Note that (4-x) -3 = 1-x
x^2 + 4x + 4 + 1 -2x + x^2 = 5
2x^2 + 2x + 4 = 5
2x^2 + 2x -1 = 0
x = [-2 +/-sqrt12]/4
= [-1 +/-sqrt3]/2
Check my math. I'm getting old
use substitution method for solving
x+y=4
(x+2)^2+(y-3)^2=5
I got the first part to solve it
x +y =4
y=4-x
the second part i got to
(x+2)^2 + ((4-x)-3)^2=5
x^2 + 2x + 2 ((4-x)-3)^2 = 5
and that is all i could get that i knew was right but i don't know how to get
2 answers
First, you have an error in multiplying.
(x+2)^2 + ((4-x)-3)^2=5
(x+2)^2 = x^2 + 4x + 4
((4-x)-3)^2 = (-x+1)^2 = x^2 - 2x + 1
The two together become
x^2 + 4x + 4 + x^2 - 2x + 1 = 5
Combining terms, you get
2x^2 + 2x + 5 = 5
2x^2 + 2x = 0
2x(x + 1) = 0
So x = 0 or -1
Find y from your shorter equation, then check by inserting both values into the longer equation.
I hope this helps. Thanks for asking.
(x+2)^2 + ((4-x)-3)^2=5
(x+2)^2 = x^2 + 4x + 4
((4-x)-3)^2 = (-x+1)^2 = x^2 - 2x + 1
The two together become
x^2 + 4x + 4 + x^2 - 2x + 1 = 5
Combining terms, you get
2x^2 + 2x + 5 = 5
2x^2 + 2x = 0
2x(x + 1) = 0
So x = 0 or -1
Find y from your shorter equation, then check by inserting both values into the longer equation.
I hope this helps. Thanks for asking.
1 answer