How to find all points having an x coordinate of 3 whose distance from the...

The following formula is used to determine the separation between two points:
${\displaystyle d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$
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Substituting these values and 5 for d, we can solve for ${\displaystyle y_1}$:
${\displaystyle 5=\sqrt{{(-1-3)}^2+{(-4-y_1)}^2}}$
First, square both sides of the equation:
${\displaystyle 5^2={\left(\sqrt{{(-1-3)}^2+{(-4-y_1)}^2}\right)}^2}$
${\displaystyle 25={(-1-3)}^2+{(-4-y_1)}^2}$
${\displaystyle 25={(-4)}^2+{(-4-y_1)}^2}$
${\displaystyle 25=16+{(-4-y_1)}^2}$
${\displaystyle 25=16+({\left(-4\right)}^2-(2\cdot -4\cdot y_1)+{y_1}^2)}$
${\displaystyle 25=16+(16-(-8y_1)+{y_1}^2)}$
${\displaystyle 25=16+16+8y_1+{y_1}^2}$
${\displaystyle 25=32+8y_1+{y_1}^2}$
${\displaystyle 25={y_1}^2+8y_1+32}$
${\displaystyle 25-25={y_1}^2+8y_1+32-25}$
${\displaystyle 0={y_1}^2+8y_1+7}$
${\displaystyle {y_1}^2+8y_1+7=0}$
This can be considered as:
${\displaystyle (y_1+7)(y_1+1)=0}$
Now that we have found the solutions, we can solve each term on the left for 0:
Solution 1:
${\displaystyle y_1+7=0}$
${\displaystyle y_1+7-7=0-7}$
${\displaystyle y_1+0=-7}$
${\displaystyle y_1=-7}$
Solution 2:
${\displaystyle y_1+1=0}$
${\displaystyle y_1+1-1=0-1}$
${\displaystyle y_1+0=-1}$
${\displaystyle y_1=-1}$
The two points satisfying this problem are:
${\displaystyle (3,-7)}$ and ${\displaystyle (3,-1)}$