SOLUTION: How would you graph y=-2/3+5 and -2x+4y=-8 with a point of (6,1)

Question 1210484: How would you graph y=-2/3+5 and -2x+4y=-8 with a point of (6,1)

Found 4 solutions by CPhill, ikleyn, timofer, josgarithmetic:
Answer by CPhill(2138)   (Show Source): You can put this solution on YOUR website!
It looks like you want to **graph two lines** and determine if the point $(6, 1)$ lies on their intersection.
The first equation, $y = -2/3 + 5$, is simply a horizontal line since it's a constant value. The second equation, $-2x + 4y = -8$, is a linear equation in standard form.
Let's first clarify and re-write the equations for graphing.
## 1. Analyze and Simplify the Equations
### Equation 1: $y = -2/3 + 5$
This equation simplifies to a single constant:
$$y = -\frac{2}{3} + \frac{15}{3}$$
$$y = \frac{13}{3} \approx 4.33$$
This is a **horizontal line** where every point has a y-coordinate of $\frac{13}{3}$.
### Equation 2: $-2x + 4y = -8$
It's easiest to graph this line using the slope-intercept form ($y = mx + b$).
1. **Isolate $y$:**
$$4y = 2x - 8$$
2. **Divide by 4:**
$$y = \frac{2}{4}x - \frac{8}{4}$$
$$y = \frac{1}{2}x - 2$$
This line has a **y-intercept ($b$) of $-2$** and a **slope ($m$) of $\frac{1}{2}$** (up 1, right 2).
---
## 2. Graphing the Lines
To graph these lines, you'll plot points and draw the lines on a coordinate plane.
### Line 1: $y = \frac{13}{3}$
1. **Plot the y-intercept:** $(\mathbf{0, \frac{13}{3}})$ or $(0, 4.33)$.
2. **Draw a horizontal line** passing through that point.
* Example points: $(3, \frac{13}{3})$, $(-3, \frac{13}{3})$.
### Line 2: $y = \frac{1}{2}x - 2$
1. **Plot the y-intercept:** $(\mathbf{0, -2})$.
2. **Use the slope** ($m = \frac{1}{2}$) to find other points:
* From $(0, -2)$, go up 1 and right 2 to find $(\mathbf{2, -1})$.
* From $(2, -1)$, go up 1 and right 2 to find $(\mathbf{4, 0})$.
3. **Draw a straight line** through these points.
---
## 3. Check the Given Point $(6, 1)$
Finally, let's see if the point $(6, 1)$ lies on **either** line.
### Check Line 1: $y = \frac{13}{3}$
Substitute $y=1$:
$$1 = \frac{13}{3}$$
This is **False**. The point $(6, 1)$ is **not** on Line 1.
### Check Line 2: $y = \frac{1}{2}x - 2$
Substitute $x=6$ and $y=1$:
$$1 = \frac{1}{2}(6) - 2$$
$$1 = 3 - 2$$
$$1 = 1$$
This is **True**. The point $(6, 1)$ **is** on Line 2.
The point $(6, 1)$ is only on the line $y = \frac{1}{2}x - 2$.
Would you like to find the exact point where these two lines intersect?
Answer by ikleyn(53419)   (Show Source): You can put this solution on YOUR website!
.
How would you graph y=-2/3+5 and -2x+4y=-8 with a point of (6,1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As the problem is formulated/presented in the post,
it is self-contradictory and has no solution.

It looks like first equation is written incorrectly.

===========================================

In his response, @josgarithmetic says "Yes", but his "Yes" does relate to another, different problem.

So, his response "Yes" goes via hidden "No", which he hesitates to pronounce.

Answer by timofer(136)   (Show Source): You can put this solution on YOUR website!
The question is missing or wrong itself.
What do you have and what do you want to ask?

Is this, "how do you graph the system of equations y=-(2/3)x+5 and -2x+4y=-8 ?" And then how do you intend that point (6,1) to be used?
Answer by josgarithmetic(39701)   (Show Source): You can put this solution on YOUR website!
Yes. Each of the equations contain the point (6,1).
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