Answer:
1) When x is 1, y is 4, and vice versa
2) When x is -6, y is 42, and when x is 7, y is 114
3) When x is -2, y is 2, and when x is 8, y is 42
Step-by-step explanation:
1)
[tex]y=x^2-6x+9[/tex]
[tex]y+x=5[/tex] which can be rearranged as [tex]y=5-x[/tex].
Substituting this lone y into the first equation, you get:
[tex]5-x=x^2-6x+9[/tex]
Move everything to one side:
[tex]x^2-5x+4=0[/tex]
Factor:
[tex](x-4)(x-1)=0[/tex]
[tex]x=1, 4[/tex]
[tex]y=4, 1[/tex]
2)
[tex]y-30=12x[/tex] which can be rearranged as [tex]y=12x+30[/tex].
[tex]y=x^2+11x-12[/tex]
Let's use elimination this time and subtract the first equation from the second:
[tex]0=x^2-x-42[/tex]
Factor:
[tex](x-7)(x+6)=0[/tex]
[tex]x=-6, 7[/tex]
[tex]y= -42, 114[/tex]
3)
[tex]y=x^2-2x+6[/tex]
[tex]y=4x+10[/tex]
Let's set the two equations equal to each other through substitution:
[tex]x^2-2x-6=4x+10[/tex]
Move everything to one side:
[tex]x^2-6x-16=0[/tex]
Factor:
[tex](x-8)(x+2)=0[/tex]
[tex]x=-2, 8[/tex]
[tex]y= 2, 42[/tex]
Hope this helps!
