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Solve the system using the substituation method...

x=5y - 2

2x + y = -4

not sure what I am doing.......Thanks.

G

Solve the system using the substituation method...

x=5y - 2

2x + y = -4

not sure what I am doing.......Thanks.

G

2x + y = -4

2(5y-2)+y=-4

10y-2+y=-4

8y=-4

--------

8 8

y=-2

ok, but when I us y = -2 on

x=5y-2

x=(5)(-2)-2

x=-12

my book says the answer should be (-2,0)

This is where I was stuck. I should have showed what I had already done.

we have: 2(5y - 2) + y = -4

ok; on the 2(5y - 2):

the TWO terms inside the brackets are multiplied by the 2; so:

2 * 5y = 10y

2 * -2 = -4

so we now have : 10y - 4

so equation is now: 10y - 4 + y = -4

now add 4 to both sides to get: 10y + y = 0

so: 11y = 0, y = 0

substitute y=0 in 2x + y = -4:

2x + 0 = -4

2x = -4

x = -4/2 = -2

DO NOT FORGET:

if a(b + c) = x

then ab + ac = x

Tatoo that on your...whatever...

G

Any tips on where you would like me to tattoo this????

G

My math book sucks....(lack of a better word). I am so confused over the systems of linear equations in two variables. How do you know what method to use to solve the equation?????

Here is the problem I am looking at going ....hah?????

-0.25x-0.05y=0.2

10x + 2y=-8

The answer in the book says it is an infinitely many solutions answere

(x,y)(y=-5x-4

Any help you can give me will be so much appreciated. I am so lost here. I will also tatoo the answer on the right side of my brain Dennis if I finally get this into that part of my brain Why I do this to myself just so I don't feel dumb to my kids is beyond me!

- Joined
- Feb 4, 2004

- Messages
- 15,943

The tutor's name is "Denis", not "Dennis".

The method you use to solve a system, unless specified in the book, is up to you.

Since this particular system involves decimals, a good first step might be to multiply through to get rid of the decimal places. See where that takes you.

Eliz.

afreemanny said:"How do you know what method to use to solve the equation?????"

Well, you really don't; it becomes (in long run) a personal choice...

"Here is the problem I am looking at going ....hah?????

-0.25x-0.05y=0.2 [1]

10x + 2y=-8 [2]

The answer in the book says it is an infinitely many solutions answere

(x,y)(y=-5x-4"

I'd use this approach (my choice, like, Eliz would probably use something

different) ; multiply [1] by -40:

10x + 2y = -8 [1]

10x + 2y = -8 [2]

Now subtract them: 0 + 0 = 0 !!

So they're the same equation;

10x + 2y = -8

2y = -10x - 8

y = -5x - 4

"Why I do this to myself just so I don't feel dumb to my kids is beyond me!"

Hmmm...methinks we've all done worse things than that

- Joined
- Feb 4, 2004

- Messages
- 15,943

afreemanny said:"How do you know what method to use to solve the equation?

My first impulse, without thinking, was to multiply the first equation by 20, giving me:Denis said:Well, you really don't; it becomes (in long run) a personal choice.... I'd use this approach (my choice, like, Eliz would probably use something different...

. . . . . -5x - 1y = 4

. . . . .10x + 2y = -8

Then I probably would have multiplied by 2, adding down to get the same result as Denis. But you could also solve the first new equation above to get y = -5x - 4, and substitute:

. . . . .10x + 2(-5x - 4) = -8

. . . . .10x - 10x - 8 = -8

. . . . .-8 = -8

Either way, you get a "trivial" result, which means "infinitely-many solutions".

As you can see, the particular method you choose doesn't matter. Use whatever occurs to you first, or whatever you're more comfortable with. It's your choice.

Eliz.

G

Sorry for the tutor being wrote as "Dennis", instead of "Denis"..... I am always in a hurry.

Thanks Eliz....I just didn't realize they were letting you use either solution. I am learning slowly with this topic, but have come around others.

I am glad with all and anybody that thinks we have all done worse things! Keeps me on my toes!!!

I appreciate everyone's help that answers on this site. You all are wonderful!