2 Answers

Ask a Question

Ashis Biswas · Answer 1 · 2023-10-29T07:57:09+0000

$\stackrel{\textit{2nd equation }* 2}{\begin{array}{rrrrr} -8x&+&2y&=&8\\ 2(4x&+&8y)&=&2(-4) \end{array}}\qquad \implies \qquad \begin{array}{rrrrr} -8x&+&2y&=&8\\ 8x&+&16y&=&-8\\\cline{1-5} 0&+&18y&=&0 \end{array} \\\\\\ 18y=0\implies y=\cfrac{0}{18}\implies \boxed{y=0} \\\\\\ \stackrel{\textit{we know that}}{4x+8y=-4}\implies 4x+8(0)=-4\implies 4x=-4\implies x\cfrac{4}{-4}\implies \boxed{x=-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{one solution}}{(-1~~,~~0)}~\hfill$

you'd want to often enough check their slopes first, if the slopes differ, they meet, if equal, they're parallel and either never meet or meet everywhere.

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

2 Answers

0 Comments

Please log in or register to add a comment.

0 Comments

Please log in or register to add a comment.

Other Questions