First question
\(line:~~~~y=x+3\)
\(curve:~~~~x^2+y^2=29\)
now we have to replace the y of curve by the y of line, therefore
\(x^2+(x+3)^2=29\)
\(x^2+x^2+6x+9=29\)
\(2x^2+6x+9-29=0\)
\(2x^2+6x-20=0\)
we can multiply each member by \(\frac{1}{2}\)
\(\boxed{\boxed{x^2+3x-10=0}}\)
Now we have to find the roots of this funtion
\(x^2+3x-10=0\)
Sum and produc or Bhaskara
Then we find two axis
\(\boxed{x_1=2~~and~~x_2=-5}\)
now we replace this values to find the Y, we can replace in curve or line equation, I'll prefer to replace it in line equation.
\(y=x+3\)
\(y_1=x_1+3\)
\(y_1=2+3\)
\(\boxed{y_1=5}\)
\(y_2=x_2+3\)
\(y_2=-5+3\)
\(\boxed{y_2=-2}\)
Therefore
\(\boxed{\boxed{P_1(2,5)~~and~~P_2(-5,-2)}}\)
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The second question give to us
\(y=ax+b\)
\(P_1(2,13)\)
\(P_2(-1,-11)\)
We just have to replace the value then we'll get a linear system.
point 1
\(13=2a+b\)
point 2
\(-11=-a+b\)
then our linear system will be
\(\begin{Bmatrix}2a+b&=&13\\-a+b&=&-11\end{matrix}\)
I'll multiply the second line by -1 and I'll add to first one
\(\begin{Bmatrix}2a+(a)+b+(-b)&=&13+(-11)\\-a+b&=&-11\end{matrix}\)
\(\begin{Bmatrix}3a&=&24\\-a+b&=&-11\end{matrix}\)
\(\begin{Bmatrix}a&=&8\\-a+b&=&-11\end{matrix}\)
therefore we can replace the value of a, at second line
\(\begin{Bmatrix}a&=&8\\-8+b&=&-11\end{matrix}\)
\(\begin{Bmatrix}a&=&8\\b&=&-11+8\end{matrix}\)
\(\boxed{\boxed{\begin{Bmatrix}a&=&8\\b&=&-3\end{matrix}}}\)
then our function will be
\(\boxed{\boxed{y=8x-3}}\)
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The third one, we have
\(line:~~~~y=3x-4\)
\(curve:~~~~y=x^2-2x-4\)
This resolution will be the same of our first question.
Let's replace the y of curve by the y of line
\(3x-4=x^2-2x-4\)
\(0=x^2-2x-4-3x+4\)
therefore
\(\boxed{\boxed{x^2-5x=0}}\)
now we have to find the roots of this function.
\(x^2-5x=0\)
put x in evidence
\(x*(x-5)=0\)
\(\boxed{x_1=0~~and~~x_2=5}\)
then
\(y=3x-4\)
\(y_1=3x_1-4\)
\(y_1=3*0-4\)
\(\boxed{y_1=-4}\)
\(y_2=3x_2-4\)
\(y_2=3*5-4\)
\(y_2=15-4\)
\(\boxed{y_2=11}\)
our points will be
\(\boxed{\boxed{P_1(0,-4)~~and~~P_2(5,11)}}\)
I hope you enjoy it ;)
