4x+3y = 2 and 5x-7y = -19
I already know that x=-1 and y=2 but if you know how, please find the answer(in an algebraic equation) to each of the expressions above and explain the processes thoroughly. Serious answers please. Thank you
2007-07-07 16:31:10 · 9 answers · asked by curiosium 1 in Science & Mathematics ➔ Mathematics
why do you multiply everything on the first equation by 7 and the second equation by 3?
2007-07-08 16:42:33 · update #1
what do you do and why so to get -43 and 43?
2007-07-08 16:44:55 · update #2
4x + 3y = 2 ==> multiply everything times 7
28x + 21y = 14
5x - 7y = -19 ==> multiply everything times 3
15x - 21y = -57
Add those two equations together:
28x + 21y = 14
15x - 21y = -57
----------------------
43x = -43
x = -1
substitute back into one of the original equations:
4(-1) + 3y = 2
-4 + 3y = 2
3y = 6
y = 2
x = -1 y = 2
I chose addition method because I find it easier to multiply than to work with a bunch of fractions.
2007-07-07 16:34:27 · answer #1 · answered by Becky M 4 · 1⤊ 0⤋
There are two ways of doing this: elimination and substitution. I prefer elimination. The goal of elimination is to be able to multiply each equation by a constant so that when you add them one variable falls out.
4x + 3y = 2
5x - 7y = -19
I am going to multiply eq 1 by 5 and eq 2 by -4 so that eq 1 has an x coefficient of 20 and eq 2 has an x coefficient of -20.
20x + 15y = 10
-20x + 28y = 76
Add the two equations together:
43y = 86
y = 2.
You can now plug in 2 in either of the original equations:
4x+3*2 = 2
4x+6 = 2
4x = -4
x = -1.
Using substitution. Solve one equation for one variable and then substitute the expression into the other equation.
2007-07-07 23:45:42 · answer #2 · answered by Michael 2 · 0⤊ 0⤋
You can solve this problem using systems or by using matrices. Do you know how to do either? It's extremely hard to explain online. Basically, with systems, you would write one equation in terms of x, like x= -3/4y + 1/2 (Get everything that is not x on the other side). Then you can substitute "x" (that equation) into the second equation. This will give you everything in terms of y. Then you can solve for y. Then substitute y back into the equation with x (-3/4y +1/2). Now you have both x and y. Sorry if it is confusing...check your textbook or another source if you need to.
You can also use matrices, if you know what they are. I think that is much easier, but it's too hard to explain online. I hope this helps at least a little!
2007-07-07 23:41:40 · answer #3 · answered by Clare 2 · 0⤊ 0⤋
These are the steps:
* 4x + 3y = 2
5x - 7y = -19
* 4x * 5 + 3y * 5 = 2 * 5 (multiply this by 5)
5x * 4 - 7y * 4 = -19 * 4 (multiply it by 4)
The results of the multiplication:
* 20x + 15y = 10 (1)
20x - 28y = -76 (2)
Take (1) minus (2), we have:
* 20x + 15y - 20x +28y = 10 +76
=> 43y = 86
=> y = 2
Replace y = 2 into (1), we have x = -1.
2007-07-07 23:40:42 · answer #4 · answered by lyhn990 4 · 1⤊ 0⤋
4x + 3y = 2
5x - 7y = - 19
28x + 21y = 14
15x - 21y = - 57----ADD
43x = - 43
x = - 1
- 4 + 3y = 2
3y = 6
y = 2
x = - 1 , y = 2
2007-07-08 07:51:24 · answer #5 · answered by Como 7 · 0⤊ 0⤋
I like substitution better because it is harder :) ;seriously lol
Elimination is easier though. Just "stack" the equations on one another. :)
2007-07-07 23:44:04 · answer #6 · answered by Anonymous · 0⤊ 0⤋
first you solve for x when y=0
then you plug the answer into the other equation to solve for y
2007-07-07 23:37:00 · answer #7 · answered by KES 1 · 0⤊ 2⤋
USE ELIMINATION ( or substitution) BUT ELIMINATION is cooler
2007-07-07 23:35:56 · answer #8 · answered by fatttypants 2 · 0⤊ 1⤋
it's way past ur bed time !
2007-07-07 23:34:59 · answer #9 · answered by Anonymous · 0⤊ 4⤋