Pleeeeeeeeeeease Help ME!!! This math question makes no sense :(?

Question

If 3x + 4y = 14, and (2x)²(x²)³y= (-5/2)x, then y = __________

This makes no sense to me please explain. And I am in geometry right now, so I havent gotten into logarithm as someone had suggested when I first posted this question a week ago.

PLEASE HELP ME!!! Thanx in advance!

Answer 1

those are two separate questions asking to solve for y, right?

first one:  all you need to do is isolate y, so, 3x + 4y = 14 => 4y = 14 - 3x, then divide both sides by 4 to get y, which would be y = (14 - 3x)/4 => 7/2 - 3x/4

second one, you can just divide both sides by all the x's on the left, so you get: y = (-5/2)x/(4x^8), and cancel the x's, so you now have: (-5/2)/(4x^7) => y = -5/8x^7.

Answer 2

4y=14 - 3x   so y = (14 - 3x)/4

and

4x²(x^6)y = 4(x^8)y = -5x/2 or y = (-5x) / (2*4(x^8)) = -5/(8(x^7))

so y = (14-3x)/4 = -5/(8(x^7))

if teacher wants a numerical answer, you have to go on and solve for x.

I see no ovious way to solve analytically, but numerically I get

x =4.666684 (approximately)

<edit>

no (3.5,0) is not right.  (0,3.5) gives you plugged into 3x+4y = 14 and it gives you 0 = 0 so it IS a solution, too.  Note that

you can not divide by (x^7) if x is zero.  So solutions are easiest to find graphically.

Answer 3

you solve for x in the first equation, then u plug taht in the second eqation and you solve for y

Answer 4

This problem caught my eye because it really is strange.

I started out by working with the 2nd equation:

(2x)²(x²)³y= (-5/2)x

(4x²)(x²)³y = (-5/2)x  

(4x²)(x)^6y = (-5/2)x

4(x)^8y = (-5/2)x

(2) 4(x)^8y = (-5/2)x (2) =

8(x)^8y = -5x

  I got this far by raising 2x to the 2nd power (=4x²).Then I raised (x²)³ to get (x)^6 b/c (x²)(x²)(x²) = (x)^6. Multiplying 4x² by (x)^6 = 4(x)^8 because when you multiply common bases, their exponents are added. Then I multiplied by 2 to get rid of the denominator on the right side of the equation.

  Usually when you are given two equations, you are supposed to solve one of the equations for one variable ... in other words you could solve eq# 1 for y:

3x + 4y = 14

4y = 14 - 3x

divide by 4

y = 14/4 - 3/4x

Then you substitute your y value into eq# 2 and solve for x, etc., etc. I tried that with this ... and got:

28(x)^8 - 6(x)^9 = -5x

  I just can't believe that this is what your teacher had in mind . Maybe you (or a curious math whiz out there) can look at my attempt and get somewhere with it ??

I'm really curious, I'd like to know the answer ...

Good luck ... sorry I couldn't be of help :(

LATER : After additional details, eq#1 can be solved to get x=4, y=1/2 ...does that help?

STILL MORE ... try graphing eq#1: y = -3/4x + 7/2 and for eq#2: y = -5/8x^7

For help graphing high powers of x check out:

there's a graph of x^6 about 1/4 down the page ... you just have to take into consideration slope = -5/8. With a negative slope the line will fall (left to right) down 5 over 8, etc

Answer 5

Basically use the sfirst equation solving for x in terms of y, then substitute your first equation into the second one. It wiil probably get a little messy sense you have such a high power for x's.

Let's try it:

3x + 4y = 14

3x=14-4y

x=(14-4y)/3 Now put this in for x in the secon equation

But first simplify the second equation to be: (2x^8)y=(-5/2)x

That becomes y=(-5/2)x/(2x^8)=-5/(4x^7)

So now we have y= -5/(4x^7) with x=(14-4y)/3

Substituting for x we get y=-5/[(4(14-4y)^7/3^7]

But that is       y = -5*(3^7)/[(4(14-4y)^7]

There is no way to solve for y without logs.....sorry

Answer 6

What you have is a system of equations.  To solve for y, you need to solve the first equation for y, (so y = (14-3x)/4), and insert that into the second one.  Then multiply that out and solve for x.  This will give you a numerical value for x.  Then use that back in the first equation to find out what y is.