Solve this system of equations algebraically: x+y=4 & x^2+y^2-6y=0

Question

whpalmer4 · Answer

$$x+y=4$$$$x^2+y^2-6y=0$$

This seems like it should be straightforward to solve by substitution.  Solve the first equation for one of the variables in terms of the other.  Then use that equation to substitute in the second equation, giving you a somewhat more complicated equation, but with only one variable instead of two.  You should be able to solve that equation.  When you do, use the result to find the value of the other variable, using the substitution equation.

I'll do an example:

$$x+y  = 2$$$$x^2 + y^2 = 4$$
You've got a line passing through a circle. The solutions will be the points where they intersect.

Solve first equation for $x$ in terms of $y$:
$$x+y = 2$$$$x = 2-y$$

Now substitute $(2-y)$ wherever you see $x$ in the second equation:$$(2-y)^2 + y^2 = 4$$$$4-4y+y^2 + y^2 = 4$$$$2y^2-4y = 0$$$$2y(y-2) = 0$$$$y = 0,~y=2$$

Now we plug those values of $y$ back in the substitution equation to find the values of $x$:

$y = 0$, $x = 2-y = 2-0 = 2$
that means one intersection takes place at $(2,0)$
$y = 2$, $x = 2-y = 2-2 = 0$
that means another intersection takes place at $(0,2)$

Here's a graph:

whpalmer4 · Answer

Your problem doesn't have a solution with nice round numbers like mine, so don't be discouraged when you start seeing things like $\sqrt{17}$...

whpalmer4 · Answer

Your problem doesn't have a solution with nice round numbers like mine, so don't be discouraged when you start seeing things like $\sqrt{17}$...