substitution method

G

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?????

Solve the system using the substituation method...
x=5y - 2

2x + y = -4

not sure what I am doing.......Thanks.
 
So take the first equation that's already solved for one of the variables, and substitute this in for that variable in the second equation.

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

Solve for y. Back-solve for x.

Eliz.
 
This is what I came up with:

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. :)
 
Check your work. What is (2)(-2)? (It isn't "2".) What is 4 ÷ 8? (It isn't "2".)

Eliz.
 
manny, you seem "lost" on this one...

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...:)
 
Thanks

Any tips on where you would like me to tattoo this????
 
I think I need a tutor

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!
 
Is this a new exercise? If so, please post new questions in new threads, not as replies to old threads.

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.
 
Re: I think I need a tutor

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 :)
Click to expand...
 
Re: I think I need a tutor

afreemanny said:
"How do you know what method to use to solve the equation?
Click to expand...
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...
Click to expand...
My first impulse, without thinking, was to multiply the first equation by 20, giving me:

. . . . . -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.
 
Thanks

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! :D
 
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