Question

I need some help with math -
A pair of equations is shown below:

y = 3x − 5
y = 6x − 8

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

Answer 1

Substitution Method

Step 1: equate the two functions
\[3x -5 = 6x - 8\]
Step 2: isolate the variable (in our case, it's x)
\[x = 1\]

The two lines would intersect at point (1, -2).

Answer 2

can u explain how to get how the lines would intersect @Cardinal_Carlo

Answer 3

@Cardinal_Carlo like how did u find where they intersect

Answer 4

@Outh-1 can u help me

Answer 5

You can find the point in one of two ways: by graphing, or by doing substitution twice.

Answer 6

@Outh-1 how do u do it by graphing

Answer 7

Graph both lines. Try to draw it out if you can.

Answer 8

So for your equations' intersection, part of the answer is in part A which we found to be x = 1, but we still need its corresponding y-value to get the intersecting point. So:
\[x = 1\]
\[y = f(1) = 3(1)-5 = 6(1) - 8 = -2\]
Now we have both x and y values, and our point of intersection is (1, -2)

Answer 9

is it supposed to come up as y = f(1) = 3(1) - 5 = 6(1) - 8 = -2 cause i dont really understand that. where do all the ones come from? sorry im really bad at math.

Answer 10

@Cardinal_Carlo ^^^

Answer 11

and where did the f come from @Cardinal_Carlo

Answer 12

Okay, so to make things simple:

Step 1 Take the solution we got from part A.
Note: Solution value is synonymous with intersection value.
\[x = 1\]
This x-value is our solution for part A and happens to also be the intersecting x-coordinate.
Step 2 Plug this value in to either equation above like so:
\[y = 3(1) - 5\]
and we'll get a y-value of:
\[y = -2\]
So now that we have both x and y values, we have the intersecting coordinates (1, -2).

P.S. Never mind the 'f' if you haven't come across functions yet in your class.

Answer 13

@Cardinal_Carlo thanks so much :)

Answer 14

You're welcome