How do you write the equation y=3x+17 in standard form?

princetonaqo3
2022-10-23
Answered

How do you write the equation y=3x+17 in standard form?

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Milton Hampton

Answered 2022-10-24
Author has **16** answers

Standard form in this context means that the variables are on one side and any number terms are on the other side.

y=3x+17

$3x-y=-17\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ standard form ax+by=c

(It will always be possible to give x as a positive)

$y=3x+17\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ slope-intercept form, y=mx+c

$3x-y+17=0\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ general form ax+by+c=0

y=3x+17

$3x-y=-17\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ standard form ax+by=c

(It will always be possible to give x as a positive)

$y=3x+17\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ slope-intercept form, y=mx+c

$3x-y+17=0\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\leftarrow$ general form ax+by+c=0

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I am given the following matrix A and I need to find a nullspace of this matrix.

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?

$A=\left(\begin{array}{cccc}2& 1& 4& -1\\ 1& 1& 1& 1\\ 1& 0& 3& -2\\ -3& -2& -5& 0\end{array}\right)$

I have found a row reduced form of this matrix, which is:

${A}^{\prime}=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

And then I used the formula ${A}^{\prime}x=0$, which gave me:

${A}^{\prime}x=\left(\begin{array}{cccc}1& 0& 3& -2\\ 0& 1& -2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)$

Hence I obtained the following system of linear equations:

$\{\begin{array}{l}{x}_{1}+3{x}_{3}-2{x}_{4}=0\\ {x}_{2}-2{x}_{3}+3{x}_{4}=0\end{array}$

So I just said that ${x}_{3}=\alpha $, ${x}_{4}=\beta $ and the nullspace is:

$nullspace(A)=\{2\beta -3\alpha ,2\alpha -3\beta ,\alpha ,\beta )\text{}|\text{}\alpha ,\beta \in \mathbb{R}\}$

Is my thinking correct?