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
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
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.
To solve the system of equations using the substitution method, you need to substitute the expression for "y" from the first equation into the second equation. Here's how to continue from where you left off:
From the first equation, you found that y = 4 - x. Now substitute this value for y in the second equation:
(x + 2)^2 + ((4 - x) - 3)^2 = 5.
Simplify the expression:
(x + 2)^2 + (1 - x)^2 = 5.
Expanding the square terms:
(x^2 + 4x + 4) + (1 - 2x + x^2) = 5.
Combine like terms:
x^2 + 4x + 4 + 1 - 2x + x^2 = 5.
Simplify further:
2x^2 + 2x + 5 = 5.
Subtract 5 from both sides:
2x^2 + 2x = 0.
Now, you can factor the equation:
2x(x + 1) = 0.
Apply the zero product property:
2x = 0 or x + 1 = 0.
Solve for x:
x = 0 or x = -1.
Now substitute the values of x back into the first equation to solve for y:
For x = 0:
0 + y = 4,
y = 4.
For x = -1:
-1 + y = 4,
y = 5.
Hence, the solution to the system of equations is (x, y) = (0, 4) and (-1, 5).