Answer :
Answer:
The solutions are the points (1,4) and (4,1)
Step-by-step explanation:
we have
[tex]6x+y=x^{2}+9[/tex] ----> equation A
[tex]x+y=5[/tex] ----> [tex]y=-x+5[/tex] ---> equation B
substitute equation B in equation A
[tex]6x+(-x+5)=x^{2}+9[/tex]
solve for x
[tex]6x-x+5=x^{2}+9[/tex]
[tex]5x+5=x^{2}+9[/tex]
[tex]x^{2}+9-5x-5=0[/tex]
[tex]x^{2}-5x+4=0[/tex]
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-5x+4=0[/tex]
so
[tex]a=1\\b=-5\\c=4[/tex]
substitute in the formula
[tex]x=\frac{-(-5)(+/-)\sqrt{-5^{2}-4(1)(4)}} {2(1)}[/tex]
[tex]x=\frac{5(+/-)\sqrt{9}} {2}[/tex]
[tex]x=\frac{5(+/-)3} {2}[/tex]
[tex]x=\frac{5(+)3} {2}=4[/tex]
[tex]x=\frac{5(-)3} {2}=1[/tex]
The solutions are x=1,x=4
Find the values of y
For x=1 -----> [tex]y=-(1)+5=4[/tex] ----> (1,4)
For x=4 -----> [tex]y=-(4)+5=1[/tex] ----> (4,1)
therefore
The solutions are the points (1,4) and (4,1)