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Q#!:solve by using elimination methods....
3x+4y=8
5x-2y=9
Q#2solve by using substituition method
15x-2y=11
7x+3y=19
solve the following equations
4x-2y=13
6x+y=12
one more equation plz to solve it out
x+0.5y=7
2x+y=19
one mooooreee plzzz
3x-4y=13
12x-16y=52
Q#4find the equation of line of the form
y+mx+b using
1:(3,0)&(0,4)intercepts method
2:(-2,-5)&m=-3/7 slope point method


taannxxx!!!urz 10 points are waiting for uu...i am badly stuck in it..urz participation will be highly appreciated InshAlllaahh!!

2007-03-10 04:31:58 · 6 answers · asked by student of honorZ 2 in Science & Mathematics Mathematics

6 answers

Ok, resolvamos paso a paso :

Q#!:solve by using elimination methods....

3x+4y=8

5x-2y=9

Let's multiply the second equation by two :

10x - 4y = 18

3x + 4y = 8

13x = 26 >>> x = 2

then replacing it on the first equation : y = 1/2

Q#2solve by using substituition method:

15x-2y=11

7x+3y=19

from the first equation : (15x - 11 ) / 2 = y

replacing it on the second equation :

7x + 3*( 15x - 11 ) / 2 = 19

7x + (45x - 33) / 2 = 19

59x - 33 = 38

59x = 71 >>> x = 71 / 59

replacing it on : (15x - 11 ) / 2 = y

15*71 / 39 - 11 = 2y >>> y = 636 / 22

solve the following equations :

4x-2y=13

6x+y=12

Let's multiply the second equation by 2 :

4x - 2y = 13

12x + 2y = 24

16x = 37 >>> x = 37 / 16

replacing "x" on the first equation :

148 / 16 -2y = 13

2y = -60 / 16 >>> y = -15 / 8

one more equation plz to solve it out

x+0.5y=7

2x+y=19

Let's multiply the first equation by -2

-2x - y = -14

2x + y = 19

0 = 5 ????, This one is not ok.

But if it is ok, the answer will be, that there are no solutions

one mooooreee plzzz

3x-4y=13`

12x-16y=52

Let' multiply the first equation by "-4"

-12x + 16y = -52

12x - 16y = 52

0 = 0

So, the equation has inifinity solutions.

Q#4find the equation of line of the form
y+mx+b using

1:(3,0)&(0,4)intercepts method

Y = mx + b

Replacing the point : 3,0

x = 3 >>> y = 0 :

0 = 3m + b

replacing the point : 0,4

4 = b

then : m = -4 / 3

The equation : Y = -4 / 3*X + 4

2:(-2,-5)&m=-3/7 slope point method

m = slope

Y = mx + b

Y = -3/7x + b

replacing the point -2, -5

-5 = 6/7 + b

b = 41 / 7

Y = -3/7*X + 41/7

2007-03-10 04:35:30 · answer #1 · answered by anakin_louix 6 · 1 1

ight, i assume the first guy is doing the same thing, ill edit when i get the answers

1) 3x +4y = 8
10x - 4y = 18 (multiplied it by to to get y's to cancel)

13x = 26 (added the equations)
x = 2 (divided by 13 on each side)

3(2) - 4y = 8
6 - 4y = 8
-4y = 2 (subtract 6 from both sides)
y = (-1/2) (divide by -4 on both sides)

now plug it in


2) for substitution solve one of the equations for y, then get lets say (2x +1, not your answer btw) and plug it in for y so 2x + 2x + 1 = 1000001 then 4x + 1 = 1000001 then 4x = 100000 then x = 250000



4)
1) (4-0)/ (0-3)=(4/-3) or (-4/3) is the slope
so...
y = (-4/3)x+b
4 = (-4/3)(0) + b (picked a point, used the one with x as zero to get rid of fraction)
4=b

so for #1 it is y = (-4/3)x + 4


2) do it the way i did #1 cept skip the first part where i found the slope (y-y)/(x-x), just go straight to "plug an chug"

2007-03-10 04:39:08 · answer #2 · answered by Ross 3 · 0 1

Dude - these are very easy, you should learn how to do them like you can say A B C. Try to think of them as solving puzzles instead of doing homework. You need to view these as fun.

4x-2y=13, so 4x=13+2y, so x=(13+2y)/4

6x+y=12, so 6((13+2y)/4)+y=12, so 1.5(13+2y)+y=12, so 19.3+3y+y=12, so 19.3+4y=12, so 4y=-7.3, so y=-1.825

then 4x-2(-1.825)=13, so 4x+3.65=13, so 4x=9.35, so x=2.3375

now check> 4(2.3375)-2(-1.825)=13?
9.35+3.65=13?
13=13 ok

2007-03-10 04:47:10 · answer #3 · answered by Anonymous · 0 0

I think you really need to learn the elimination method. Look at the examples in your textbook.

2007-03-10 04:38:44 · answer #4 · answered by Gnomon 6 · 0 1

Try working backwards
Q1.x=2 y=0.5

2007-03-10 04:39:24 · answer #5 · answered by Confused 6 · 0 2

hey budy

YOU WANT THE WHOLE HOMEWORK DONE RIGHT??
FREAKIN' LAZY !!

2007-03-10 04:46:48 · answer #6 · answered by Anonymous · 0 0

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