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: \(-x + y = 2\)
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:
- \(-x + y = 2\)
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 \(y\):
In this exercise, the substitution method was used by first solving the linear equation for \(y\):
- \(y = x + 2\)
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:
Finding the values of \(x\) involves setting each factor to zero:
- \(-x^2 + x = 0\)
Finding the values of \(x\) involves setting each factor to zero:
- \(x = 0\)
- \(-x + 1 = 0\), which solves to \(x = 1\)
Other exercises in this chapter
Problem 17
For the following exercises, find the determinant. \(\left|\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|\)
View solution Problem 17
For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$$
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Use the matrices below to perform scalar multiplication. \(A=\left[\begin{array}{rr}4 & 6 \\ 13 & 12\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 21 & 1
View solution Problem 17
For the following exercises, use the matrices below to perform scalar multiplication. $$ A=\left[\begin{array}{cc}{4} & {6} \\ {13} & {12}\end{array}\right], B=
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