Find the intercepts of $y=-\frac{6}{5}x+6$

pokvarilaap
2022-09-05
Answered

Find the intercepts of $y=-\frac{6}{5}x+6$

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Mario Monroe

Answered 2022-09-06
Author has **12** answers

You just need to set the y to 0 and use it in the equation

$0=-\frac{6}{5}x+6$

$\frac{6}{5}x=6$

Frome where x=5

$0=-\frac{6}{5}x+6$

$\frac{6}{5}x=6$

Frome where x=5

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I already know how the set of solutions of system of linear equations over real numbers infinite field $PP$ is expressed.

When there is only single solution then it is just a vector of scalars, where each scalar is a real number.

When there are more than 1 solution, and actually infinite solutions, then it is just a parametric linear vector space in the form:$\mathrm{\forall}t\in \mathbb{R}:\overrightarrow{p}+\overrightarrow{v}\cdot t$ where $\overrightarrow{p}\in {\mathbb{R}}^{n}l{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\overrightarrow{v}\in {\mathbb{R}}^{n}$ where n denotes the number of real variables in each linear equation thus $n\in \mathbb{N}$

But my question is how the set of solutions of system of linear equations over the finite field$\mathbb{Z}}_{2$ or galois field GF(2) is expressed?

I already know that when there is only single solution then it is also just a vector of scalars, but where each scalar is a binary number either zero or one in$\mathbb{Z}}_{2$ where $\mathbb{Z}}_{2}=\{0,1\$ , but when there are more than 1 solution, but always finite number of solutions, then how are they expressed?

Is this similar to how real solutions are expressed by parametric linear vector space by modulo 2? Or something else? I don't know. I am trying to google the answer for this question for days but I don't find the answer anywhere. It seems like nobody talks about this topic.

How?

When there is only single solution then it is just a vector of scalars, where each scalar is a real number.

When there are more than 1 solution, and actually infinite solutions, then it is just a parametric linear vector space in the form:

But my question is how the set of solutions of system of linear equations over the finite field

I already know that when there is only single solution then it is also just a vector of scalars, but where each scalar is a binary number either zero or one in

Is this similar to how real solutions are expressed by parametric linear vector space by modulo 2? Or something else? I don't know. I am trying to google the answer for this question for days but I don't find the answer anywhere. It seems like nobody talks about this topic.

How?

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If we want to graph a horizontal line, we will do the following:

$y=0x+3$

No matter the domain for x, the range for y will always be 3. Therefore, we have a horizontal line.

$y=0\left(0\right)+3=(0,3)$

$y=0\left(1\right)+3=(1,3)$

$y=0\left(2\right)+3=(2,3)$

Now the formula to graph a vertical line looks like this:

$x=3$

Well, wait a second. Where is the y? I would like to see the y in the equation. But it is missing. How can I write the equation for a vertical line that includes the y variable? This is all I can think of:

$x=0y+3$

And with the following domain:

$x=0\left(0\right)+3$

$x=0\left(1\right)+3$

$x=0\left(2\right)+3$ Is this correct? Is it ok to reverse the x and y, as I just did above? Or does this not make it a slope-intercept equation anymore? It should still be a linear equation, since the variables are raised to the first power, in my opinion. But the slope-intercept form looks like this: y = mx + b. So I am not sure if this is still a slope-intercept equation.

No matter the domain for x, the range for y will always be 3. Therefore, we have a horizontal line.

Now the formula to graph a vertical line looks like this:

Well, wait a second. Where is the y? I would like to see the y in the equation. But it is missing. How can I write the equation for a vertical line that includes the y variable? This is all I can think of:

And with the following domain: