SOLUTION: Graph -8y+10x=16 by finding three solutions. Write the equation in y=mx+b form first. Then choose three x-values and find the corresponding y-values

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Question 1205545: Graph -8y+10x=16 by finding three solutions. Write the equation in y=mx+b form first. Then choose three x-values and find the corresponding y-values
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

Graph by finding three solutions.
Write the equation in






make table
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plot these points and draw a line through



Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
equation is -8y + 10x = 16

convert to y = mx + b format by doing the following:

start with -8y + 10x = 16
add 8y to both sides of the equation and subtract 16 from both sides of the equation to get:
10x - 16 = 8y
flip sides of the equation to get:
8y = 10x - 16
divide both sides by 8 to get:
y = 10/8 * x - 16/8
simplify to get:
y = 1.25 * x - 2

-2 is the y-intercept
1.25 is the slope.

solve for the x-intercept as follows:
start with y = 1.25 * x - 2
set y = 0 to get:
0 = 1.25 * x - 2
add 2 to both sides of the equation to get:
2 = 1.25 * x
solve for x to get:
x = 2/1.25 = 1.6

the x-intercept is equal to 1.6
that's the value of x when y is equal to 0.

the y-intercept is equal to -2.
that's the value of y when x is equal to 0.

the slope of 1.25 says that y goes up 1.25 units when x goes to the right 1 unit.
there are two points on the graph that can be used to show this.
they are (0,-2) and (4,3)
x goes to the right 4 units.
y goes up 5 units.
the slope is the change in y divided by the change in x = 5/4 = 1.25

the graph looks like this:

the x-intercept is shown on the graph as (0,-2)
the y-intyercept is shown on the graph as (1.6,0)

the points are shown in (x,y) format.
x ix the x-value of the point.
y is the y-value of the point.

here's what the graph looks like.



the fact that the two equations create the same figure on the graph indicates that they are equivalent.


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