How do you get it? I've tried for 30 minutes now...
y=x-3
x^2+y^2=65
Yes I do know substitution, plus add, and the third way to find intersections.
Not a bad effort. Now, you don't want to end up with your last line unless you have a 0 on the right hand side as that allows you to equate each factor equal to 0 which we don't have here.
What you should do is try to get your expression in the form of a quadratic ($\displaystyle ax^{2} + bx + c = {\color{red}0}$). The 0 is important as this is the key for using the quadratic formula or to factor for the reason mentioned above:
$\displaystyle x^{2} + x^2 - 6x + 9 = 65$
$\displaystyle 2x^{2} - 6x - 56=0$
$\displaystyle x^{2} - 3x - 28 = 0$ (divided both sides by 2)
Can you see what to do now?
Hello, AlphaRock! Your work looks fine so far (except for the 2 in the last step). Now let us solve it!
Are you familiar with solving quadratic equations? There are a few methods you can use:
First, you can try factoring:
$\displaystyle x^2-3x=28\Rightarrow x^2-3x-28=0$
$\displaystyle \Rightarrow x^2+(4-7)x+(-7)\cdot4=0$
$\displaystyle \Rightarrow (x+4)(x-7)=0$
Next, if you can't figure out how to factor the expression, or if the expression is simply too messy to easily factor, you can use the quadratic formula, $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $\displaystyle a, b,\text{ and }c$ are the coefficients of the 2nd-degree, 1st-degree, and constant terms, respectively.
In this case, we have
$\displaystyle x^2-3x=28\Rightarrow x^2-3x-28=ax^2+bx+c=0$
So, in the formula, $\displaystyle a=1, b=-3,\text{ and }c=-28$.
Alternatively, you can complete the square. For example:
$\displaystyle x^2-3x=28\Rightarrow x^2-3x+\frac94=28+\frac94$
$\displaystyle \Rightarrow\left(x-\frac32\right)^2=\frac{121}4$
$\displaystyle \Rightarrow x-\frac32=\pm\sqrt{\frac{121}4}$
$\displaystyle \Rightarrow x=\frac32\pm\frac{11}2$
I hope that helps!
Edit: Beaten by o_O! I'm too slow with my LaTeX!
Ah, here's a more harder question:
4x+y+11=0 and (x-5)^2+(y-3)^2=16
What do I do to solve this question...? I'm clueless on what ot do next.
Here's what I did:
y=(-11-4x)
(x-5)^2 + (-11-4x-3)^2 = 16
x^2-10x+25+(-14-4x)^2 = 16
x^2-10x+25 + 196 + 16x^2 + 112x = 16
17x^2+102x+221-16 = 16-16
17x^2+102x+205=0
>>Imaginary. What do I do...? THis is all my work.
Your work is okay so far. If you continue, using the quadratic formula, you should get
$\displaystyle x=-\frac{2\sqrt{221}\,\text{i}+51}{17},\ \frac{2\sqrt{221}\,\text{i}-51}{17};\
y=\frac{8\sqrt{221}\,\text{i}+17}{17},\ -\frac{8\sqrt{221}\,\text{i}-17}{17}$
Edit: But, make sure you have the right problem, because I notice that if you change the addition in the second equation to a subtraction the two graphs will intersect.