How to solve for $X$ given $A = X + Y$ and $B = 1 + YX$?

Question

So I have the following system of equations:

$A = X + Y$

$B = 1 + YX$

And I'm asked to solve for $X$ and $Y$. My question is how do I do so? I was taught that I should first try to isolate $X$ and get it in terms of $A$ and $B$. So my attempt was the following:

$A = X + Y$

$B = 1 + YX$

into

$-A = -X-Y$

$B = 1 + YX$

into

$-AX = -x^2-YX$

$B = 1 + YX$

so adding $-AX + B$ together I get

$B-Ax=-x^2+1$

which does not isolate $X$ in terms of $A$ and $B$ as there's still an $X$ on the LHS

So this is where I get stuck. Can someone please help me figure out the necessary algebra? Is my way of going at this completely wrong? I have the same issue isolating $Y$. Any help is greatly appreciated.

Answer 1

$X = A - Y$ and $Y = \frac{B - 1}{X}$ ($X\neq 0$), then $X^2 - AX - B + 1 = 0$. Solving quadratic equation, you get $X = \frac{A \pm \sqrt{A^2 + 4(B - 1)}}{2}$

Answer 2

From the first equation, you get that $Y=A-X$. Now, from the second equation you get that$$B=1+(A-X)X=1-AX+X^2.$$Now, use the quadratic formula.

Answer 3

If we assume that $x=0$ is not a solution, we can write $$x+y-a=0\iff x^2-ax+xy=x^2-ax+b-1=0.$$

Then

$$\left(x-\frac a2\right)^2=1-b+\frac{a^2}2$$

from which you can draw $x$ ($2$ solutions).

Answer 4

If you familiar Vieta's fomulas

$$\begin{cases}X+Y= A\\ XY=B-1\end{cases}$$

Then, Vieta gives us

$$t^2-At+(B-1)=0$$

where $t_1=X,~t_2=Y.$