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Section 9.5 Nonlinear Systems

Subsection 1. Write a system of equations

Problems that involve two variables can sometimes be described by a system of equations.

Subsubsection Examples

Example 9.56.

The area of a rectangle is 874 square millimeters, and its perimeter is 122 millimeters. Write a system of equations for the dimensions of the rectangle.

Solution

We use the formulas for area and perimeter, where \(w\) stands for the width of the rectangle, and \(l\) stands for its length.

\begin{align*} lw \amp = 874\\ 2l+2w \amp = 122 \end{align*}
Example 9.57.

A small plane flies at a constant speed. In 5 hours, it travels 750 miles with a tailwind, but it travels only 600 miles in 5 hours against the wind. Write a system of equations for the speed of the plane and the speed of the wind.

Solution

We let \(v\) stand for the speed of the plane, and \(w\) stand for the speed of the wind. We use the formula \(RT=D\text{.}\)

\begin{align*} (v+w)5 \amp = 750\\ (v-w)5 \amp = 600 \end{align*}

Subsubsection Exercises

Write a system of equations for the problem.

Darryl plans to mix some 50% solution with some 75% solution to make 20 liters of 60% solution. How much of each should he use?

Answer
\begin{gather*} x+y=20\\ 0.5x+0.75y=12 \end{gather*}

The perimeter of a rectangle is 38 inches. If we triple the width and decrease the length by 8 inches, we increase the area of the rectangle by 40%. What are the dimensions of the original rectangle?

Answer
\begin{gather*} 2l+2w=38\\ 3w(l-8)=1.4lw \end{gather*}

The express train travels 15 miles per hour faster than the local. The local takes 10 minutes longer to travel 30 miles than the express takes. Find the speed of each.

Answer
\begin{gather*} v=15+l\\ \dfrac{30}{l} = \dfrac{30}{v}+\dfrac{1}{6} \end{gather*}

Delbert made $65 interest on his money market account this year. Francine invested $100 less than Delbert at 0.5% higher interest rate and earned $66. How much did Delbert invest, and at what interest rate?

Answer
\begin{gather*} Pr=65\\ (P-100)(r+0.005) = 66 \end{gather*}

Subsection 2. Find the vertex and intercepts of a parabola

We review some earlier skills for graphing parabolas.

Subsubsection Example

Example 9.62.

Find the vertex and intercepts of the parabola \(~y=-x^2-2x+8\text{.}\)

Solution

To find the \(x\)-intercepts of the graph, we set \(y=0\) and solve for \(x\text{.}\) We begin by factoring \(-1\text{,}\) the coefficient of \(x^2\text{,}\) from the equation.

\begin{align*} -(x^2+2x-8) \amp = 0\\ -(x+4)(x-2) \amp = 0\\ x=-4,~x \amp = 2 \end{align*}

The \(x\)-intercepts are \((-4,0)\) and \((2,0)\text{.}\) The \(y\)-intercept, \((0,8)\text{,}\) is given by the constant term. We find the \(x\)coordinate of the vertex by using the formula

\begin{equation*} x_v = \dfrac{-b}{2a} = \dfrac{1(-2)}{2(-1)} = -1 \end{equation*}

To find the \(y\)-coordinate, we substitute \(\alert{-1}\) into the equation.

\begin{equation*} y_v = -(\alert{-1})^2-2(\alert{-1})+8 = 9 \end{equation*}

The vertex is the point \((-1,9)\text{.}\) The graph of the parabola is shown below.

triangle

Subsubsection Exercises

Find the intercepts and the vertex of the parabola.

\(y=x^2-8x+15\)

Answer
\((5,0),~(3,0),~(0,15),~(4,-1)\)

\(y=8-2x^2\)

Answer
\((2,0),~(-2,0),~(0,8),~(0,8)\)

\(y=3x^2+6x+3\)

Answer
\((-1,0),~(-1,0),~(0,3),~(-1,0)\)

\(y=x^2-4x-1\)

Answer
\((2+\sqrt{5},0),~(2-\sqrt{5},0),~(0,-1),~(2,-5)\)

Subsection 3. Use substitution

We can use substitution to solve some systems of equations.

Subsubsection Example

Example 9.67.

Use substitution to write an equation in one variable.

\begin{align*} x^2-8x+y^2+2y \amp = 23\\ x+2y \amp = 12 \end{align*}
Solution

We solve the second equation for \(x\) to get \(x = \alert{12-2y}\text{,}\) and then substitute this expression into the first equation.

\begin{align*} (\alert{12-2y})^2-8(\alert{12-2y})+y^2+2y \amp = 23\\ 144-48y+4y^2-96+16y+y^2+2y \amp = 23\\ 5y^2-30y+25 \amp = 0 \end{align*}

Subsubsection Exercises

Use substitution to write an equation in one variable.

\begin{align*} x^2-2x+y^2+4y \amp = 20\\ y-x \amp = 4 \end{align*}
Answer
\(2x^2+10x+12=0\)
\begin{align*} x^2-y^2 \amp = 16\\ xy \amp = 6 \end{align*}
Answer
\(x^4-16x^2-36=0\)
\begin{align*} x^2-12x+y^2+2 \amp = 0\\ y \amp = 4x-7 \end{align*}
Answer
\(17x^2-68x+51=0\)
\begin{align*} x^2-4x+y^2+2y \amp = 5\\ x+2y \amp = -5 \end{align*}
Answer
\(5y^2+30y+40=0\)