Math / Algebra

QuestionFor the equation $y=-\frac{x}{5}+2$ a) Complete the table: \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline-5 & $\square$ \\ \hline 0 & $\square$ \\ \hline 5 & $\square$ \\ \hline \end{tabular} b) Use the appropriate tool to graph the given equation.

Studdy Solution

STEP 1

What is this asking? We need to find the $y$ values for some given $x$ values in a linear equation and then graph the line. Watch out! Don't mix up your $x$ and $y$ values!
Also, make sure your graph matches the table.

STEP 2

1. Substitute and Solve
2. Graph It

STEP 3

Alright, let's kick things off with $x = -5$ !
We've got our equation: $y = -\frac{x}{5} + 2$ .
Let's plug in our $x$ value: $y = -\frac{-5}{5} + 2$ .

STEP 4

A negative divided by a negative is a positive, so $-\frac{-5}{5}$ becomes $\frac{5}{5}$ , which simplifies to 1.
So, $y = 1 + 2 = \textbf{3}$ .
When $x = -5$ , $y = \textbf{3}$ !

STEP 5

Next up, $x = 0$ !
Substituting into our equation: $y = -\frac{0}{5} + 2$ .

STEP 6

Zero divided by anything is just zero, so we have $y = 0 + 2 = \textbf{2}$ .
When $x = 0$ , $y = \textbf{2}$ !

STEP 7

Last one for the table: $x = 5$ !
Plugging it in: $y = -\frac{5}{5} + 2$ .

STEP 8

$-\frac{5}{5}$ simplifies to $-1$ , giving us $y = -1 + 2 = \textbf{1}$ .
So when $x = 5$ , $y = \textbf{1}$ !

STEP 9

Now for the fun part!
We've got our points: $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ .
Let's plot these bad boys on a graph.

STEP 10

Remember, the first number in each pair is our $x$ value (how far we go left or right), and the second number is our $y$ value (how far we go up or down).

STEP 11

Once you've plotted the three points, grab your ruler (or any straight edge) and draw a line right through them!
Extend the line beyond the points in both directions, because the equation represents an infinite line, not just the three points we calculated.

STEP 12

Our completed table:
$\begin{array}{cc} x & y \\ -5 & 3 \\ 0 & 2 \\ 5 & 1 \\ \end{array}$ And we've got our graph with the line passing through $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ !

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QuestionFor the equation $y=-\frac{x}{5}+2$ a) Complete the table: \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline-5 & $\square$ \\ \hline 0 & $\square$ \\ \hline 5 & $\square$ \\ \hline \end{tabular} b) Use the appropriate tool to graph the given equation.

Studdy Solution

STEP 1

What is this asking? We need to find the $y$ values for some given $x$ values in a linear equation and then graph the line. Watch out! Don't mix up your $x$ and $y$ values!
Also, make sure your graph matches the table.

STEP 2

1. Substitute and Solve
2. Graph It

STEP 3

Alright, let's kick things off with $x = -5$ !
We've got our equation: $y = -\frac{x}{5} + 2$ .
Let's plug in our $x$ value: $y = -\frac{-5}{5} + 2$ .

STEP 4

A negative divided by a negative is a positive, so $-\frac{-5}{5}$ becomes $\frac{5}{5}$ , which simplifies to 1.
So, $y = 1 + 2 = \textbf{3}$ .
When $x = -5$ , $y = \textbf{3}$ !

STEP 5

Next up, $x = 0$ !
Substituting into our equation: $y = -\frac{0}{5} + 2$ .

STEP 6

Zero divided by anything is just zero, so we have $y = 0 + 2 = \textbf{2}$ .
When $x = 0$ , $y = \textbf{2}$ !

STEP 7

Last one for the table: $x = 5$ !
Plugging it in: $y = -\frac{5}{5} + 2$ .

STEP 8

$-\frac{5}{5}$ simplifies to $-1$ , giving us $y = -1 + 2 = \textbf{1}$ .
So when $x = 5$ , $y = \textbf{1}$ !

STEP 9

Now for the fun part!
We've got our points: $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ .
Let's plot these bad boys on a graph.

STEP 10

Remember, the first number in each pair is our $x$ value (how far we go left or right), and the second number is our $y$ value (how far we go up or down).

STEP 11

Once you've plotted the three points, grab your ruler (or any straight edge) and draw a line right through them!
Extend the line beyond the points in both directions, because the equation represents an infinite line, not just the three points we calculated.

STEP 12

Our completed table:
$\begin{array}{cc} x & y \\ -5 & 3 \\ 0 & 2 \\ 5 & 1 \\ \end{array}$ And we've got our graph with the line passing through $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ !

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QuestionFor the equation y=−x5+2y=-\frac{x}{5}+2y=−5x​+2 a) Complete the table: \begin{tabular}{|c|c|} \hlinexxx & yyy \\ \hline-5 & □\square□ \\ \hline 0 & □\square□ \\ \hline 5 & □\square□ \\ \hline \end{tabular} b) Use the appropriate tool to graph the given equation.

Studdy Solution

STEP 1

What is this asking? We need to find the yyy values for some given xxx values in a linear equation and then graph the line. Watch out! Don't mix up your xxx and yyy values!Also, make sure your graph matches the table.

STEP 2

1. Substitute and Solve2. Graph It

STEP 3

Alright, let's **kick things off** with x=−5x = -5x=−5!We've got our **equation**: y=−x5+2y = -\frac{x}{5} + 2y=−5x​+2.Let's **plug in** our xxx value: y=−−55+2y = -\frac{-5}{5} + 2y=−5−5​+2.

STEP 4

A negative divided by a negative is a positive, so −−55-\frac{-5}{5}−5−5​ becomes 55\frac{5}{5}55​, which simplifies to **1**.So, y=1+2=3y = 1 + 2 = \textbf{3}y=1+2=3.When x=−5x = -5x=−5, y=3y = \textbf{3}y=3!

STEP 5

Next up, x=0x = 0x=0!Substituting into our **equation**: y=−05+2y = -\frac{0}{5} + 2y=−50​+2.

STEP 6

Zero divided by anything is just **zero**, so we have y=0+2=2y = 0 + 2 = \textbf{2}y=0+2=2.When x=0x = 0x=0, y=2y = \textbf{2}y=2!

STEP 7

Last one for the table: x=5x = 5x=5!Plugging it in: y=−55+2y = -\frac{5}{5} + 2y=−55​+2.

STEP 8

−55-\frac{5}{5}−55​ simplifies to −1-1−1, giving us y=−1+2=1y = -1 + 2 = \textbf{1}y=−1+2=1.So when x=5x = 5x=5, y=1y = \textbf{1}y=1!

STEP 9

Now for the fun part!We've got our **points**: (−5,3)(-5, 3)(−5,3), (0,2)(0, 2)(0,2), and (5,1)(5, 1)(5,1).Let's **plot** these bad boys on a graph.

STEP 10

Remember, the first number in each pair is our xxx value (how far we go **left or right**), and the second number is our yyy value (how far we go **up or down**).

STEP 11

Once you've plotted the **three points**, grab your ruler (or any straight edge) and **draw a line** right through them!Extend the line beyond the points in both directions, because the equation represents an **infinite line**, not just the three points we calculated.

STEP 12

Our completed table:xy−530251\begin{array}{cc} x & y \\ -5 & 3 \\ 0 & 2 \\ 5 & 1 \\ \end{array}x−505​y321​And we've got our graph with the line passing through (−5,3)(-5, 3)(−5,3), (0,2)(0, 2)(0,2), and (5,1)(5, 1)(5,1)!

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QuestionFor the equation $y=-\frac{x}{5}+2$ a) Complete the table: \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline-5 & $\square$ \\ \hline 0 & $\square$ \\ \hline 5 & $\square$ \\ \hline \end{tabular} b) Use the appropriate tool to graph the given equation.

What is this asking? We need to find the $y$ values for some given $x$ values in a linear equation and then graph the line. Watch out! Don't mix up your $x$ and $y$ values!
Also, make sure your graph matches the table.

1. Substitute and Solve
2. Graph It

Alright, let's kick things off with $x = -5$ !
We've got our equation: $y = -\frac{x}{5} + 2$ .
Let's plug in our $x$ value: $y = -\frac{-5}{5} + 2$ .

A negative divided by a negative is a positive, so $-\frac{-5}{5}$ becomes $\frac{5}{5}$ , which simplifies to 1.
So, $y = 1 + 2 = \textbf{3}$ .
When $x = -5$ , $y = \textbf{3}$ !

Next up, $x = 0$ !
Substituting into our equation: $y = -\frac{0}{5} + 2$ .

Zero divided by anything is just zero, so we have $y = 0 + 2 = \textbf{2}$ .
When $x = 0$ , $y = \textbf{2}$ !

Last one for the table: $x = 5$ !
Plugging it in: $y = -\frac{5}{5} + 2$ .

$-\frac{5}{5}$ simplifies to $-1$ , giving us $y = -1 + 2 = \textbf{1}$ .
So when $x = 5$ , $y = \textbf{1}$ !

Now for the fun part!
We've got our points: $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ .
Let's plot these bad boys on a graph.

Remember, the first number in each pair is our $x$ value (how far we go left or right), and the second number is our $y$ value (how far we go up or down).

Once you've plotted the three points, grab your ruler (or any straight edge) and draw a line right through them!
Extend the line beyond the points in both directions, because the equation represents an infinite line, not just the three points we calculated.

Our completed table:
$\begin{array}{cc} x & y \\ -5 & 3 \\ 0 & 2 \\ 5 & 1 \\ \end{array}$ And we've got our graph with the line passing through $(-5, 3)$ , $(0, 2)$ , and $(5, 1)$ !