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