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**lisathedork****Member**- Registered: 2013-10-24
- Posts: 4

My progress so far, am I'm doing this right?

10. y = (2/3)x - 1

y = -x + 4

2/3x 1 = -x + 4

2/3x + x = 4 + 1

5/3x = 5

5x = 15

X=3

Y=-x + 4

Y = -3 + 4

Y=1

11. x + y = 0

3x + y = -4

X + y (3x + y) = 0 (-4)

X + y 3x y = 0 + 4

-2x = 4

X= -2

X+ y = 0

-2 + y = 0

Y = 2

12. 4x + 3y = -15

y = x + 2

4x +3(x +2) = -15

4x + 3x + 6 = -15

7x = -21

X = -3

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Can someone help me guide me through this?

13. x + 2y = -4

4y = 3x + 12

14. y = 2x

x + y = 3

15. x = 3 - 3y

x + 3y = -6

16. y = -2x + 1

y = x - 5

17. y = (1/2)x - 3

y = (3/2)x 1

18. x + y = 2

4y = -4x + 8

19. Why are the substitution and elimination methods necessary when we already know how to solve systems of linear equations using the graphing method?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,270

Hi;

You are correct for #10.

For #11 you are correct but next time remember that in mathematics x and X are not the same. To prevent confusion use the same case throughout. Also it would have been faster to subtract the first equation from the second.

#12 you are correct but you have forgotten to solve for y.

Can someone help me guide me through this?

You are doing fine. Try the next one.

Why are the substitution and elimination methods necessary when we already know how to solve systems of linear equations using the graphing method?

Graphing is adequate for 2 equations in two unknowns ( x and y) but real world problems are much, much larger. That is the reason we need other methods to solve them.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,335

lisathedork wrote:

19. Why are the substitution and elimination methods necessary when we already know how to solve systems of linear equations using the graphing method?

If the true answer to a question was x = 5.99997, you'd be unlikely to achieve this accuracy from a graph and be certain you'd got the right answer. More likely, you'd think the answer was x = 6.

Those other methods allow you to work to any level of accuracy. Sometimes elimination is quicker, but some equations can only be solved by substitution. There are some other methods as well, which you can find out by searching the MIF pages.

eg. Q13. I'd make x the subject of the first and then substitute into the second.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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