$\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}$

if we were to solve both equations for "y", we'd get

$10x-18y=2\implies 10x-2=18y\implies \cfrac{10x-2}{18}=y\implies \cfrac{5}{9}x-\cfrac{1}{9}=y \\\\\\ -5x+9y=-1\implies 9y=5x-1\implies y=\cfrac{5x-1}{9}\implies y = \cfrac{5}{9}x-\cfrac{1}{9}$

notice, the 1st equation is really the 2nd in disguise, since both lines are just pancaked on top of each other, every point in the lines is a solution or an intersection, and since both go to infinity, well, there you have it.

8 0

1 year ago

Please help I'm in need.

Snezhnost [94]

The last one is correct!

8 0

2 years ago

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