YAMathT: Not sure how to solve this algebra problem

duragezic

Lifer
Oct 11, 1999
11,234
4
81
I had a quiz in a class today but had trouble at the end with solving the equations for the two unknowns. The numbers are from memory but I'm just curious how to do it.

It is probably easy as hell but I just drew a blank and my algebra isn't too hot though I've tried to rely less on my calculator. Haha funny story about that: in one of my first EE classes, intro to signal processing, this guy who sat next to me was a senior EE but re-taking it. He had said one day "this is what makes me an engineer maaan", as he raises his 89 in the air with a big smile. :D Anyway...

I needed to solve for two variables, which I will call x and y to make it easy.

eqn1:
182e-6 = .5 * x * (3 - y)^2

eqn2:
235e-6 = .5 * x * (4 - y)^2

eqn3:
433e-6 = .5 * x * (5 - y)^2


I guess I would only need 2 equations if there are only 2 unknowns. My ideas:

1) divide one equation by the other, but IIRC there is something like I can't just cancel out the square cause I'd lose the sign. I then end up with two values, and eliminate the value that doesn't make sense, leaving me with the value for x (or y) which I would then substitute into any equation to solve for y (or x).

2) just re-arrange any equation and solve for x (or y) in terms of the other variable, plug that into another equation and solve for y (or x), then with that value I could get x (or y) as a number.

Would either of these work? I'm not sure if where I said "any equation" is correct. Or is there just a common, better way of doing something like this?


my elegant solution ;) (not important to my question)
Since I was pressed for time doing the rest of the problem, and unsure how to solve those equations, I did a good ol' guess and check.

It was solving for values for a MOSFET given a bunch of other ones, and since I knew a common threshold voltage for a n-MOSFET was 1.0 V, I started with it and by my third guess, the resulting i_D was near exact one of the givens. w00t!

So I think I got them right but maybe not, or I got lucky, and I should know how to do this anyway.
 

Kelemvor

Lifer
May 23, 2002
16,930
7
81
Normally for 2 variables you solve one equation to isolate one variable. Then you plug that into the other equation so you end up with only one variable. THen when you get that, you go back and figure out the other one.
 

HN

Diamond Member
Jan 19, 2001
8,186
4
0
in eq1, solve for x
plug that into eq2 and solve for e
plug that into eq3 and solve for y. at this point you can solve for the value of y.

then go in reverse to solve for the other two variables.

or you might just have to use the quadratic formula.

*edit*
:laugh: nevermind me :laugh:
 

txrandom

Diamond Member
Aug 15, 2004
3,773
0
71
Originally posted by: HN
in eq1, solve for x
plug that into eq2 and solve for e
plug that into eq3 and solve for y. at this point you can solve for the value of y.

then go in reverse to solve for the other two variables.

e means exponent, why would you solve for that?
 

Tiamat

Lifer
Nov 25, 2003
14,074
5
71
Originally posted by: HN
in eq1, solve for x
plug that into eq2 and solve for e
plug that into eq3 and solve for y. at this point you can solve for the value of y.

then go in reverse to solve for the other two variables.

or you might just have to use the quadratic formula.

I think "e" in this case means multiply by 10 to the power of the following number rather than a variable.


3 eqn and 2 unknowns means you have a redundancy somewhere. You only need 2 eqns for 2 unknowns.

I pm'd you my results. I got y as a negative value, and x as a positive, but in the order of E-6.
 

HN

Diamond Member
Jan 19, 2001
8,186
4
0
Originally posted by: txrandom
Originally posted by: HN
in eq1, solve for x
plug that into eq2 and solve for e
plug that into eq3 and solve for y. at this point you can solve for the value of y.

then go in reverse to solve for the other two variables.

e means exponent, why would you solve for that?

heh, back to reading algebra formulas for me :eek:
 

duragezic

Lifer
Oct 11, 1999
11,234
4
81
I don't have an 89! Sometimes my 83+ can pull through when I have no clue at all, and it does have a simultaneous eqn solver, but it doesn't help here.

Ok, I think I tried the 1st suggestion on the quiz, but the algebra got kinda weird. It went like this:

1) Solving eqn. 1 for x:
x = 364e-6 / (3 - y)^2

2) plug that expression for x into eqn. 2 which results in:
235e-6 = .5 * 364e-6 / (3 - y)^2 * (4 - y)^2

3) solving for that I did: multiply both sides by 2, divide both sides by 364e-6 which leaves:
1.2912 = (4 - y)^2 / (3 - y)^2

4) unsure what to do here so I brought the (3 - y)^2 to left side, distributed the square and multiplied those three terms by 1.2912, then distributed the (4 - y)^2 on the right side, so it's:
11.6209 - 7.747y + 1.2912y^2 = 16 - 8y - y^2

5) move y's to one side, numbers to other, leaving:
-4.3791 = -.253y - .2912y^2

6) pull out common y, divide by remaining, left with:
y = -4.3791 / (-.253 - .2912y)

Those last couple of steps probably aren't right. The numbers seem pretty effed up by that point, and doesn't really seem like I'm getting anywhere moving crap or distributing or whatever. But if that is even right at step 6, I know I could use the simple 0= solver on my 83, but I'm not sure how to split up the terms to isolate y from the denominator, in which case I could then solve for y, then get x.

On the right track, or way off? :)

 

Cogman

Lifer
Sep 19, 2000
10,277
125
106
To me, at least, it looks like you are on the right track, I haven't actually checked your math, but the method is correct. Next you can use the third equation to isolate y. One nit pick, if each equation has .5 x in them then you really don't need to solve for x, just .5 x then replace it everywhere. The hard part is going to be isolating and solving for y, good luck on that one :)
 

esun

Platinum Member
Nov 12, 2001
2,214
0
0
This is my thinking:

For ease of expression, let A = 182e-6, B = 235e-6, C = 433e-6.

Then we have (moving all those 0.5's over):

2A = x * (3 - y)^2 (1)
2B = x * (4 - y)^2 (2)
2C = x * (5 - y)^2 (3)

Solve (1) for x:

x = 2A / (3 - y)^2 (4)

Plug x into (2):

B/A = (4 - y)^2 / (3 - y)^2

2 square roots:

B/A = (4 - y1) / (3 - y1)
B/A * (3 - y1) = 4 - y1
3B/A - 4 = y1(B/A - 1)
y1 = (3B/A - 4) / (B/A - 1)

-B/A = (4 - y2) / (3 - y2)
-B/A * (3 - y2) = 4 - y2
4 - 3B/A = -y2(B/A + 1)
3B/A - 4 = y2(B/A + 1)
y2 = (3B/A - 4) / (B/A + 1)

Use (4) to solve for the respective x1, x2 values for y1, y2. Then plug into (3) and see which pair works. That's your solution.

EDIT: BTW, if you didn't have to show your work (i.e. this was a problem just to solve numerically on your calculator), you could've just graphed the three equations (on your TI-83) and found the intersection point of all 3. That would give you the solution very quickly.
 

duragezic

Lifer
Oct 11, 1999
11,234
4
81
Thanks for the help. The way I was going in my last post, I must've made a mistake with something simple as the numbers are way off when I use the solver on my calculator.

The graphing idea is good... I'm sure it would've been fine to use as long as I sketched the graph or said what I did (it's not a math class). But now that I think of it, I can't graph more than one variable with my 83+. I guess if I went along the same lines as with algebra and solved for one variable in terms of the other, it'd be pretty much doing the same thing except avoiding the algebra by using the graphing to find the value. Cool.

Still though, the various techniques or rules involved with solving it from beginning to end with algebra are something I should know better!

I'll try it again with esun's suggested way, as I KNOW I'll have hours of free time at "work" tomorrow. :)
 

LordMorpheus

Diamond Member
Aug 14, 2002
6,871
1
0
the system is over-defined. Find a solution for any two of those equations. If it satisfies the third, then you have your answer. If not, try other solutions to those first two equations. If none fit the third equation, there is no answer.
 

Itchrelief

Golden Member
Dec 20, 2005
1,399
0
71
Originally posted by: duragezic

4) unsure what to do here so I brought the (3 - y)^2 to left side, distributed the square and multiplied those three terms by 1.2912, then distributed the (4 - y)^2 on the right side, so it's:
11.6209 - 7.747y + 1.2912y^2 = 16 - 8y - y^2

5) move y's to one side, numbers to other, leaving:
-4.3791 = -.253y - .2912y^2

Looks like here you could move everything to one side, 0.2912y^2 + 0.253y -4.3791 = 0

and plug into the quadratic equation

Sorry, haven't checked your work up to that point, and honestly, even if I did, I'd probably F it up anyways.
 

zerocool1

Diamond Member
Jun 7, 2002
4,487
1
81
femaven.blogspot.com
Originally posted by: txrandom
Originally posted by: HN
in eq1, solve for x
plug that into eq2 and solve for e
plug that into eq3 and solve for y. at this point you can solve for the value of y.

then go in reverse to solve for the other two variables.

e means exponent, why would you solve for that?

he's using the carot (^) so that means exponent. (e = 2.71828183) e is a constant
 

Demon-Xanth

Lifer
Feb 15, 2000
20,551
2
81
2*182e-6/((3-Y)^2) = x
2*235e-6/((4-y)^2) = x
2*433e-6/(5-y)^2) =x

Now that you got a fun filled variety of ways you can plug in for x.
2*182e-6/((3-Y)^2) = 2*235e-6/((4-y)^2)
182e-6/((3-Y)^2) = 235e-6/((4-y)^2)
182e-6*235e-6 = 1/(((4-y)^2)*((3-Y)^2))
1/(182e-6*235e-6) = (4-y)^2*(3-Y)^2

I'll let you simplify the rest. :)
Once you figure out Y, plug it back into the original equasion and find X.
 

PlatinumGold

Lifer
Aug 11, 2000
23,168
0
71
Originally posted by: duragezic
I had a quiz in a class today but had trouble at the end with solving the equations for the two unknowns. The numbers are from memory but I'm just curious how to do it.

It is probably easy as hell but I just drew a blank and my algebra isn't too hot though I've tried to rely less on my calculator. Haha funny story about that: in one of my first EE classes, intro to signal processing, this guy who sat next to me was a senior EE but re-taking it. He had said one day "this is what makes me an engineer maaan", as he raises his 89 in the air with a big smile. :D Anyway...

I needed to solve for two variables, which I will call x and y to make it easy.

eqn1:
182e-6 = .5 * x * (3 - y)^2

eqn2:
235e-6 = .5 * x * (4 - y)^2

eqn3:
433e-6 = .5 * x * (5 - y)^2


I guess I would only need 2 equations if there are only 2 unknowns. My ideas:

1) divide one equation by the other, but IIRC there is something like I can't just cancel out the square cause I'd lose the sign. I then end up with two values, and eliminate the value that doesn't make sense, leaving me with the value for x (or y) which I would then substitute into any equation to solve for y (or x).

2) just re-arrange any equation and solve for x (or y) in terms of the other variable, plug that into another equation and solve for y (or x), then with that value I could get x (or y) as a number.

Would either of these work? I'm not sure if where I said "any equation" is correct. Or is there just a common, better way of doing something like this?


my elegant solution ;) (not important to my question)
Since I was pressed for time doing the rest of the problem, and unsure how to solve those equations, I did a good ol' guess and check.

It was solving for values for a MOSFET given a bunch of other ones, and since I knew a common threshold voltage for a n-MOSFET was 1.0 V, I started with it and by my third guess, the resulting i_D was near exact one of the givens. w00t!

So I think I got them right but maybe not, or I got lucky, and I should know how to do this anyway.



not good at math, used to be but haven't done it in too long and forgotten almost everything i know about it, but. . .

eqn1:
182e-6 = .5 * x * (3 - y)^2

eqn2:
235e-6 = .5 * x * (4 - y)^2

eqn3:
433e-6 = .5 * x * (5 - y)^2

in the above 3 eqns is the .5 necessary? as it is the same .5X in all three eqns couldn't it just be written X?

so

eqn1:
182e-6 = (3 - y)^2(X)

X = (182e-6) / (3-y)^2

eqn2:
235e-6 = (4 - y)^2(X)

235e-6 = (4-y)^2 * ((182e-6) / (3-y)^2)

eqn3:
433e-6 = (5 - y)^2(X)

after this i'm lost. i'm not good at quadradics, but i guess you could square the y's and do the quadradic.

 

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,606
166
111
www.slatebrookfarm.com
relatively simple to solve, but is this an exercise in you algebraic manipulation skills? Or, do you just need the answer?

Since you mentioned you have a TI83,
Solve each equation for x as you've already done.

Enter them as equations to graph (pretend that x is y and y is x)
Graph. The solution is the point where they overlap. (but x and y are reversed.)
The TI83 can solve for the point of intersection.

edit: note: it's still relatively simple to solve for y to avoid x and y being reversed, however it's quicker to solve for x.
 

Paperdoc

Platinum Member
Aug 17, 2006
2,292
270
126
In your post at 6:20 pm on 7/25, step 4 gets you to an equation with quadratic terms on both sides. Gather them and use the standard rule for the solution to a quadratic equation.
You had:
11.6209 - 7.747y + 1.2912y^2 = 16 - 8y - y^2
BUT that should have been a +y^2 at the end, not -. So, gathering:

(1.2912 - 1) y^2 + (8 - 7.747) y +(11.6209 - 16) = 0
0.2912 y^2 + 0.253 y - 4.3791 = 0

The solution to ANY quadtratic equation of the form:
ay^2 + by + c = 0

is given by:

y = (-b ± SQRT(b^2 - 4*a*c)) / 2*a
= (-0.253 ± SQRT(0.253*0.253 -4 * 0.2912 * (-4.3791))) / 2 * 0.2912
= (-0.253 ± 2.2726) / 0.5824
= either 3.4677 or - 4.3365
From the first equation, x = 364 e-6 / (3-y)^2
and hence, if y = 3.4677, then x = 1664 e-6,
OR if y = -4.3365, then x = 6.763 e-6

Should be only one of those sets of x and y fits all three equations.

From your post and others, it appears you did not know the general solution to a quadratic equation. That is the key to solving this one. HN was right to point that out, and Itchrelief took you most of the way, too, but said "plug that in ..." etc, assuming you did know the general solution.
 

Fenixgoon

Lifer
Jun 30, 2003
31,528
9,886
136
personally, i would solve for X symbolically, then substitute it in any of the other equations, and bust out a quadratic formula to solve for Y. then, plug the two values of Y back in the equations, solve for X. plug everything in to make sure both X and Y values are true for all 3 equations.