Solve the following system of equations: $ \begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases} $

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

$$ x=\sqrt{61}-1 $$ $$ y=\sqrt{61}+1 $$ or $$ x=\sqrt{61}+1 $$ $$ y=\sqrt{61}-1 $$

$$ x=0 $$ $$ y=\sqrt{61} $$ or $$ x=\sqrt{61} $$ $$ y=0 $$

$$ x=\sqrt{61}-2.5 $$ $$ y=\sqrt{61}+2.5 $$ or $$ x=\sqrt{61}+2.5 $$ $$ y=\sqrt{61}-2.5 $$

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

Solve the following system of equations: $ \begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases} $

$$ x=\frac{\sqrt{61}}{2}-2.5 $$ $$ y=\frac{\sqrt{61}}{2}+2.5 $$ or $$ x=\frac{\sqrt{61}}{2}+2.5 $$ $$ y=\frac{\sqrt{61}}{2}-2.5 $$

$$ x=\sqrt{61}+5 $$ $$ y=\sqrt{61}-5 $$ or $$ x=\sqrt{61}-5 $$ $$ y=\sqrt{61}+5 $$

$$ x=0 $$ $$ y=\sqrt{61} $$ or $$ x=\sqrt{61} $$ $$ y=0 $$

$$ x=\sqrt{3}+1 $$ $$ y=\sqrt{3}-1 $$ or $$ x=\sqrt{3}-1 $$ $$ y=\sqrt{3}+1 $$

Solution of a system of equations - one of them is linear and the other quadratic

Solving a system of equations when one of them is linear and the other is quadratic

When we have a system of equations where one of the equations is linear and the other quadratic
we will use the substitution method:

We will isolate one variable from an equation, place in the second equation the value of the expression of the variable we have isolated, and in this way, we will obtain an equation with one variable. We will solve for $X$ or $Y$ and then place it in one of the original equations to find the complete point. The point we discover will be the point of intersection of the line with the parabola, and it will also be the solution of the system of equations.

Detailed explanation

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Test yourself on linear-quadraric systems of equations!

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Let's look at an example

Given the following system of equations: $y=x^2+2x+7$ , $2x+y=4$

Solution:
Let's isolate one of the variables from one of the equations:
$2x+y=4$
$y=-2x+4$

Now, let's substitute the value we isolated into the second equation. We will obtain:

$-2x+4=x^2+2x+7$

Let's combine like terms and transpose members. We will obtain:
$-X^2-4x-3=0$
We will solve it with the help of the quadratic formula and will obtain:
$X=-1 ,-3$
Now we will gradually place an $X$ we found into one of the equations to find the complete points: Let's start with $X=-1$

$2\times(-1)+y=4$
$-2+y=4$
$y=6$
Let's write this solution $(-1,6)$
let's move to the second $X$ we found $X=-3$ and we will obtain:
$2*(-3)+y=4$
$-6+y=4$
$y=10$ let's write this solution $(-3,10)$