-12x=24 \\ -8x=24-48 \\ -8x=-24 \\ x=\frac{-24}{-8}=3 \\ \\ y=8-2\cdot3=8-6=2 \end{gathered}[/tex]So the solution to this system of equations is the pair (3, 2).
If we double the second equation, we will get an equivalent equation, that is, an equation were the same pair is also a solution:
[tex]\begin{gathered} \mleft\{\begin{aligned}4x+6y=24 \\ 4x+2y=16\end{aligned}\mright. \\ \text{Subtracting the equations:} \\ 4x+6y-(4x+2y)=24-16 \\ 6y-2y=8 \\ 4y=8 \\ y=2 \\ \\ 4x+2\cdot2=16 \\ 4x=16-4 \\ 4x=12 \\ x=3 \end{gathered}[/tex]Also, when we double the equation, the new equation is represented by the same line or function, if we graph the equation or if we isolate the variable y:
[tex]\begin{gathered} 2x+y=8 \\ y=8-2x \\ \\ 4x+2y=16 \\ 2y=16-4x \\ y=\frac{16-4x}{2} \\ y=8-2x \end{gathered}[/tex]So if we multiply an equation by any constant value, the solution will not change.
