if possible, solve the nonlinear system of equations.
x^2+y^2=58
y-x=4
from the 2nd ---> y = x+4
sub into the 1st
x^2 + (x+4)^2 = 58
2x^2 + 8x+16-58 = 0
x^2 + 4x - 21 = 0
(x+7)(x-3) = 0
x = -7 or x = 3
if x = -7, y = -7+4 = -3
if x = 3 , y = 3+4 = 7
x=3, y=7
or
x=-7, y = -3
thank you x)
To solve the given nonlinear system of equations, we can use the substitution method or the elimination method. Let's go with the substitution method.
Step 1: Solve one of the equations for a variable.
Let's solve the second equation, y - x = 4, for y:
y = x + 4
Step 2: Substitute the expression obtained in Step 1 into the other equation.
Substitute y in the first equation with (x + 4):
x^2 + (x + 4)^2 = 58
Step 3: Simplify and solve for x.
Expand the equation by squaring the binomial and combine like terms:
x^2 + (x^2 + 8x + 16) = 58
2x^2 + 8x + 16 = 58
Rearrange the equation and simplify:
2x^2 + 8x + 16 - 58 = 0
2x^2 + 8x - 42 = 0
Step 4: Solve the quadratic equation.
To solve the quadratic equation, we can either factorize or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the equation 2x^2 + 8x - 42 = 0:
a = 2, b = 8, c = -42
Substitute the values into the quadratic formula:
x = (-8 ± √(8^2 - 4*2*(-42))) / (2*2)
x = (-8 ± √(64 + 336)) / 4
x = (-8 ± √400) / 4
x = (-8 ± 20) / 4
Simplify both solutions:
x₁ = (-8 + 20) / 4 = 3
x₂ = (-8 - 20) / 4 = -7
Step 5: Find the corresponding y-values.
To find the y-values, substitute the x-values obtained in Step 4 into any of the original equations. Let's use the second equation, y - x = 4.
For x = 3:
y - 3 = 4
y = 7
For x = -7:
y - (-7) = 4
y + 7 = 4
y = -3
Therefore, the solutions to the given nonlinear system of equations are:
(x₁, y₁) = (3, 7)
(x₂, y₂) = (-7, -3)
The solutions to the equations are x = 3, y = 7 and x = -7, y = -3.