Problem 17

Question

Use any method to solve the system of nonlinear equations. $$ \begin{array}{r} -x^{2}+y=2 \\ -x+y=2 \end{array} $$

Step-by-Step Solution

Verified
Answer
The solutions are \((0, 2)\) and \((1, 3)\).
1Step 1: Solve the Linear Equation for y
Start by solving the linear equation \(-x + y = 2\) for \(y\). Add \(x\) to both sides to get: \(y = x + 2\).
2Step 2: Substitute y in the Nonlinear Equation
Substitute \(y = x + 2\) into the nonlinear equation \(-x^2 + y = 2\). This changes the equation to \(-x^2 + (x + 2) = 2\).
3Step 3: Simplify and Solve the Equation
Simplify the equation \(-x^2 + x + 2 = 2\) by subtracting 2 from both sides to get \(-x^2 + x = 0\). Factor out an \(x\) to obtain \(x(-x + 1) = 0\).
4Step 4: Solve for x
Set each factor of the equation \(x = 0\) and \(-x + 1 = 0\) equal to zero to solve for \(x\). This gives the solutions \(x = 0\) and \(x = 1\).
5Step 5: Find y for each x value
Use the expression \(y = x + 2\) to find \(y\) for each value of \(x\). For \(x = 0\), substitute into \(y = x + 2\) to get \(y = 2\). For \(x = 1\), substitute into \(y = x + 2\) to get \(y = 3\).
6Step 6: Verify the Solutions in the Original Equations
For each solution pair, substitute back into both original equations to verify. Check \((0, 2)\): in \(-x + y = 2\), \(-0 + 2 = 2\) is true, and in \(-x^2 + y = 2\), \(-0^2 + 2 = 2\) is true. Check \((1, 3)\): in \(-x + y = 2\), \(-1 + 3 = 2\) is true, and in \(-x^2 + y = 2\), \(-(1)^2 + 3 = 2\) is true.

Key Concepts

System of EquationsLinear EquationsSubstitution MethodFactorization
System of Equations
A system of equations refers to a set of two or more equations with the same set of variables. Solving these systems means finding the values of the variables that satisfy all the equations simultaneously. In this exercise, the system consists of two equations:
  • The first equation is nonlinear: \(-x^2 + y = 2\)
  • The second equation is linear: \(-x + y = 2\)
To solve a system, you can use methods like substitution, elimination, or graphing. Solving systems of equations is fundamental in mathematics as it has applications in various scientific and engineering fields. Understanding how different types of equations in a system can interact helps in approaching complex real-world problems.
Linear Equations
A linear equation is an equation in which each term is either a constant or the product of a constant and a single variable. These equations form a straight line when graphed on a coordinate plane. In this problem, the linear equation is given by:
  • \(-x + y = 2\)
This equation is simple to solve for one variable in terms of the other, typically resulting in expressions like: \(y = x + 2\). Linear equations are essential for solving more complex systems, especially when used in conjunction with nonlinear equations. They provide a straightforward pathway to isolate and substitute variables.
Substitution Method
The substitution method is a common technique for solving systems of equations. It involves solving one equation for one of the variables and then substituting this expression into another equation.
In this exercise, the substitution method was used by first solving the linear equation for \(y\):
  • \(y = x + 2\)
Next, this expression was substituted into the nonlinear equation, transforming it into a single equation with one variable. This simplification allows for easier solving, making it particularly useful when dealing with systems where one equation is linear and the other is nonlinear.
Factorization
Factorization is a mathematical process used to express an equation as a product of its factors. In this exercise, once the linear expression was substituted into the nonlinear equation, the equation was simplified and set to zero:
  • \(-x^2 + x = 0\)
The next step involved factorizing this expression into: \(x(-x + 1) = 0\).
Finding the values of \(x\) involves setting each factor to zero:
  • \(x = 0\)
  • \(-x + 1 = 0\), which solves to \(x = 1\)
This factorization is crucial because it breaks down the problem into simpler parts, making it easier to identify the solution points for the original system of equations.