Not sure if this the right place to post the comment but yea...
On pg. 397 1-12..problem #7
It starts 5x+y=3 And 10x+2y=0
I took the first equation multiplied it by 2 and i got 10x+2y+6
So 10x+2y=0 And 10x+2y=6
Whats the next step?
Also when your substituting can you use the number if its negative and a positive or do you have to mulitply it by negative one to make them both positive?
And you made a good first step. Why did you make that step? So that either the 'x' term or the 'y' term would be set up for elimination when you subtract, right?
Then, lo and behold, you subtracted and both variables were gone! And you ended up with 0 = -6 right?
Does 0 = -6? (pause for Jeopardy theme music). No matter what 'x' and 'y' values we use, 0 is never = -6. Hmmmm...
Sounds like we don't have a solution, doesn't it? Well, isn't that one of the possibilities? (Check out the "Unit 7 Skills Review" on mathchamber!)
If you convert the equations to slope-intercept form, you'll get 1) y = -5x and 2) y = -5x + 3
What do these two lines have in common? What's different? Graph 'em. I'll bet you see something SPECIAL!
The best way to check what's happening is to graph the lines. Then it will become clear to you.
Riddle: When is -12 equal to -12? Answer: Always!
(ok, not such a great riddle)
But think about. Is there any value of 'x' or 'y' that will stop -12 from equaling -12. NO!!
So this means there are MANY x's and y's that make this system of equations TRUE.
Did you graph the lines yet (or convert them to slope-intercept form). What you should have found is that they are the SAME LINE! So the solution is the INFINITELY MANY POINTS ON THE LINE.
5 comments:
Not sure if this the right place to post the comment but yea...
On pg. 397 1-12..problem #7
It starts 5x+y=3
And 10x+2y=0
I took the first equation multiplied it by 2 and i got 10x+2y+6
So 10x+2y=0
And 10x+2y=6
Whats the next step?
Also when your substituting can you use the number if its negative and a positive or do you have to mulitply it by negative one to make them both positive?
Yup, you're at the right place!
And you made a good first step. Why did you make that step? So that either the 'x' term or the 'y' term would be set up for elimination when you subtract, right?
Then, lo and behold, you subtracted and both variables were gone! And you ended up with 0 = -6 right?
Does 0 = -6? (pause for Jeopardy theme music). No matter what 'x' and 'y' values we use, 0 is never = -6. Hmmmm...
Sounds like we don't have a solution, doesn't it? Well, isn't that one of the possibilities? (Check out the "Unit 7 Skills Review" on mathchamber!)
If you convert the equations to slope-intercept form, you'll get
1) y = -5x and
2) y = -5x + 3
What do these two lines have in common? What's different? Graph 'em. I'll bet you see something SPECIAL!
Let me know if this helped!
-Mr. C.
As for your second question, I don't really understand what you're asking... maybe an example would help.
Basically, you want to have an equation in 'y=' or 'x=' form so that you can use the "right side" expression as a substitute.
For example, if your system was:
3x+2y=14
y=2x
you could use the second equation to substitute ('2x' for 'y') into the first one, giving you:
3x+2(2x)=14
Then you would solve for 'x' from there.
Not sure if that helps you.
-Mr. C.
I dont get wat to do with the problem, using substitution, when u get -12=-12 and 6=6 ????? :(
Hi Ryan,
The best way to check what's happening is to graph the lines. Then it will become clear to you.
Riddle: When is -12 equal to -12?
Answer: Always!
(ok, not such a great riddle)
But think about. Is there any value of 'x' or 'y' that will stop -12 from equaling -12. NO!!
So this means there are MANY x's and y's that make this system of equations TRUE.
Did you graph the lines yet (or convert them to slope-intercept form). What you should have found is that they are the SAME LINE! So the solution is the INFINITELY MANY POINTS ON THE LINE.
Let me know if this helped.
-Mr. C.
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