A pair of equations is shown below:
y = 8x − 9
y = 4x − 1
Part A: Explain how you will solve the pair of equations by substitution or elimination. Show all the steps and write the solution. (5 points)
Part B: If the two equations are graphed, at what point will the lines representing the two equations intersect? Explain your answer. (5 points)

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- ccchristian1

- schrodinger

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- anonymous

I have no clue ;-;
Try tagging so of the mods and ambassadors.

- ccchristian1

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## More answers

- anonymous

- UsukiDoll

k that's enough.. anyway for a we can just use substitution.

- UsukiDoll

suppose we have two equations
y=8x-9 and y=4x-1. If we subsitute y =4x-1 into the first equation, we obtain
4x-1=8x-9
we need all the variable terms on the left side of the equation and all of the constant terms on the right side of the equation. Therefore, subtract 8x from both sides and add one on both sides
4x-8x-1=8x-8x-9
-4x-1 = -9
-4x-1+1 = -9+1
-4x = -8
dividing -4 on both sides we obtain x = 2
-4x = -8
----- -----
-4 -4
x = 2
now to check to see if we got the right answer we substitute x = 2 back into the original equations
y=8x-9 and y=4x-1.
y = 8(2) - 9 = 16-9 =7
y=4(2)-1 = 8-1 = 7
Both of our y values are 7
7 = 8x -9
7 = 16-9
7=7
7 = 4x-1
7=8-1
7=7
Since we have 7 on both sides when x = 2, we have the correct answer.

- ccchristian1

wow thank you

- UsukiDoll

as for part B
I think I have already done it.. when two equations intersect, they have a point in common
If we let x = 0 1 2 for y=8x-9, we have
x y =8x-9
0 -9
1 8-9 = -1
2 16-9 = 7
Similarly, for y =4x-1
x y =4x-1
0 -1
1 4-1 =3
2 8-1 = 7
both of these equations have the point (2,7) in common. So the two lines will meet at point (2,7) . I would draw on OS, but lately their drawing system has been going nuts.

- ccchristian1

okay thank you

- ccchristian1

thanks so much

- UsukiDoll

:)

- UsukiDoll

you can use what I typed on here xD!

- ccchristian1

yeah i am lol

- ccchristian1

you got more time, i will post the other question in a different post so you can earn snother medal

- ccchristian1

another medal*

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