Question

Solve and classify the system using substitution or elimination

Answer 1

$\bf \begin{cases} -\cfrac{2}{3}x+\cfrac{7}{3}y=-\cfrac{5}{3}\\\\ \cfrac{4}{5}x-\cfrac{14}{5}y=2 \end{cases}\qquad \qquad \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{3}}{-2x+7y=-5}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{4x-14y=10} \end{cases}$

if you ever wonder why we multiply both sides by the LCD, well is just to do away with the denominators.

now, notice something, if we multiply the first equation by say -2, we end up with

-2( -2x + 7y ) = -2( -5 )

4x - 14y = 10

well well well, low and behold the second equation is really the first equation in disguise, so both equations are really the same thing, namely is the same exact line, and so their graph is just one line pancaked on top of the other, which means every single point in the lines are solutions namely infinitely many solutions.

that said, let's proceed with the substitution

$\bf \stackrel{\textit{solving for \underline{y} in the first equation}}{-2x+7y=-5\implies 7y=2x-5}\implies y=\cfrac{2x-5}{7} \\\\\\ \stackrel{\textit{now plugging in that \underline{y} in the second equation}}{4x-14\left( \cfrac{2x-5}{7} \right)=10}\implies 4x-2(2x-5)=10 \\\\\\ 4x-4x+10=10\implies \stackrel{\stackrel{\textit{infinitely many solutions}}{\downarrow }}{0=0}$