# Math Help - HELP with a IMPOSSIBLE question

1. ## HELP with a IMPOSSIBLE question

 ive uploaded a question thats im really stuck on... ive got the answer but i really wanna know how its worked out aswell please can someone explain thanks

2. ## Re: HELP with a IMPOSSIBLE question

Hey Gary

What you have is a system of equations. You have two equations and two unknowns, which makes it not very impossible to solve. I will put stars ( $*$) where it might not be clear what I am doing.

Equation A: $x^2-2xy+4y=-60$
Equation B: $x+y=18$

What you want to do is define one of the variables $(x$ or $y)$ in one of the equations and plug it into the other equation in order to solve. When you evaluate equation B, you will notice that you easily can define by manipulating a bit: $x=18-y$. If you plug the found $x$ into equation A, you get the following:

$(18-y)^2-2y(18-y)+4y=-60 *$

This can obviously be a nicer looking expression:

Step one: $18^2+y^2-36y-36y+2y^2+4y=-60$
Step two: $324+3y^2-68y=-60 *$

As you might notice, we now have a quadratic equation of sorts, which we must solve for $y$ using the proper method. A quadratic equation has the form $ax^2+bx+c=0$ and therefore we must set our equation equal to 0. This we can do by adding 60 on both sides:

$324+3y^2-68y + 60=-60+60$ will become $3y^2-68y+384=0$

To proceed, we must find the discriminant $D$ to plug into the solution formula used for quadratic equations. If the equation is to have a real solution, $D$ must be 0 or larger. If it's zero, one solution exists. If it's larger than zero, two solutions exist. The discriminant is generally defined $D=b^2-4ac$

in this case we have:

$D=(-68)^2-4 \cdot 3 \cdot 384=16$

There are two solutions to the equation. The solution formula is, generally

$x= \frac{-b \pm \sqrt{D}}{2a}$

In this case:

$y= \frac{68 \pm \sqrt{16}}{2 \cdot 3}= \frac{32}3 \lor 12$

Now that $y$ is known in numerical values, also $x$ can be defined numerically. Remember equation B? We simply plug the found $y$ values in instead of $y$ and get a very simple equation which will yield some numerical $x$ values:

$x+y=18 \Rightarrow x+ \frac{32}{3} =18 \lor x+12=18$,

thus we get $x=\frac{22}{3} \lor 6$

I am to tired to actually think about the other content of your problem, but I guess you can figure it out from here. Good night. I hope my reply was helpful and I didn't obscure too much in my drunkenness

$*$ Remember: $(a-b)^2=(a-b)(a-b)=a^2+b^2-2ab$
$* 18^2=324$

3. ## Re: HELP with a IMPOSSIBLE question

can someone explain as its done please xx

4. ## Re: HELP with a IMPOSSIBLE question

Uh, I spent 30 minutes writing a solution with explanations for you tonight. What is it exactly you don't understand?

I can put my explanation a little bit differently, if that will help.

First of all, as I said, you have a system of equations. You have two equations and two unknowns, and that makes it possible to define one of the unknowns in one of the equations and then substitute it for the same unknown $($in this case, I chose $x)$ in the other equation, thus eliminating it $($in this case still $x)$ from that equation, making it possible to solve for the other unknown.

The equations are:

Equation A: $x^2-2xy+4y=-60$
Equation B: $x+y=18$

If you solve equation B for $x$, $x$ gets defined thus:

$x+y=18 \iff x=18-y$

Now you can plug that definition of $x$ into equation A and proceed as I already described in the first post. If this still doesn't make any sense to you, you can watch this video, where the same thing is done with a simpler expression:

Secondly, the system of equations will yield a new equation which is a quadratic equation, as it can be written in the form $ax^2+bx+c=0$. This equation cannot simply be solved as a "simple" linear equation, and therefore you should solve it using the quadratic formula, which is both explained in my post, and in this video. Quadratic equations can also be solved by factoring or by plugging them into calculators, but if you want to learn the fundamental technique, learn the formulas as described:

Having found the numerical value of $y$ by solving the quadratic equation, both equations shed $y$ as an unknown, and they can be solved as any other equation. The only twist here is, that the quadratic equation yields two values of $y$, and therefore also $x$ will have to be defined with the respect to the two solutions for $y$.