Graph Linear Equations
- Thread starter tbailey
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HallsofIvy
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First, you can start by writing it correctly! (1/2)x+ (1/3)y= 2. What you wrote would normally be interpreted as 1/(2x)+ 1/(3y)= 2 which is not, however, a linear function.
Now, I presume you know that linear functions are so called because there graphs are straight lines- and that a line is determined by two points. What you can do is pick two values for x or y, so that you can solve for the corresponding value of the other variable. For example, if you take x= 0, just because it is easy to use, you get (1/3)y= 2. Solve that for y and mark that point on a graph. If you take y= 0, you get (1/2)x= 2. Solve that for x and mark that point on a graph. Finally, draw the straight line through those two points.
Now, I presume you know that linear functions are so called because there graphs are straight lines- and that a line is determined by two points. What you can do is pick two values for x or y, so that you can solve for the corresponding value of the other variable. For example, if you take x= 0, just because it is easy to use, you get (1/3)y= 2. Solve that for y and mark that point on a graph. If you take y= 0, you get (1/2)x= 2. Solve that for x and mark that point on a graph. Finally, draw the straight line through those two points.
Hello, tbailey!
Multiply by 6: .\(\displaystyle 3x + 2y \:=\:12\)
If \(\displaystyle y = 0\!:\;3x + 0 \,=\,12 \quad\Rightarrow\quad x \,=\,4\)
. . We have a point \(\displaystyle (4,0)\)
If \(\displaystyle x = 0\!:\;0 + 2y \,=\,12 \quad\Rightarrow\quad y \,=\,6\)
. . We have a point \(\displaystyle (0,6)\)
Plot the points and draw the line.
Multiply by 6: .\(\displaystyle 3x + 2y \:=\:12\)
If \(\displaystyle y = 0\!:\;3x + 0 \,=\,12 \quad\Rightarrow\quad x \,=\,4\)
. . We have a point \(\displaystyle (4,0)\)
If \(\displaystyle x = 0\!:\;0 + 2y \,=\,12 \quad\Rightarrow\quad y \,=\,6\)
. . We have a point \(\displaystyle (0,6)\)
Plot the points and draw the line.
Hold on there!
Despite the fact that I don't type it as the OP did (and that I don't want others to type it as the OP did),
the Order of Operations handles it correctly. I just entered it exactly given by tbailey in www.quickmath.com
in the plotting or graphing equations section.
It graphed the correct line (using appropriate x any y parameters).
Edit:
This is to note that I have read daon2's post which follows this one.
Last edited:
Some sources use adjacency as a type of do-first multiplication. I do not want to argue about this, just noting that this could snowball into a similar recent "famous debate."
Ex: Some interpret 1/2(2) as 1/(2*2). Others, including myself, may interpret it as (1/2)*2, treating the '/' sign as the operator: a/b = a*b^(-1)
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An alternate approach. Given:
\(\displaystyle \dfrac{1}{2}x+\dfrac{1}{3}y=2\)
Divide through by 2, to get the line in the two-intercept form:
\(\displaystyle \dfrac{x}{4}+\dfrac{y}{6}=1\)
Now, we know we have the two intercepts (0,6) and (4,0). Plot these points and extend a line through them.
\(\displaystyle \dfrac{1}{2}x+\dfrac{1}{3}y=2\)
Divide through by 2, to get the line in the two-intercept form:
\(\displaystyle \dfrac{x}{4}+\dfrac{y}{6}=1\)
Now, we know we have the two intercepts (0,6) and (4,0). Plot these points and extend a line through them.