Solving Simultaneous Equations involving algebraic equations/?

Question

4/x+1/y=5/3
1/x-3/y=3/2

Could you pls explain step by step the process to solve the simulataneous equations.

Answer 1

4/x+1/y=5/3
1/x-3/y=3/2

The method to sovling simlutaneous equations is to add them together in such a way as to eliminate one of the variables then solve for the remaining variable. To do this with your equations, multiply the second equation through by -4 and add the two equations together.

4/x+1/y=5/3
-4/x+12/y=-12/2

1/y+12/y=5/3-12/2, now multiply by y

1+12=(5/3)y-6y

13=(-13/3)y

39=-13y

y = -3, so if y is -3 then substitute that into one of the equations to find x.

4/x+1/(-3)=5/3, mulitply by x

4-(1/3)x=(5/3)x
4=2x

x=2, y=-3

Answer 2

First step -- isolate one of the variables.
In this case, add 3/y to both sides of the lower equation.
This leaves you with that equation reading 1/x = 3/2 + 3/y.
Invert both sides, leaving x = 1/(3/2 + 3/y).

Plug this value for x into the top equation, which then becomes
4/(1/(3/2 + 3/y)) + 1/y = 5/3.
But 1/(1/(3/2 + 3/y)) = 3/2 + 3/y, so multiplying this by 4 gives
12/2 + 12/y + 1/y = 5/3.
Putting all of the y's on one side and all of the numbers on the other gives 12/y + 1/y = 5/3 - 12/2.
adding these gives 13/y = (-6+5/3) = -18/3 + 5/3 = -13/3.
Thus y = -3.

Plugging this value for y into the top equation yields
4/x + 1/(-3) = 5/3
adding 1/3 to each side yields 4/x = 5/3 + 1/3 . or 6/3, or 2.
Since 4/x = 2, x must equal 2.

Checking, using both values,
4/2 + 1/(-3) = 5/3, and 2 - 1/3 = 5/3
1/2 -(3/-3) = 3/2, and 1/2 + 1 = 2.

Answer 3

I found this website elesson.co.uk I am learning algebra again. They use animated teaching method. You get very similar to a tutor teaching. It is great even I understand it. Please have a look. There is a great explanation about simultaneous equations on there. I have not got there yet.

Looking at your question and what I looked up on elesson it seems that you need to get y's the same so I would times the top equation by 3 and get 12/x + 3/y = 5. Now you can get rid of y's. I will look again and tell you what to do.

Right. Add the two equations
12/x + 3/y = 5 and
1/x - 3/y=3/2
you'll lose +3/y with -3/y and get
13/x = 5 3/2 0r 13/2
13/x = 13/2 so x must be 2

I 'll go to have another look.
You have to use x in one of equations. Let's use it in 12/x+3/y =5
you get 12/2 + 3/y = 5
that's 6 + 3/y = 5

So 3/y = -1 So y = -3

I hope this is OK.

Answer 4

You have to make them so that they both either the x's or the y's can be cancelled out and then add/subtract the two equations to/from each other or sometimes you can just add/subtract the two equations from each other, if they already have x's or y's in common. This one doesn't

e.g. second equation x 4. So your two equations are:

4/x + 1/y = 5/3
4/x - 12/y = 12/2

You can now take away equation 2 from equation 1.

(4/x + 1/y = 5/3) - (4/x - 12/y = 12/2)

So for this you would cancel out the x's (4/x - 4/x - 0). Now you do (1/y - -12/y, which would be 1/y + 12/y) then (5/3 - 12/2).

Once you have this you will swap the answer with the y. e.g:

2/y = 4/6 would mean you would do:

2/(4/6) = y

You can then work out the sum to work out y.

Answer 5

Here are your 2 equations that I've somewhat rearranged and labeled for convenience in referring to them:
A) 4 {1/x} + 1[1/y] = 5/3
B) {1/x} - 3[1/y] = 3/2

In your equations it's expedient to treat {1/x} and [1/y] -- think bracket and brace (respectively) as your variables; then once they're determined, just take the reciprocals to get x and y. From A and B I'm choosing to eliminate the bracket variable {1/x}. Which variable you choose to eliminate doesn't matter. I'm gonna multiply equation B thru, by -4 and get what I'll label -4B:

-4{1/x} + 12[1/y] = - 6

Now add this equation to equation A (Your {1/x} is gone!)

13[1/y] = 5/3 + (-6) or -13/3 then divide each side by 13 and get

[1/y] = -1/3 which is more convenient in cking than the obvious y = -3. Plug either info into either equation A or B and determine that {1/x} = 1/2 which means
x = 2

Answer 6

4/x + 1/y = 5/3 ---(1)
1/x - 3/y = 3/2 ---(2)
(2)x4: 4/x - 12/y = 6 ---(3)
(3)-(1): -13/y = 13/3 =? y = -3

Substitute y = -3 into (1).
4/x + 1/-3 = 5/3
4/x = 5/3 + 1/3
4/x = 2
x = 4/2
x = 2

Answer 7

This website does not support the characters necessary to show the mathematical expressions you need. It is very clumsy. You should be working with other college students who are studying mathematics or engineering (third year) for help, the Laplace transforms are way beyond what most people here can handle. Besides, I can't remember how to do it and my notes are buried in boxes. Sorry !

Answer 8

4/x+1/y=5/3 (Eq. #1)
1/x-3/y=3/2 (Eq. #2)

4/x + 1/y = 5/3 (Eq. #1)
4y + x = 5xy/3
x = 5xy/3 - 4y
x = y(5x/3 - 4)
y = x/(5x/3 - 4)

Now substitute results from Eq. #1 into Eq. #2:
1/x-3/y=3/2 (Eq. #2)
1/x - 3/[x/(5x/3 - 4)] = 3/2
1/x - [3 * (5x/3 - 4)]/x = 3/2
1/x - 5 + 12/x = 3/2
13/x = 13/2
26 = 13x
2 = x

Now substitue to calculate y:
y = x/(5x/3 - 4)
y = 2/(5*2/3 - 4)
y = 2/(10/3 - 4)
y = 2/(-2/3)
y = -3

So the solution is (2,-3) using substitution

Answer 9

let s=1/x ........(a)
and t=1/y........(b)

substitute into original equations
4s+t=5/3.......(1)
s-3t=3/2....... (2)
multiply(1) by 3
12s+3t=5......(3)
add(2)and(3)
13s=13/2
>>>s=1/2
substitute back into (1)
2+t=5/3
>>>t=5/3-2= -1/3

from(a),s=1/x
1/2=1/x>>>>x=2
from(b),t=1/y
-1/3=1/y>>>>y= -3

hence,x=2,y= -3

i hope that this helps

Answer 10

4/x+1/y=5/3.............(e1)
1/x-3/y=3/2...............(e2)

(e1)*3 +(e2)

12/x+3/y=15/3
1/x-3/y=3/2

so 3/y and -3/y cancelled

12/x+1/x=15/3+3/2
12+1/x=(30+9)/6
13/x=39/6
39x=(13*6)
x=78/39
x=2

put value of x in (e2)

1/2-3/y=3/2
-3/y=3/2-1/2
-3/y=2/2
2y=-(3*2)
2y=-6
y=-6/2
y=-3

that's the simplest way i can explain good bye