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Chapter 5 Review and Tests

Review

Definitions, Concepts, and Formulas

Examples

5.1 Systems of Equations in Two Variables

A system of equations is a set of equations with common variables.

A solution of a system of equations in two variables x and y is an ordered pair of numbers (x1,y1) $(x_{1}, y_{1})$ that satisfies all equations in the system. The solution set of a system of two linear equations of the form ax+by=c $a x + b y = c$ is the point(s) of intersection of the graphs of the equations.

Systems with no solution are called inconsistent, and systems with at least one solution are consistent. A system of two linear equations in two variables may have one solution (independent equations), no solution (inconsistent system), or infinitely many solutions (dependent equations).

Three methods of solving a system of equations are:

Graphical Method
1. Graph both equations on the same coordinate axes.
2. Find the point(s) of intersection of the two graphs.
Substitution Method
1. Solve one equation for one variable in terms of the other variable.
2. Substitute the result from (i) into the other equation.
Elimination (or Addition) Method
1. Multiply one or both equations by appropriate numbers to get two new equations in which the coefficients of the variable to be eliminated are opposites.
2. Add the resulting equations in (i) to get an equation in one variable.
3. Solve the resulting equation in (ii).

Solve graphically:

${x + y x - y = = 5 - 1$ ${\begin{array}{rcl} x + y & = & 5 \\ x - y & = & - 1 \end{array}$

Solution

From the graphs, the point of intersection of the two lines is (2, 3). So, x=2 $x = 2$ and y=3 $y = 3$ .
Use the substitution method to solve:

${2 x 3 x + - y 2 y = = 38 (1) (2)$ ${\begin{array}{rcl} 2 x & + & y & = & 3 & (1) \\ 3 x & - & 2 y & = & 8 & (2) \end{array}$

Solution

Solve the first equation for y and then substitute the result in the second equation.

${2 x + y 3 x - 2 y = = 3, 8 then y = 3 - 2 x$ ${\begin{array}{lrc} 2 x + y & = & 3, & then & y = 3 - 2 x \\ 3 x - 2 y & = & 8 \end{array}$

$3 x - 2 (3 - 2 x) 7 x x = = = 8142 Substitute y = 3 - 2 x in (2) . Simplify . Solve for x .$ $\begin{array}{rcl} 3 x - 2 (3 - 2 x) & = & 8 & Substitute y = 3 - 2 x in (2) . \\ 7 x & = & 14 & Simplify . \\ x & = & 2 & Solve for x . \end{array}$

To find y, we substitute x=2 $x = 2$ in the equation y=3−2x $y = 3 - 2 x$ , to get:

$y = 3 - 2 x = 3 - 2 (2) = 3 - 4 = - 1 .$ $y = 3 - 2 x = 3 - 2 (2) = 3 - 4 = - 1 .$

The solution is (2,−1) $(2, - 1)$ .
Use the elimination method to solve:

${3 x + 2 y 2 x - 3 y = = 512 (1) (2)$ ${\begin{array}{rcl} 3 x + 2 y & = & 5 & (1) \\ 2 x - 3 y & = & 12 & (2) \end{array}$

Solution

Let’s eliminate y. Multiply equation (1) by 3 and equation (2) by 2, to get:

${9 x + 6 y = 15 4 x - 6 y = 24 13 x = 39 x = 3 Multiply equation (1) by 3. Multiply equation (2) by 2 . Add . Solve for x .$ $\begin{array}{rcl} {\begin{cases} 9 x + 6 y = 15 \\ 4 x - 6 y = 24 \end{cases} & \begin{array}{l} Multiply equation (1) by 3. \\ Multiply equation (2) by 2 . \end{array} \\ \begin{array}{l} 13 x = 39 \\ x = 3 \end{array} & \begin{array}{l} Add . \\ Solve for x . \end{array} \end{array}$

Substitute x=3 $x = 3$ in either of the original equations and solve for y:

$3 x + 2 y 3 (3) + 2 y y = = = 55 - 2 (1) Substitute x = 3 in (1) . Solve for y .$ $\begin{array}{rcl} 3 x + 2 y & = & 5 & (1) \\ 3 (3) + 2 y & = & 5 & Substitute x = 3 in (1) . \\ y & = & - 2 & Solve for y . \end{array}$

The solution is (3, −2). $(3, - 2) .$

5.2 Systems of Linear Equations in Three Variables

A linear equation in n variables x1,x2,…,xn $x_{1}, x_{2}, \dots, x_{n}$ is an equation that can be written in the form

a 1 x 1 + a 2 x 2 + \dots + a n x n = b,

$a_{1} x_{1} + a_{2} x_{2} + \dots + a_{n} x_{n} = b,$

where b and the coefficients a1, a2,…, an $a_{1}, a_{2}, \dots, a_{n}$ are real numbers.

A system of linear equations can be solved by transforming it into an equivalent system that is easier to solve.

The following operations produce equivalent systems:

Interchange the position of any two equations.
Multiply any equation by a nonzero constant.
Add a nonzero multiple of one equation to another.

The Gaussian elimination method is a procedure for converting a system of linear equations into an equivalent system in triangular form. See summary on page 530.

A linear system may have (i) only one solution, (ii) infinitely many solutions, or (iii) no solution.

Solve the system:

$⎧ ⎩ ⎨ ⎪ ⎪ x + y + 2 z 3 x + 2 y + z x + 2 y + 3 z = = = 91014 (1) (2) (3)$ ${\begin{array}{rcl} x + y + 2 z & = & 9 & (1) \\ 3 x + 2 y + z & = & 10 & (2) \\ x + 2 y + 3 z & = & 14 & (3) \end{array}$

Multiply equation (1) by −3 $- 3$ and add to (2), to get

$- y - 5 z = - 17 (4)$ $- y - 5 z = - 17 (4)$

Multiply equation (4) by −1 $- 1$ , to get

$y + 5 z = 17 (5)$ $y + 5 z = 17 (5)$

Multiply equation (1) by −1 $- 1$ and add it to (3), to get

$y + z = 5 (6)$ $y + z = 5 (6)$

Multiply equation (5) by −1 $- 1$ and add it to (6), to get

$or - 4 z z = = - 12 (7) 3 (8) Divide both sides of (7) by - 4.$ $\begin{array}{rcl} - 4 z & = & - 12 (7) \\ or & z & = & 3 (8) & Divide both sides of (7) by - 4. \end{array}$

The original system is equivalent to:

$x + y y + + 2 z 5 z z = = = 9173 (1) (5) (8)$ $\begin{array}{rcl} x & + & y & + & 2 z & = & 9 & (1) \\ y & + & 5 z & = & 17 & (5) \\ z & = & 3 & (8) \end{array}$

Substitute z=3 $z = 3$ in (5), to get y=2 $y = 2$ . Now, substitute z=3 $z = 3$ and y=2 $y = 2$ in (1), to get x=1 $x = 1$ .

The solution of the system is x=1, y=2, $x = 1, y = 2,$ and z=3, $z = 3,$ or the ordered triple (1, 2, 3).

5.3 Partial-Fraction Decomposition

In partial-fraction decomposition, we reverse the addition of rational expressions. A rational expression P(x)Q(x) $\frac{P (x)}{Q (x)}$ is called a proper fraction if deg P(x)<deg Q(x). $deg P (x) < deg Q (x) .$

There are four cases to consider in decomposing P(x)Q(x) $\frac{P (x)}{Q (x)}$ into partial fractions:

Q(x) has only distinct linear factors. (See page 540.)
Q(x) has repeated linear factors. (See page 543.)
Q(x) has distinct irreducible quadratic factors. (See page 545.)
Q(x) has repeated irreducible quadratic factors. (See page 546.)

Find the partial-fraction decomposition of:

$3 x + 11 x 2 - 2 x - 3 .$ $\frac{3 x + 11}{x^{2} - 2 x - 3} .$

Solution

$3 x + 11 x 2 - 2 x - 3 = 3 x + 11 ( x - 3 ) ( x + 1 ) = A x - 3 + B x + 1$ $\frac{3 x + 11}{x^{2} - 2 x - 3} = \frac{3 x + 11}{(x - 3) (x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1}$

Multiply both sides by (x−3)(x+1) $(x - 3) (x + 1)$ to get the identity:

$3 x + 11 = A (x + 1) + B (x - 3) (1)$ $3 x + 11 = A (x + 1) + B (x - 3) (1)$
1. Let x=3 $x = 3$ in (1), to get:
  
  $or 3 (3) + 11 20 = = A (3 + 1) + B (3 - 3) 4 A or A = 5$ $\begin{array}{rcl} 3 (3) + 11 & = & A (3 + 1) + B (3 - 3) \\ or & 20 & = & 4 A or A = 5 \end{array}$
2. Let x=−1 $x = - 1$ in (1), to get:
  
  $or 3 (- 1) + 11 8 = = A (- 1 + 1) + B (- 1 - 3) - 4 B or B = - 2$ $\begin{array}{rcl} 3 (- 1) + 11 & = & A (- 1 + 1) + B (- 1 - 3) \\ or & 8 & = & - 4 B or B = - 2 \end{array}$
Substitute A=5 $A = 5$ and B=−2, $B = - 2,$ to get

$3 x + 11 x 2 - 2 x - 3 = 5 x - 3 + - 2 x + 1 .$ $\frac{3 x + 11}{x^{2} - 2 x - 3} = \frac{5}{x - 3} + \frac{- 2}{x + 1} .$

5.4 Systems of Nonlinear Equations

A system of equations (inequalities) in which at least one equation (inequality) is nonlinear is called a system of nonlinear equations (inequalities). The methods of substitution, elimination, or graphing are used to solve a nonlinear system.

Solve the system:

${x 2 + y 2 x 2 - 3 y = = 25 - 3 (1) (2)$ ${\begin{array}{rcl} x^{2} + y^{2} & = & 25 & (1) \\ x^{2} - 3 y & = & - 3 & (2) \end{array}$

Add −1 $- 1$ times equation (2) to (1), to get

$or or is y 2 + 3 y y 2 + 3 y - 28 (y - 4) (y + 7) y = 4 or y = = = = 2800 - 7 Subtract 28 from both sides . Factor . Zero-product property$ $\begin{array}{lrcll} y^{2} + 3 y & = & 28 \\ or & y^{2} + 3 y - 28 & = & 0 & Subtract 28 from both sides . \\ or & (y - 4) (y + 7) & = & 0 & Factor . \\ is & y = 4 or y & = & - 7 & Zero-product property \end{array}$
1. Substitute y=4 $y = 4$ in equation (2), to get:
  
  $x 2 - 3 (4) = - 3 or x 2 = 9 or x = \pm 3$ $x^{2} - 3 (4) = - 3 or x^{2} = 9 or x = \pm 3$
  
  So, (3, 4) and (−3, 4) $(- 3, 4)$ are possible solutions of the system.
2. Substitute y=−7 $y = - 7$ in equation (2), to get
  
  $x 2 - 3 (- 7) = - 3 or x 2 = - 24 (Not possible) .$ $x^{2} - 3 (- 7) = - 3 or x^{2} = - 24 (Not possible) .$
  
  You can check that the only solutions are (3, 4) and (−3, 4) $(- 3, 4)$ .

5.5 Systems of Inequalities

A statement of the form ax+by<c $a x + b y < c$ is a linear inequality in the variables x and y. The symbol < $<$ may be replaced by ≤, >, or ≥. $\leq, >, or \geq .$ A procedure for graphing a linear inequality in two variables is described on page 560.

The graph of the solution set of a system of inequalities is obtained by (i) graphing each inequality of the system in the same coordinate plane and then (ii) finding the region that is common to every graph in the system.

Nonlinear systems of inequalities are solved essentially the same way as linear systems.

Graph the solution set of

{y y \geq \leq x 2 2 x + + 14 (1) (2)

${\begin{array}{rcl} y & \geq & x^{2} & + & 1 & (1) \\ y & \leq & 2 x & + & 4 & (2) \end{array}$

Solution

	Inequality (1)	Inequality (2)
Step 1 Change each inequality to equality.	$y = x 2 + 1$ $y = x^{2} + 1$	$y = 2 x + 4$ $y = 2 x + 4$
Step 2 Graph each equality of Step 1.	The graph is a parabola with vertex at (0, 1), opens upward.	The graph is a line through the points (0, 4) and (−2, 0). $(- 2, 0) .$
Step 3 Test point.	Testing (0, 0) in (1) gives 0≥1 $0 \geq 1$ (a false statement). The solution is inside of the parabola.	Testing (0, 0) in (2) gives 0≤4, $0 \leq 4,$ a true statement. The solution lies on the side of the line that contains (0, 0).
Step 4 Shade each region; the solution is the region common to all the regions.	See the figure.

5.6 Linear Programming

In a linear programming model, a quantity f (to be maximized or minimized) that can be expressed in the form f=ax+by $f = a x + b y$ is called an objective function of the variables x and y The constraints are the restrictions placed on the variables x and y that can be expressed as a system of linear inequalities.

A procedure for solving a linear programming problem is given on page 572.

Find the maximum value of f=12x+3y $f = 12 x + 3 y$ , subject to the constraints:

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x 3 x + + y y x y \leq \leq \geq \geq 5900

${\begin{array}{rcl} x & + & y & \leq & 5 \\ 3 x & + & y & \leq & 9 \\ x & \geq & 0 \\ y & \geq & 0 \end{array}$

Solution

The feasible region is shaded in the figure.

Corner Point	Value of f=12x+3y $f = 12 x + 3 y$
(0, 0)	f=12(0)+3(0)=0 $f = 12 (0) + 3 (0) = 0$
(3, 0)	f=12(3)+3(0)=36 $f = 12 (3) + 3 (0) = 36$
(2, 3)	f=12(2)+3(3)=33 $f = 12 (2) + 3 (3) = 33$
(0, 5)	f=12(0)+3(5)=15 $f = 12 (0) + 3 (5) = 15$

The maximum value of f is 36 at (3, 0).

Review Exercises

Building Skills

In Exercises 1–18, solve each system of equations by using the method of your choice. Identify systems with no solution and systems with infinitely many solutions.

{3xx−+y2y==−53 ${\begin{array}{lcr} 3 x & - & y & = & - 5 \\ x & + & 2 y & = & 3 \end{array}$
{x+3y+6y==04x−2 ${\begin{array}{rcl} x + 3 y + 6 & = & 0 \\ y & = & 4 x - 2 \end{array}$
{2x+4y3x+6y==310 ${\begin{array}{rcl} 2 x + 4 y & = & 3 \\ 3 x + 6 y & = & 10 \end{array}$
{x2x−−y2y==29 ${\begin{array}{rcl} x & - & y & = & 2 \\ 2 x & - & 2 y & = & 9 \end{array}$
⎧⎩⎨3x12x−+y13y==32 ${\begin{array}{rcl} 3 x & - & y & = & 3 \\ \frac{1}{2} x & + & \frac{1}{3} y & = & 2 \end{array}$
{0.02y−0.03x1.5x−2y==−0.043 ${\begin{array}{rcl} 0.02 y - 0.03 x & = & - 0.04 \\ 1.5 x - 2 y & = & 3 \end{array}$
⎧⎩⎨⎪⎪x2x3x+−−3yy3y+++zz2z===0510 ${\begin{array}{lcr} x & + & 3 y & + & z & = & 0 \\ 2 x & - & y & + & z & = & 5 \\ 3 x & - & 3 y & + & 2 z & = & 10 \end{array}$
⎧⎩⎨⎪⎪2xx+y3y−z+2z===1151 ${\begin{array}{rcl} 2 x & + & y & = & 11 \\ 3 y - z & = & 5 \\ x & + 2 z & = & 1 \end{array}$
⎧⎩⎨⎪⎪x3y2x++−y2z3z===107 ${\begin{array}{rcl} x & + & y & = & 1 \\ 3 y & + & 2 z & = & 0 \\ 2 x & - & 3 z & = & 7 \end{array}$
⎧⎩⎨⎪⎪2xx2x−−+3y3y3y+++z2z2z===2−13 ${\begin{array}{rcl} 2 x & - & 3 y & + & z & = & 2 \\ x & - & 3 y & + & 2 z & = & - 1 \\ 2 x & + & 3 y & + & 2 z & = & 3 \end{array}$
⎧⎩⎨⎪⎪xx3x++−y5yy++−z5zz===1−14 ${\begin{array}{rcl} x & + & y & + & z & = & 1 \\ x & + & 5 y & + & 5 z & = & - 1 \\ 3 x & - & y & - & z & = & 4 \end{array}$
⎧⎩⎨⎪⎪x2xx+++3y6yy−−+2z4zz===−431 ${\begin{array}{rclr} x & + & 3 y & - & 2 z & = & - 4 \\ 2 x & + & 6 y & - & 4 z & = & 3 \\ x & + & y & + & z & = & 1 \end{array}$
{x2x++4y5y++3z4z==14 ${\begin{array}{rcl} x & + & 4 y & + & 3 z & = & 1 \\ 2 x & + & 5 y & + & 4 z & = & 4 \end{array}$
{3xx−−y2y++2z3z==92 ${\begin{array}{rcl} 3 x & - & y & + & 2 z & = & 9 \\ x & - & 2 y & + & 3 z & = & 2 \end{array}$
⎧⎩⎨⎪⎪3xx3x−++y2yy===2910 ${\begin{array}{rcl} 3 x & - & y & = & 2 \\ x & + & 2 y & = & 9 \\ 3 x & + & y & = & 10 \end{array}$
⎧⎩⎨⎪⎪x3x2x−++2y4y2y===1117 ${\begin{array}{rcl} x & - & 2 y & = & 1 \\ 3 x & + & 4 y & = & 11 \\ 2 x & + & 2 y & = & 7 \end{array}$
⎧⎩⎨⎪⎪x2xx+−+yy4y+++zz2z===345 ${\begin{array}{rcrlr} x & + & y & + & z & = & 3 \\ 2 x & - & y & + & z & = & 4 \\ x & + & 4 y & + & 2 z & = & 5 \end{array}$
⎧⎩⎨⎪⎪xx2x−++y2yy+++z3z4z===426 ${\begin{array}{rcl} x & - & y & + & z & = & 4 \\ x & + & 2 y & + & 3 z & = & 2 \\ 2 x & + & y & + & 4 z & = & 6 \end{array}$

In Exercises 19–24, graph each system of inequalities.

⎧⎩⎨⎪⎪xx+−yyx≤≤≥110 ${\begin{array}{rcl} x & + & y & \leq & 1 \\ x & - & y & \leq & 1 \\ x & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪2x4x+−3y3yx≤≤≥6120 ${\begin{array}{rcl} 2 x & + & 3 y & \leq & 6 \\ 4 x & - & 3 y & \leq & 12 \\ x & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪4x2xx++−y5y2y≤≥≥7−1−5 ${\begin{array}{rcl} 4 x & + & y & \leq & 7 \\ 2 x & + & 5 y & \geq & - 1 \\ x & - & 2 y & \geq & - 5 \end{array}$
⎧⎩⎨⎪⎪7xx2x−+−2yy3y+++6149≤≥≥000 ${\begin{array}{rcrlr} 7 x & - & 2 y & + & 6 & \leq & 0 \\ x & + & y & + & 14 & \geq & 0 \\ 2 x & - & 3 y & + & 9 & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪x3x7x+−+yyy4y−−++39323≤≤≥≥0000 ${\begin{array}{rcrlr} x & + & y & - & 3 & \leq & 0 \\ 3 x & - & y & - & 9 & \leq & 0 \\ y & + & 3 & \geq & 0 \\ 7 x & + & 4 y & + & 23 & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪x5xx7x++−+5yy5yy−−−+1171725≤≤≤≥0000 ${\begin{array}{rcrl} x & + & 5 y & - & 11 & \leq & 0 \\ 5 x & + & y & - & 7 & \leq & 0 \\ x & - & 5 y & - & 17 & \leq & 0 \\ 7 x & + & y & + & 25 & \geq & 0 \end{array}$

In Exercises 25–28, solve each linear programming problem.

Maximize z=2x+3y, $z = 2 x + 3 y,$ subject to the constraints x≥0, y≥0, 3x+7y≤21. $x \geq 0, y \geq 0, 3 x + 7 y \leq 21 .$
Minimize z=−5x+2y, $z = - 5 x + 2 y,$ subject to the constraints x≥−2, y≤1, x−2y+2≤0. $x \geq - 2, y \leq 1, x - 2 y + 2 \leq 0 .$
Minimize z=x+3y, $z = x + 3 y,$ subject to the constraints 2x−3y+2≥0, 4x+y−10≤0, x−3y−9≤0, $2 x - 3 y + 2 \geq 0, 4 x + y - 10 \leq 0, x - 3 y - 9 \leq 0,$ 3x+y+3≥0. $3 x + y + 3 \geq 0 .$
Maximize z=2x+5y, $z = 2 x + 5 y,$ subject to the constraints x+3y−3≤0, 3x−y−9≤0, x+4y+10≥0, $x + 3 y - 3 \leq 0, 3 x - y - 9 \leq 0, x + 4 y + 10 \geq 0,$ 3x−2y+2≥0. $3 x - 2 y + 2 \geq 0 .$

In Exercises 29–34, solve each nonlinear system of equations.

{xx2+−3y3x==17y+3 ${\begin{array}{rcl} x & + & 3 y & = & 1 \\ x^{2} & - & 3 x & = & 7 y + 3 \end{array}$
{4xy2−−y2y==3x−2 ${\begin{array}{rcl} 4 x & - & y & = & 3 \\ y^{2} & - & 2 y & = & x - 2 \end{array}$
{x5x2−+yy2==424 ${\begin{array}{rcl} x & - & y & = & 4 \\ 5 x^{2} & + & y^{2} & = & 24 \end{array}$
{x2x2−+2y3y2==729 ${\begin{array}{rcl} x & - & 2 y & = & 7 \\ 2 x^{2} & + & 3 y^{2} & = & 29 \end{array}$
{x2+xy2y2==29 ${\begin{array}{rcl} x y & = & 2 \\ x^{2} & + & 2 y^{2} & = & 9 \end{array}$
{2x2+xy3y2==321 ${\begin{array}{rcl} x y & = & 3 \\ 2 x^{2} & + & 3 y^{2} & = & 21 \end{array}$

In Exercises 35–40, graph each nonlinear system of inequalities.

{xx2−+yy2≤≤113 ${\begin{array}{rcl} x & - & y & \leq & 1 \\ x^{2} & + & y^{2} & \leq & 13 \end{array}$
{xx2−+yy2≥≤113 ${\begin{array}{rcl} x & - & y & \geq & 1 \\ x^{2} & + & y^{2} & \leq & 13 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪xx2−+yy2xy≤≤≥≥11300 ${\begin{array}{rcl} x & - & y & \leq & 1 \\ x^{2} & + & y^{2} & \leq & 13 \\ x & \geq & 0 \\ y & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪xx2−+yy2xy≥≥≤≥11300 ${\begin{array}{rcl} x & - & y & \geq & 1 \\ x^{2} & + & y^{2} & \geq & 13 \\ x & \leq & 0 \\ y & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪x2+yy2xy≤≤≥≥3x2500 ${\begin{array}{rcl} y & \leq & 3 x \\ x^{2} & + & y^{2} & \leq & 25 \\ x & \geq & 0 \\ y & \geq & 0 \end{array}$
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪x2+yy2xy≥≤≥≥3x2500 ${\begin{array}{rcl} y & \geq & 3 x \\ x^{2} & + & y^{2} & \leq & 25 \\ x & \geq & 0 \\ y & \geq & 0 \end{array}$

In Exercises 41–46, find the partial-fraction decomposition of each rational expression.

x+4x2+5x+6 $\frac{x + 4}{x^{2} + 5 x + 6}$
x+14x2+3x−4 $\frac{x + 14}{x^{2} + 3 x - 4}$
3x2+x+1x(x−1)2 $\frac{3 x^{2} + x + 1}{x {(x - 1)}^{2}}$
3x2+2x+3(x2−1)(x+1) $\frac{3 x^{2} + 2 x + 3}{(x^{2} - 1) (x + 1)}$
x2+2x+3(x2+4)2 $\frac{x^{2} + 2 x + 3}{{(x^{2} + 4)}^{2}}$
2xx4−1 $\frac{2 x}{x^{4} - 1}$

Applying the Concepts

Investment. A speculator invested part of $15,000 in a high-risk venture and received a return of 12% at the end of the year. The rest of the $15,000 was invested at 4% annual interest. The combined annual income from the two sources was $1300. How much was invested at each rate?
Agriculture. A farmer earns a profit of $525 per acre of tomatoes and $475 per acre of soybeans. His soybean acreage is 5 acres more than twice his tomato acreage. If his total profit from the two crops is $24,500, how many acres of tomatoes and how many acres of soybeans does he have?
Geometry. The area of a rectangle is 63 square feet, and its perimeter is 33 feet. What are the dimensions of the rectangle?
Numbers. Twice the sum of the reciprocals of two numbers is 13, and the product of the numbers is 19. $\frac{1}{9} .$ Find the numbers.
Geometry. The hypotenuse of a right triangle is 17. If one leg of the triangle is increased by 1 and the other leg is increased by 4, the hypotenuse becomes 20. Find the sides of the triangle.
Paper route. Chris covers her paper route, which is 21 miles long, by 7:30 a.m. each day. If her average rate of travel were 1 mile faster each hour, she would cover the route by 7 a.m. What time does she start in the morning?
Agriculture. A rectangular pasture with an area of 6400 square meters is divided into three smaller pastures by two fences parallel to the shorter sides. The width of two of the smaller pastures is the same, and the width of the third is twice that of the others. Find the dimensions of the original pasture if the perimeter of the larger of the subdivisions is 240 m.
Leasing. Budget Rentals leases its compact cars for $23.00 per day plus $0.17 per mile. Dollar Rentals leases the same car for $24.00 per day plus $0.22 per mile.
1. Find the cost functions describing the daily cost of leasing from each company.
2. Graph the two functions in part (a) on the same coordinate axes.
3. From which company should you lease the car if you plan to drive (i) 50 miles per day; (ii) 60 miles per day; (iii) 70 miles per day?
Break-even analysis. Auto-Sprinkler Corp. manufactures seven-day, 24-hour variable timers for lawn sprinklers. The corporation has a monthly fixed cost of $60,000 and a production cost of $12 for each timer manufactured. Each timer sells for $20.
1. Write the cost function and the revenue function for selling x timers per month.
2. Graph the two functions in part (a) on the same coordinate plane and find the break-even point graphically.
3. Find the break-even point algebraically.
4. How many timers should be sold in a month to realize a profit (before taxes) of 15% of the cost?
Equilibrium quantity and price. The demand equation for a product is p=4000x, $p = \frac{4000}{x},$ where p is the price per unit and x is the number of units of the product. The supply equation for the product is p=x20+10. $p = \frac{x}{20} + 10 .$
1. Find the equilibrium quantity.
2. Find the equilibrium price.
3. Graph the supply and demand functions on the same coordinate plane and label the equilibrium point.
Apartment lease. Alisha and Sunita signed a lease on an apartment for nine months and split the rent evenly. At the end of six months, Alisha got married and moved out. She paid the landlady an amount equal to the difference between double-occupancy rental and single-occupancy rental for the remaining three months, and Sunita paid the single-occupancy rate for the same three months. If the nine-month rental cost Alisha $2250 and Sunita $3150, what were the single and double monthly rates?
Seating arrangement. The 600 graduating seniors at Central State College are seated in rows, each of which contains the same number of chairs, and every chair is occupied. If five more chairs were in each row, everyone could be seated in four fewer rows. How many chairs are in each row?
Passing a final exam. Twenty-six students in a college algebra class took a final exam on which the passing score was 70. The mean score of those who passed was 78, and the mean score of those who failed was 26. The mean of all scores was 72. How many students failed the exam?
Finding numbers. The average of a and b is 2.5, the average of b and c is 3.8, and the average of a and c is 3.1. Find the numbers a, b, and c.
May–December match. Steve and his wife, Janet, have the same birthday. When Steve was as old as Janet is now, he was twice as old as Janet was then. When Janet becomes as old as Steve is now, the sum of their ages will be 119. How old are Steve and Janet now?
Percentage increase. Let x and y be positive real numbers. Increasing x by y percent gives 46, and increasing y by x percent gives 21. Find x and y.
Criminals rob a bank. Three criminals—Butch, Sundance, and Billy—robbed a bank and divided the loot in the following manner: Because Butch planned the job and drove the getaway car, he got 75% as much as Sundance and Billy put together. Because Sundance was an expert safecracker, he got $500 more than 50% of Billy’s and Butch’s cut put together. Billy got $1000 less than three times the difference of Butch and Sundance’s cut. How much did they steal, and what was each criminal’s take?
Curve fitting. Find all of the curves of the form y=ax2+bx+c $y = a x^{2} + b x + c$ that contain the points (0, 1), (1, 0), and (−1, 6). $(- 1, 6) .$
Using exponents. Given that 2x=8y $2^{x} = 8^{y}$ and 9y=3x−2, $9^{y} = 3^{x - 2},$ find x and y.
Area of an octagon. A square of side length 8 has its corners cut off to make it a regular octagon. Find the area of the octagon. (See the figure.)
Building houses. A builder has 42 units of material and 32 units of labor available during a given period. She builds two-story houses or one-story houses or some of both. Suppose she makes a profit of $10,000 on each two-story house and $4000 on each one-story house. A two-story house requires 7 units of material and 1 unit of labor, whereas a one-story house requires 1 unit of material and 2 units of labor. How many houses of each type should she build to make a maximum profit? What is the maximum profit?
Minimizing cost. An animal food is to be a mixture of two products X and Y. The content and cost of 1 pound of each product is given in the following table.

Product Protein Grams Fat Grams Carbohydrates Grams Cost

X 180 2 240 $0.75

Y 36 8 200 $0.56

How much of each product should be used to minimize the cost if each bag must contain at least 612 grams of protein; at least 22 grams of fat; and, at most, 1880 grams of carbohydrates?
Lobster prices. The Captain’s Catch classifies its lobsters as small, medium, and large. A package of one large, two medium, and four small lobsters sells for $344; a package of two large and three small lobsters sells for $255; and a package of two large, two medium, and five small lobsters sells for $449. What is the individual price of each size lobster?
Ancient mathematics. The Nine Chapters in the Mathematical Art, a Chinese work written about 250 b.c. contains the following problem: Three sheafs of good crop, two sheafs of mediocre crop, and one sheaf of bad crop are sold for 39 dou. Two sheafs of good, three mediocre, and one bad are sold for 34 dou. And one good, two mediocre, and three bad are sold for 26 dou. What price is received for a sheaf of each of good crop, mediocre crop, and bad crop?

Product	Protein Grams	Fat Grams	Carbohydrates Grams	Cost
`X`	180	2	240	$0.75
`Y`	36	8	200	$0.56

Practice Test A

In Problems 1–7, solve each system of equations.

{2x2x−+yy==44 ${\begin{array}{rcl} 2 x & - & y & = & 4 \\ 2 x & + & y & = & 4 \end{array}$
{x3x++2y6y==824 ${\begin{array}{rcl} x & + & 2 y & = & 8 \\ 3 x & + & 6 y & = & 24 \end{array}$
{−2x4x+−y2y==44 ${\begin{array}{rcl} - 2 x & + & y & = & 4 \\ 4 x & - & 2 y & = & 4 \end{array}$
{3x2x+−3y2y==−15−10 ${\begin{array}{rcl} 3 x & + & 3 y & = & - 15 \\ 2 x & - & 2 y & = & - 10 \end{array}$
⎧⎩⎨⎪⎪⎪⎪5x32x3+−y2y8==143 ${\begin{array}{rcl} \frac{5 x}{3} & + & \frac{y}{2} & = & 14 \\ \frac{2 x}{3} & - & \frac{y}{8} & = & 3 \end{array}$
{3x−y+y4==x20 ${\begin{array}{rcl} y & = & x^{2} \\ 3 x & - & y & + & 4 & = & 0 \end{array}$
{2x2+x3x−−3y3y==−48 ${\begin{array}{rcl} x & - & 3 y & = & - 4 \\ 2 x^{2} & + & 3 x & - & 3 y & = & 8 \end{array}$
Two gold bars together weigh a total of 485 pounds. One bar weighs 15 pounds more than the other.
1. Write a system of equations that describes these relationships.
2. How much does each bar weigh?
Convert the given system to triangular form.

$⎧ ⎩ ⎨ ⎪ ⎪ 2 x 4 x 2 x + + + y 6 y 2 y + + + 2 z z 7 z = = = 415 - 1$ ${\begin{array}{rcl} 2 x & + & y & + & 2 z & = & 4 \\ 4 x & + & 6 y & + & z & = & 15 \\ 2 x & + & 2 y & + & 7 z & = & - 1 \end{array}$
Use back-substitution to solve the given linear system.

$⎧ ⎩ ⎨ ⎪ ⎪ x - 6 y 9 y + - 3 z 5 z 2 z = = = - 2 210$ ${\begin{array}{rclr} x & - & 6 y & + & 3 z & = & - 2 \\ 9 y & - & 5 z & = & 2 \\ 2 z & = & 10 \end{array}$

In Exercises 11–13, solve the system of equations.

⎧⎩⎨⎪⎪x2xx+−+y2yy++−z2zz===8412 ${\begin{array}{rclr} x & + & y & + & z & = & 8 \\ 2 x & - & 2 y & + & 2 z & = & 4 \\ x & + & y & - & z & = & 12 \end{array}$
⎧⎩⎨⎪⎪x3x−3y3y+++z2zz===−15−8 ${\begin{array}{rcl} x & + & z & = & - 1 \\ 3 y & + & 2 z & = & 5 \\ 3 x & - & 3 y & + & z & = & - 8 \end{array}$
⎧⎩⎨⎪⎪2xxx−+−yy5y+++zz5z===2−17 ${\begin{array}{rclr} 2 x & - & y & + & z & = & 2 \\ x & + & y & + & z & = & - 1 \\ x & - & 5 y & + & 5 z & = & 7 \end{array}$
A vending machine’s coin box contains nickels, dimes, and quarters. The total number of coins in the box is 300. The number of dimes is three times the number of nickels and quarters together. Assuming that $30.65 is in the box, find the number of nickels, dimes, and quarters that it contains.

For Problems 15 and 16, write the form of the partial-fraction decomposition of the given rational expression. You do not need to solve for the constants.

2x(x−5)(x+1) $\frac{2 x}{(x - 5) (x + 1)}$
−5x2+x−8(x−2)(x2+1)2 $\frac{- 5 x^{2} + x - 8}{(x - 2) {(x^{2} + 1)}^{2}}$
Find the partial-fraction decomposition of the rational expression

$x + 3 ( x + 4 ) 2 ( x - 7 ) .$ $\frac{x + 3}{{(x + 4)}^{2} (x - 7)} .$
Graph the inequality 3x+y<6. $3 x + y < 6 .$
Graph the solution set of the system of inequalities.

${y y - - x x 2 > > 00$ ${\begin{array}{rcl} y & - & x & > & 0 \\ y & - & x^{2} & > & 0 \end{array}$
Maximize z=2x+y, $z = 2 x + y,$ subject to the constraints x≥0, y≥0, x+3y≤3. $x \geq 0, y \geq 0, x + 3 y \leq 3 .$

Practice Test B

In Problems 1–7, solve each system of equations.

{2xx−−3yy==167 ${\begin{array}{rcl} 2 x & - & 3 y & = & 16 \\ x & - & y & = & 7 \end{array}$
1. {(5, −2)} ${(5, - 2)}$
2. {(8, 0)} ${(8, 0)}$
3. {(0, −7)} ${(0, - 7)}$
4. {(4, −3)} ${(4, - 3)}$
{6x3x−−9y5y==−2−6 ${\begin{array}{rcl} 6 x & - & 9 y & = & - 2 \\ 3 x & - & 5 y & = & - 6 \end{array}$
1. {(−3, 0)} ${(- 3, 0)}$
2. {(0, 56)} ${(0, \frac{5}{6})}$
3. {(443, 10)} ${(\frac{44}{3}, 10)}$
4. {(1, 1)}
{2x4x−−5y10y==918 ${\begin{array}{rcl} 2 x & - & 5 y & = & 9 \\ 4 x & - & 10 y & = & 18 \end{array}$
1. {(0, 92)} ${(0, \frac{9}{2})}$
2. {(5y2+92, y)} ${(\frac{5 y}{2} + \frac{9}{2}, y)}$
3. {(x, 5x2+92)} ${(x, \frac{5 x}{2} + \frac{9}{2})}$
4. ∅ $\emptyset$
{3x−6x+−5y10y==12 ${\begin{array}{rcl} 3 x & + & 5 y & = & 1 \\ - 6 x & - & 10 y & = & 2 \end{array}$
1. {(0, −15)} ${(0, - \frac{1}{5})}$
2. {(2, −1)} ${(2, - 1)}$
3. ∅ $\emptyset$
4. {(−5y3 +13 , y)} ${(- \frac{5 y}{3} + \frac{1}{3}, y)}$
⎧⎩⎨⎪⎪⎪⎪x5x4+−2y5y3==1−512 ${\begin{array}{rcl} \frac{x}{5} & + & \frac{2 y}{5} & = & 1 \\ \frac{x}{4} & - & \frac{y}{3} & = & - \frac{5}{12} \end{array}$
1. {(−15, 10)} ${(- 15, 10)}$
2. {(1, 2)}
3. {(−10, −15)} ${(- 10, - 15)}$
4. {(1, 0)}
⎧⎩⎨x−2x−+3y6y==12−1 ${\begin{array}{rcl} x & - & 3 y & = & \frac{1}{2} \\ - 2 x & + & 6 y & = & - 1 \end{array}$
1. {(12, 0)} ${(\frac{1}{2}, 0)}$
2. {(2, 12)} ${(2, \frac{1}{2})}$
3. ∅ $\emptyset$
4. {(3y+12, y)} ${(3 y + \frac{1}{2}, y)}$
{3x3x2++4y16y2==1248 ${\begin{array}{l} 3 x & + & 4 y & = & 12 \\ 3 x^{2} & + & 16 y^{2} & = & 48 \end{array}$
1. {(0, 4), (3, 0)}
2. {(0, −4)} ${(0, - 4)}$
3. {(0, −4),(3, 0)} ${(0, - 4), (3, 0)}$
4. {(4, 0), (2, 32)} ${(4, 0), (2, \frac{3}{2})}$
Student tickets for a dance cost $2, and nonstudent tickets cost $5. Three hundred tickets were sold, and the total ticket receipts were $975. How many of each type of ticket were sold?
1. 150 student tickets
  
  150 nonstudent tickets
2. 200 student tickets
  
  100 nonstudent tickets
3. 210 student tickets
  
  90 nonstudent tickets
4. 175 student tickets
  
  125 nonstudent tickets
Convert the given system to triangular form.

$⎧ ⎩ ⎨ ⎪ ⎪ x 2 x x + + + 3 y 5 y 2 y + + + 3 z 4 z 2 z = = = 456$ ${\begin{array}{rcl} x & + & 3 y & + & 3 z & = & 4 \\ 2 x & + & 5 y & + & 4 z & = & 5 \\ x & + & 2 y & + & 2 z & = & 6 \end{array}$
1. $\begin{array}{rcl} x & + & 3 y & + & 3 z & = & 4 \\ y & + & 2 z & = & 8 \\ z & = & 2 \end{array}$
2. $\begin{array}{rcl} x & + & 3 y & + & 3 z & = & 4 \\ 2 y & + & z & = & 1 \\ z & = & 5 \end{array}$
3. $\begin{array}{rcl} x & + & 3 y & + & 3 z & = & 4 \\ y & + & 2 z & = & 3 \\ z & = & 5 \end{array}$
4. $\begin{array}{rcl} x & + & 3 y & + & 3 z & = & 4 \\ 2 y & + & z & = & - 2 \\ z & = & 5 \end{array}$
Use back-substitution to solve the linear system.

${\begin{array}{rcl} 2 x & + & y & + & z & = & 0 \\ 10 y & - & 2 z & = & 4 \\ 3 z & = & 9 \end{array}$
1. ${(- 2, 1, 3)}$
2. ${(0, - 3, 3)}$
3. {(0, 2, 8)}
4. ${(- 1, 1, 1)}$
Solve the given system of equations or state that the system is inconsistent.

${\begin{array}{rcl} 2 x & + & 13 y & + & 6 z & = & 1 \\ 3 x & + & 10 y & + & 11 z & = & 15 \\ 2 x & + & 10 y & + & 8 z & = & 8 \end{array}$
1. ${(3, - 1, - 4)}$
2. ${(1, - 1, 2)}$
3. $\emptyset$
4. (1, 6, 8)}
Solve the given system of equations or state that the system is inconsistent.

${\begin{array}{rcl} 4 x & - & y & + & 7 z & = & - 2 \\ 2 x & + & y & + & 11 z & = & 13 \\ 3 x & - & y & + & 4 z & = & - 3 \end{array}$
1. ${(- 3 z + 1, - 5 z + 11, z)}$
2. ${(- 14, - 14, 5)}$
3. ${(- \frac{19}{5}, 3, \frac{8}{5})}$
4. $\emptyset$
Solve the given system of equations or state that the system is inconsistent.

${\begin{array}{rcrlr} x & - & 2 y & + & 3 z & = & 4 \\ 2 x & - & y & + & z & = & 1 \\ x & + & y & - & 2 z & = & - 3 \end{array}$
1. {(1, 6, 5)}
2. {(0, 1, 2)}
3. ${(x, 5 x + 1, 3 x + 2)}$
4. $\emptyset$
Forty-six students will go to France, Italy, or Spain for six weeks during the summer. The number of students going to France or Italy is four more than the number going to Spain. The number of students going to France is two less than the number going to Spain. How many students are going to each country?
1. France: 23
  
  Italy: 21
  
  Spain: 2
2. France: 19
  
  Italy: 6
  
  Spain: 21
3. France: 21
  
  Italy: 6
  
  Spain: 19
4. France: 2
  
  Italy: 21
  
  Spain: 23

For Problems 15 and 16, write the form of the partial-fraction decomposition of the given rational functions. You do not need to solve for the constants.

$\frac{x}{(x + 2) (x - 7)}$
1. $\frac{A}{x + 2} + \frac{B x}{x - 7}$
2. $\frac{A x}{x + 2} + \frac{B x}{x - 7}$
3. $\frac{A}{x + 2} + \frac{B}{x - 7}$
4. $\frac{A}{x + 2} + \frac{B x + C}{x - 7}$
$\frac{7 - x}{(x - 3) {(x + 5)}^{2}}$
1. $\frac{A}{x - 3} + \frac{B}{x + 5} + \frac{C}{{(x + 5)}^{2}}$
2. $\frac{A}{x - 3} + \frac{B}{x + 5} + \frac{C x + D}{{(x + 5)}^{2}}$
3. $\frac{A}{x + 3} + \frac{B}{{(x + 5)}^{2}}$
4. $\frac{A}{x + 3} + \frac{B x + C}{{(x + 5)}^{2}}$
Find the partial-fraction decomposition of the rational expression

$\frac{x^{2} + 15 x + 18}{x^{3} - 9 x} .$
1. $\frac{26}{x - 9} - \frac{2}{x}$
2. $\frac{4}{x + 3} - \frac{2}{x} - \frac{1}{x - 3}$
3. $\frac{4}{x - 3} - \frac{2}{x} - \frac{1}{x + 3}$
4. $\frac{3 x - 15}{x^{2} - 9} - \frac{2}{x}$
Which of the graphs below left is the graph of $x + 3 y > 3$ ?
Which of the graphs below is the graph of the solution set of the following system of inequalities?

${\begin{array}{rcl} y & - & x^{2} & + & 4 & \geq & 0 \\ 3 x & - & y & \leq & 0 \end{array}$
Maximize $z = 3 x + 21 y,$ subject to the constraints

$x \geq 0, y \geq 0, 2 x + y \leq 8, 2 x + 3 y \leq 16 .$
1. 84
2. 112
3. $\frac{16}{3}$
4. 12

Cumulative Review Exercises Chapters P–5

In Exercises 1–8, solve each equation or inequality.

$\frac{1}{x - 1} + \frac{4}{x - 4} = \frac{5}{x - 5}$
$2 {(x + \frac{1}{x})}^{2} - 7 (x + \frac{1}{x}) + 5 = 0$
$\sqrt{3 x - 5} = x - 3$
$\frac{x - 1}{x + 3} \leq 0$
$x^{2} - 9 x + 20 > 0$
$2^{x - 1} = 5$
$\log_{x} 16 = 4$
$\log (x - 3) + \log (x - 1) = \log (2 x - 5)$

In Exercises 9–12, use transformations to graph each function.

$f (x) = | x + 1 | - 2$
$f (x) = - {(x - 2)}^{2} + 3$
$f (x) = 2^{x - 1} - 3$
$f (x) = \sqrt{x - 1} + 3$
Let $f (x) = 2 x - 2 .$
1. Find $f^{- 1} (x) .$
2. Graph f and $f^{- 1}$ on the same coordinate plane.
Let $f (x) = x^{3} - x^{2} + x - 6 .$
1. List all possible rational zeros of f.
2. Use synthetic division to test the possible rational zeros and find a real zero.
3. Use the zero from part (b) to find all of the (real or complex) zeros of f(x).
Expand and simplify: $\log_{3} (9 x^{4})$ .
A radioactive substance decays so that the amount present in t years is given by $A = A_{0} e^{- k t},$ where $A_{0}$ is the initial amount. Find the half-life of the substance if
1. $k = 0.05 .$
2. $k = 0.0002 .$
If f is a one-to-one function and $f (2) = 9,$ find $f^{- 1} (9) .$
Find the time required to double your money if it is compounded continuously at 7.5%.

In Exercises 19 and 20, solve the system of equations.

${\begin{array}{rcl} 5 x & - & 2 y & + & 25 & = & 0 \\ 4 y & - & 3 x & - & 29 & = & 0 \end{array}$
${\begin{array}{rcrlr} 2 x & - & y & + & z & = & 3 \\ x & + & 3 y & - & 2 z & = & 11 \\ 3 x & - & 2 y & + & z & = & 4 \end{array}$

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Table of Contents for Chapter 5 Review and Tests

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Table of Contents for
Chapter 5 Review and Tests