Question

find four different solutions of x+12y=6​

Answer 1

Given

• Equation → x+12y=6

To Find

• Four Different Solutions

____________________

Put y = 0

$\sf ~~~\implies~~~ x+12 \times 0=6$

$\sf ~~~\implies~~~ x=6$

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Put x = (-6)

$\sf ~~~\implies~~~ - 6+12y=6$

$\sf ~~~\implies~~~ 12y=6 + 6$

$\sf ~~~\implies~~~ 12y=12$

$\sf ~~~\implies~~~ y=1$

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Put x = 0

$\sf ~~~\implies~~~ 0+12y=6$

$\sf ~~~\implies~~~ y= \dfrac{6}{12}$

$\sf ~~~\implies~~~ y= \dfrac{1}{2}$

$\sf ~~~\implies~~~ y= 0.5$

____________________

Put x = 2

$\sf ~~~\implies~~~ 2+12y=6$

$\sf ~~~\implies~~~ 12y=6 - 2$

$\sf ~~~\implies~~~ 12y = 4$

$\sf ~~~\implies~~~ y = \dfrac{4}{12}$

$\sf ~~~\implies~~~ y = \dfrac{1}{3}$

$\\\begin{gathered}~~~~~~~~~~~~~~~~~\pink{\begin{gathered}~~~~~\sf{\begin{array}{ | c | c | c | c | c|}\hline \sf x & 6&-6&0&2\\\hline \sf y & 0&1&0.5&1/3\\ \hline \end{array}}\end{gathered}}\end{gathered}$

____________________

• Four Solution are : =
• (6,0) , (-6,1), (0,0.5), (2,1/3).

Answer 2

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$\begin{gathered}\\\begin{gathered}~~~~~~~~~~~~~~~~~\red{\begin{gathered}~~~~~\tt{\begin{array}{ | c | c | c | c | c|}\hline \tt x & 6&-6&0&2\\\hline \tt y & 0&1&0.5&1/3\\ \hline \end{array}}\end{gathered}}\end{gathered}\\\\\end{gathered}$

$\huge\fcolorbox{black}{magenta}{ ★refer \: the \: attachment᭄✍︎}$

${\huge{\pink{\underline{\underline{thanks♪～}}}}}$

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