(i) 7x^2 + y^2 = 64

(ii) x + y = 4

I’m just going to take the second one and rearrange it,

x + y = 4 => y = 4 – x…(iii)

Now I’m going to take (iii), and substitute it into (i) to solve for x,

7x^2 + y^2 = 64

=> 7x^2 + (4 – x)^2 = 64

=> 7x^2 + (16 – 8x + x^2) = 64

=> 7x^2 + 16 – 8x + x^2 = 64

=> 7x^2 + x^2 – 8x + 16 – 64 = 0

=> 8x^2 – 8x – 48 = 0

=> 8(x^2 – x – 6) = 0

=> 8(x – 3)(x + 2) = 0

=> x = 3 or -2

Now substitute each of these x values into (ii), to get the corresponding y values,

x + y = 4 = 3 + y = 4 => y = 1

OR

x + y = 4 => -2 + y = 4 => y = 6

In other words, there are two solutions for this system of equations,

When x = 3, y = 1, OR

When x = -2, y = 6.

Note: If you try to graph the first line, it should be something like an “upside-down” parabola, and the second line should just be a straight line with negative slope. The straight line will intersect the parabola at two points: (3,1), and (-2,6); hence the two solutions to the system.

i tried it a few times till i suddenly remembered it(i’m 20, it’s been a while since i did these equations..) the reason you can’t make it, and that i wasted so much time getting the wrong answers, is that we forgot a new equation that comes into play when you have squared numbers sometimes. now, in a free translation from hebrew, it’s called squared equation, but before we get to it, we need to arrange the two equations first. so what you do is, you take the second one, and isolate Y.

1) x + y = 4 —-> y = 4 – x

now you position this instead of the Y in the first equation

2) 7x^2 + y^2 = 64 —–> 7x^2 + (4-x)^2 = 64

now arrange it

7x^2 + (4-x)^2 = 64 —-> 7x^2 + 16 – 8x + x^2 = 64 —–>

8x^2 – 8x + 16 = 64 ok, this was simple algebric arraingment, that i hope you understand. now, comes into play the new equation, which requires a certain arraingment of the equation at hand.

example – X^2 + X + number = zero(must be zero)

so we arrainge our equation so –

>>>>>8x^2 – 8x – 48 = 0<<<<< now, we 3 players, A, B, C. A is the squared number always, B is the factor number, and C is the plain number. so in our case, A = 8 B = (-8) C = (-48) these are very importent and you should right them aside, for they are the factors for the squared equation which is problematic to explain here because it has 2 signs that i don't know how to type, one is plus/minus(it means you it twice, once as plus, and once as minus, you're supposed to get 2 answers) the other one is square root. so let's decide that # would be plus/minus, and what will be in these <> will be in the square root.

-B # < b^2 -4*A*C> divided by 2*A. equals X like this

-B # < b^2 -4*A*C>

————————– = X

2*A

so you put the numbers in for the A, B, and C, and get two answers because of the plus/minus, and these are X one, and X two. you put these two eich in it’s turn in one of the original equations, and get the Y one, and Y two(repectivly of course)

now, in certain cases, as i suspect in this, since the equations represent a certain parbula or line on a graph, on of the answer is supposed to none admissible. so if that is the case, you check if the answers make sense in the other equations, and dismiss the answer that doesn’t make sense, and pick the answer that does make sense. now to example this on our equations lets do it.

we said, A=8 B=(-8) C=-48

-(-8)%<8^2-4*8*(-48)>

—————————– = X

2*8

8%<64+ 1536>

——————— = X (64+1536= 1600)

16 ( <1600> = 40

8%40

———– = X

16

48 32

—– = 3 = X one —— = 2 = X two

16 16

so now we have two X’s and we put them in the y + x = 4, to get the y’s, and then we put both of them in squared equation to see if it makes sense, and we get that the x =2 y=2 doesn’t make sence and we dismiss it, and the x=3 y=1 does make since and pick that.

that’s it, it turned out to be long, but i did my best to try and explain it artiquletly. i hope you understood it, and that it helps you, and that you do well on your homework. bye.