This is a substitution problem.
Let's take "b" for an example:
Substitute 3y - 4 in the second equation for x.
4y - 2(3y - 4) = 13
4y - 6y - 8 = 13
-2y - 8 = 13
-2y = 21
y = -21/2 or 10.5
Now plug this value for y into the first equation of "b"
x = 3 * (10.5) - 4
x = 31.5 - 4
x = 27.5
------------
*Always* check your answers by plugging the results into the original equation.
4y - 2x = 13 ?
[4 * (10.5)] - [2(27.5)] = 13 ?
42 - 55 = 13 ?
13 = 13 <== Checks!
Do the same for "a", except that you have to make the coefficient of x or y, (whichever you choose), set to 1.
Good luck.
2006-09-21 16:02:06
·
answer #1
·
answered by Mitch 7
·
0⤊
1⤋
Simultaneous Equations are so fun!
Ok for Qa)
6x - 3y = 18 - (1)
4x + 3y = 12 - (2)
Let (1) - (2)
therefore:
6x - 3y = 18
(+) 4x + 3y = 12
------------------------
10x + 0 = 30
now, 10x = 30
therefore x = 30/10 or 3
knowing x = 3, you can now substiute it in one of the original equations!
eg:
4(3) + 3y = 12
12 + 3y = 12
3y = 12 - 12
3y = 0
therefore y = 0 (because anything times by 0 always equal to zero)
For Q b)
x = 3y - 4 -(1)
4y - 2x = 13 -(2)
1st step in all simultaneous equations are to put them in order
so, (1) can be re-arranged as:
-3y + x = -4
But as u can see, we can't make an unknown disappear (make it equal to zero by adding or subtraction) so, we find the best way to do so.. this is... to multiply (1) by a factor of 2
therefore:
(-3y + x = -4 ) times 2
=> -6y + 2x = -8 -(3)<-- let this become 3!
therefore it can be solved!
allow (3) - (2)
-6y + 2x = -8
(+) 4y - 2x = 13
--------------------------
-2y + 0 = 5
therefore 2y = -5
y = -5/2 or -2.5
now that you know what the y-value is. u can substitude it with the original equation.
Therefore substitute y= -2.5 into (1)
x = 3(-2.5) - 4
x = -7.5 - 4
x = -11.5
these questions can also be done via Matrix. If you are curious you can msg me!
2006-09-21 16:12:08
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
6x - 3y = 18
4x + 3y = 12
From 6x - 3y = 18
-3y = 18 - 6x
3y = -18 + 6x *sub into 4x + 3y = 12
4x + (-18 + 6x) = 12
4x - 18 + 6x = 12
10x = 12 + 18
10x = 30
x = 3 *sub into 3y = -18 + 6x
3y = -18 + 6(3)
3y = -18 + 18
3y = 0
y = 0
to check ;
6x - 3y
= 6(3) - 3(0)
= 18
4x + 3y
= 4(3) + 3(0)
= 12
===============================================
x = 3y - 4
4y - 2x = 13
4y - 2x = 13
4y - 2(3y - 4) = 13
4y -6y + 8 = 13
-2y = 13 - 8
-2y = 5
y = - 5/2 *sub into x = 3y - 4
x = 3(-5/2) - 4
x = -15/2 - 4
x = -15/2 - 8/2
x = -23/2
to check ;
4y - 2x
= 4(-5/2) - 2(-23/2)
= -20/2 + 46/2
= 26/2
= 13
2006-09-21 16:14:19
·
answer #3
·
answered by evelyn_01 3
·
0⤊
0⤋
2+2=5
2006-09-21 15:57:18
·
answer #4
·
answered by Ronnoc 3
·
0⤊
2⤋
In part a, if you look at the equations, you'll notice the -3y and +3y. If you add the two equations, the y-terms will drop out and give you 10x = 30. Once you have your answer for x, substitute it in one of the equations to find y.
In part b, the easiest thing to do is substitute 3y - 4 for the x in the second equation.
4y - 2(3y - 4) = 13. Now you just have one variable. Once you solve for y, substitute your answer in the first equation to find x.
2006-09-21 15:52:36
·
answer #5
·
answered by PatsyBee 4
·
1⤊
1⤋
okay, here is steps for the first problem:
1. line up equation and subtract
6x-3y=18
-4x+3y=12
so after subtracting the two problems, the -3y and 3y cancel eachother out. the equation now looks like this:
2x=6
so now you slove like a regular algebraic problem:
2x divided by 2 = 6 divided by 2 which is equal to 3
so x=3
Now for B:
first you need to move the x over in the first equation so the problem will now look like this:3y-x=4
then to cancel out the x's you need to mulitply the entire top equation by 2 and solve like above
2006-09-21 16:09:28
·
answer #6
·
answered by kelsey 3
·
0⤊
0⤋
For the first one, add the 2 equations to gether to get 10x=30. Use this to find x, than use x to find y in one of the equations.
For the second, replace the x in the second equation with (3y-14)
since that equals x, to get 4y - 2(3y - 14) = 13
4y - 6y + 28 = 13
Solve this for y, then plug it in to find x.
2006-09-21 15:54:01
·
answer #7
·
answered by hayharbr 7
·
0⤊
2⤋
WHAT has to be solved? Throwing some random equations at people doesn't give you an answer right away. It's pretty nice if you'd put the question with it too, you see.
Good luck with it!
2006-09-21 15:50:06
·
answer #8
·
answered by Jonneke . 2
·
0⤊
2⤋
hi,
Go to http://www.purplemath.com/
2006-09-21 15:52:40
·
answer #9
·
answered by kimmepool 2
·
0⤊
1⤋
a) x = 7.6, y = -2.8
b) x = -11.5, y = -2.5
2006-09-21 15:51:52
·
answer #10
·
answered by teddy 2
·
0⤊
2⤋