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System of equations: Idk how to do the highlighted questions

Guest May 21, 2017
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 #1
avatar+77243 
+1

 

Note  that they are only asking us to sketch the graphs

 

Here's the solution for  48    [ intersection points of a circle and a line ]  :  https://www.desmos.com/calculator/iipiffr3er

 

Here's 52 :  https://www.desmos.com/calculator/wqgjvkt18q

 

 And 53 :  https://www.desmos.com/calculator/zvh42ht4u5

 

The last two are tangent lines to circles

 

 

cool cool cool

CPhill  May 21, 2017
 #2
avatar+1224 
+1

You need not graph these equations to solve them (but the directions tell you to do so). In case you are in the situation where you are, this is what you can do:

 

\(x^2+y^2=13\)

\(y=-x+5\)

 

If I were you, I would solve this by substitution. In other words, since y=-x+5, you can substitute that into the y in the other equation, so you would get. Then, simplify:

 

\(x^2+(-x+5)^2=13\)

\(x^2+x^2-10x+25=13\) 

\(2x^2-10x+12=0\)

 

To make things simpler, factor out a 2 because that is the GCF of the left side and then factor (if possible):

 

\(2(x^2-5x+6)=0\)      x^2-5x+6 happens to be factorable

\(2(x-3)(x-2)=0\)     Set each factor set to 0 and solve.

\(x-3=0 \hspace{1cm} x-2=0\)

\(x=3\hspace{1cm}x=2\)

 

Now, plug each x-value to solve for the y-values. I'll pick the second equation as that one has simpler calculations overall:

 

\(y=-(3)+5\hspace{1cm}y=-(2)+5\)

\(y=2\hspace{1cm}y=3\)

 

Therefore, your solution set is:

 

\((2,3)\) and \((3,2)\)

 

Let's do that for the other problems for extra practice.

 

Let's list the original equations for problem 52:

 

\((x+1)^2+(y-1)^2=18\)

\(y=x+8\)

 

Substitute x+8 into the first equation and simplify:

 

\((x+1)^2+((x+8)-1)^2=18\)

\((x+1)^2+(x+7)^2=18\)

\(x^2+2x+1+x^2+14x+49=18\)     Combine like terms

\(2x^2+16x+32=0\)     Factor out a GCF of 2 to simplify things

\(2(x^2+8x+16)=0\) 

 

Now, I could factor x^2+8x+16, but this trinomial is special: it is a perfect square trinomial. I know this because I can work backwards. The x^2 term and 16 term are both perfect squares. The square root of x^2 equals x, and the square root of 16 equals 4. 4x*2=8x. If all three conditions are met (like it is here), then it is a perfect square trinomial:

 

\(2(x+4)(x+4)=0\)

\(x+4=0\)

\(x=-4\)

 

In this case, x only has one answer. This is actually easier than the previous problem because we only have to find one y-value instead of two. Now, plug in the x-value into the second equation:

 

\(y=x+8\) 

\(y=-4+8\)

\(y=4\)

 

Therefore, the solution set is:


\((-4,4)\)

 

YOu can apply this method for the next one, too...

 

\(\)

TheXSquaredFactor  May 21, 2017

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