# Solve The System Using Substitution

• Jan 30th 2013, 12:47 PM
alexthesauceboss
Solve The System Using Substitution
Hello MHF,

In this problem, it asks to "solve the system by the method of substitution."

and it gives me the following two equations:
{-7/2x - y = -18
{8x^2 - 2y^3 = 0

I pretty much know what I have to do, but there are a few things stopping me from completing the problem...I will try to demonstrate what I mean with an attached picture.
• Jan 30th 2013, 01:14 PM
ebaines
Re: Solve The System Using Substitution
You seem to have tried to say that:

$(\frac 7 2 x - y)^3 = (\frac 7 2 x)^3 + (-y)^3$,

but that's not right. If you multiply $(A+B)^3$ out you find it's not $A^3 = B^3$,but rather: $(A+B)^3 = (A+B)(A+B)(A+B) = A^3 + 3A^2B + 3AB^2 + B^3$.

I would suggest rather than use that approach try this: from the first equation you can get y by itself:

$\frac {-7} 2 x - y = -18 \ \to\ y = \frac {-7} 2 x + 18$.

Now substitute this value for 'y' into the second equation:

$8x^2 -2y^3 = 0 \ \to \ 8x^2 -2(\frac {-7} 2 x + 18)^3 = 0$

Can you take it from here?
• Jan 30th 2013, 01:24 PM
alexthesauceboss
Re: Solve The System Using Substitution
Quote:

Originally Posted by ebaines
You seem to have trid to say that:

$(\frac 7 2 x - y)^3 = (\frac 7 2 x)^3 + (-y)^3$,

but that's not right. If you multiply $(A+B)^3$ out you find it's not $A^3 = B^3$,but rather: $(A+B)^3 = (A+B)(A+B)(A+B) = A^3 + 3A^2B + 3AB^2 + B^3$.

I woud suggest rather than use that approach try this: from the first equation you can get y by itself:

$\frac {-7} 2 x - y = -18 \ \to\ y = \frac {-7} 2 x + 18$.

Now substitute this value for 'y' into the second equation:

$8x^2 -2y^3 = 0 \ \to \ 8x^2 -2(\frac {-7} 2 x + 18)^3 = 0$

Can you take it from here?

Now I end up with 8x^2 + 343/4x = 11,664.

What do I do to solve that?
• Jan 30th 2013, 07:26 PM
alexthesauceboss
Re: Solve The System Using Substitution
*Bump* Request for help on this problem still. I do not understand how to simplify and solve for x, even after I have substituted it in to the equation with the y value.
• Jan 31st 2013, 06:18 AM
ebaines
Re: Solve The System Using Substitution
This is gonna get complicated.

Starting from $8x^2 -2 (\frac {-7} 2 x +18)^3 =0$ you can multiply out the cubic to get this:

$8x^2 -2(\frac {-343}{8} x^3 + 3 \frac {882} 4 x^2 - 3 \frac {2268} 2 x + 5832) = 0$

After a bit of rearrangement:

$\frac {343} 4 x^3 - 1315 x^2 + 6804 x -11664 = 0$

Now at this point you could try to find the roots of this cubic, but it's a bit of a mess. I would suggest instead that you plot this and you will find that it has one real root, and that root is not too far from 0. By inspection you can see that if x is an integer the first term will be a fraction UNLESS x is multiple of 4, whereas the other terms are all integers. I think that's enough hints - try it and see what you get.