Tyler runs 8 2/3 miles in 4/3 of an hour. How long does it take each of them to run 10 miles?
1 answer:
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Hey Jackson!
-4x + 3y = 3
y = 2x + 1
Ok so what we need to do is solving y= 2x + 1 for y.
So let's start by using the substitution method :)
Substitute 2x + 1 for y in -4x + 3y = 3
-4x+ 3y = 3
-4x + 3(2x + 1) = 3
-4x + (3)(2x) + (3)(1) = 3
-4x + 6x + 3 = 3
2x + 3 = 3
Subtract 3 on both sides
2x + 3 - 3 = 3 - 3
2x = 0
x = 0/2
x = 0
So now since we find the number for x, we gonna use it to help us find the value for y.
To find y, we need to substitute 0 for x in y = 2x + 1
y = 2x + 1
y = 2(0) + 1
y = 0 + 1
y = 1
Thus,
The answer is: y = 1 and x = 0
How to graph?
You need to go on the thing where they put the numbers. y is located on top which has the positive numbers. So when you get there, make a line that comes from the top right side all the way to the bottom left sides. Remember that y = 1 so the line must pass through 1
I am not good with explanation. So I'll leave the graph down below then you'll see what I am talking about :)
Let me know if you have any questions. As always, it is my pleasure to help students like you!
-4x + 3y = 3
y = 2x + 1
Ok so what we need to do is solving y= 2x + 1 for y.
So let's start by using the substitution method :)
Substitute 2x + 1 for y in -4x + 3y = 3
-4x+ 3y = 3
-4x + 3(2x + 1) = 3
-4x + (3)(2x) + (3)(1) = 3
-4x + 6x + 3 = 3
2x + 3 = 3
Subtract 3 on both sides
2x + 3 - 3 = 3 - 3
2x = 0
x = 0/2
x = 0
So now since we find the number for x, we gonna use it to help us find the value for y.
To find y, we need to substitute 0 for x in y = 2x + 1
y = 2x + 1
y = 2(0) + 1
y = 0 + 1
y = 1
Thus,
The answer is: y = 1 and x = 0
How to graph?
You need to go on the thing where they put the numbers. y is located on top which has the positive numbers. So when you get there, make a line that comes from the top right side all the way to the bottom left sides. Remember that y = 1 so the line must pass through 1
I am not good with explanation. So I'll leave the graph down below then you'll see what I am talking about :)
Let me know if you have any questions. As always, it is my pleasure to help students like you!