Algebra Review IV: Linear Equations and Inequalities

$\begin{array}{ll}LinearEquations\hfill & NonlinearEquations\hfill \\ 2x=6\hfill & x^2+1=6\hfill \\ \begin{array}{l}0.3(w-5)+4\\ =6.1+0.1w\end{array}\hfill & \frac{3}{x}+4=5\hfill \\ \begin{array}{c}\frac{1}{2}(x+2)=\frac{3}{7}\\ (2x-1)=4\end{array}\hfill & \sqrt{\frac{p(1-p)}{25}=0.1}\hfill \end{array}$

1. Simplify both sides of the equation. This includes removing all parentheses and combining all like terms. If the equation has fractions, clearing these will generally simplify the remaining steps. To clear the fractions, multiply both sides of the equation by a number divisible by all denominators of the fractions.
2. Place all terms involving the variable on one side of the equation; place everything else on the other side of the equation.
3. Combine like terms or factor the variable out of all terms.
4. Divide both sides by the coefficient of the variable.

• STEP 1. Simplify both sides of the equation.

  $\begin{array}{ll}\frac{1}{2}(x+2)+\frac{3}{7}(2x-1)=4\hfill & \begin{array}{c}\text{Clear}\text{the}\text{fractions}:\\ \text{Multiply}\text{both}\text{sides}\\ \text{of}\text{the}\text{equation}\text{by}14,\\ \text{the}\text{smallest}\text{number}\\ \text{divisible}\text{by}\text{both}2\\ \text{and}7.\end{array}\hfill \\ 14\left(\frac{1}{2}(x+2)+\frac{3}{7}(2x-1)\right)=14\left(4\right)\hfill & \begin{array}{c}\text{Multiply}\text{each}\text{term}\text{on}\text{the}\\ \text{left}-\text{hand}\text{side}\text{by}14.\\ \left(\text{Use}\text{the}\text{distributive}\text{law}\text{.}\right)\end{array}\hfill \\ 14\left(\frac{1}{2}(x+2)\right)+14\left(\frac{3}{7}(2x-1)\right)=14\left[4\right]\hfill & \begin{array}{c}\text{Perform}\text{the}\text{multiplications}\\ \text{by}14.\end{array}\hfill \\ 7(x+2)+6(2x-1)=56\hfill & \begin{array}{c}\text{Apply}\text{the}\text{distributive}\\ \text{law}\text{to}\text{remove}\text{the}\text{parentheses}.\end{array}\hfill \\ 7x+14+12x-6=56\hfill & \begin{array}{c}\text{Simplify}\text{by}\text{combining}\\ \text{terms}\text{involving}x\text{and}\\ \text{constant}\text{terms}.\end{array}\hfill \\ 19x+8=56\hfill & \hfill \end{array}$

• STEP 2. There is only one term involving $x$, which is on the left-hand side. Place all terms not involving $x$ to the right-hand side of the equation.

  $\begin{array}{ll}19x+8=56\hfill & \begin{array}{c}\text{Subtract}8\text{from}\text{both}\text{sides}\\ \text{of}\text{the}\text{equation}.\end{array}\hfill \\ 19x+8-8=56-8\hfill & \hfill \end{array}$

• STEP 3. Combine like terms.

  $19x=48$

• STEP 4. Solve for $x$.

  $\begin{array}{ll}\frac{19x}{19}=\frac{48}{19}\hfill & \begin{array}{c}\text{Divide}\text{both}\text{sides}\text{of}\text{the}\\ \text{equation}\text{by}19,\text{the}\\ \text{coefficient}\text{of}x.\end{array}\hfill \\ x=\frac{48}{19}\hfill & \hfill \end{array}$

• STEP 1. Simplify both sides of the equation.

  $\begin{array}{ll}0.3(w-5)+4=6.1+0.1w\hfill & \begin{array}{c}\text{Remove}\text{parentheses}\text{on}\\ \text{the}\text{left}-\text{hand}\text{side}\text{by}\\ \text{multiplication}\text{.}\end{array}\hfill \\ 0.3w-1.5+4=6.1+0.1w\hfill & \begin{array}{c}\text{Combine}\text{the}\text{constant}\\ \text{terms}\text{on}\text{the}\text{left}-\text{hand side}\text{.}\end{array}\hfill \\ 0.3w+2.5=6.1+0.1w\hfill & \hfill \end{array}$

• STEP 2. Place all terms with $w$ on the left side of the equation and all other terms on the right side.

  $\begin{array}{ll}-0.3w+2.5=6.1+0.1w\hfill & \begin{array}{c}\text{Add}-0.1w-2.5\text{to}\text{both}\\ \text{sides}\text{of}\text{the}\text{equation}\text{.}\end{array}\hfill \\ \begin{array}{c}0.3w+2.5-|0.1w-2.5\\ =6.1+0.1w-0.1w-2.5\end{array}\hfill & \hfill \end{array}$

• STEP 3. Combine like terms.

  $0.2w=3.6$

• STEP 4. Solve for $w$.

  $\begin{array}{ll}\frac{0.2w}{0.2}=\frac{3.6}{0.2}\hfill & \begin{array}{c}\text{Divide}\text{both}\text{sides}\text{of}\text{the}\\ \text{equation}\text{by}\text{the}\text{coefficient}\text{of}w.\end{array}\hfill \\ w=18\hfill & \hfill \end{array}$

1. $2x=6$
2. $40-2x=19+5x$
3. $2.8-2(x-1.3)=1.6x$
4. $\frac{2}{3}x-6=4-\frac{1}{2}x$
5. $4(1-5x)=2(x-1)+6x$
6. $1-4\left(2x\right)=6-(x+6)$
7. $\frac{1}{5}(2-x)+1=\frac{1}{10}x$
8. $\frac{1}{4}x+\frac{1}{2}(x-4)=3-\frac{1}{6}(x-2)$

1. Plot the points (1, 3), (1, $-2$), (1, 1), and (1, $-4$). What pattern do you notice about these points?
2. Plot the points (2, $-2$), (4, $-2$), ($-2$, $-2$), and (1, $-2$). What pattern do you notice about these points?
3. Plot the points ($-4$, $-4$), ($-2$, $-2$), (0, 0), (1, 1), and (3, 3). What pattern do you notice about these points?

$D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\begin{array}{c}D=\sqrt{{(6-1)}^2+{(5-2)}^2}=\sqrt{5^2+3^2}=\sqrt{25+9}\\ =\sqrt{34}\approx 5.83\end{array}$

$\text{Midpoint}=\frac{2+12}{2}=\frac{17}{2}=7$

$\left(\frac{x_1+x_2}{2},\frac{{\text{y}}_{\text{1}}+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{1+6}{2},\frac{2+5}{2}\right)=\left(\frac{7}{2},\frac{7}{2}\right)$

$\begin{array}{c}D=\sqrt{{\left(\frac{7}{2}-1\right)}^2+{\left(\frac{7}{2}-2\right)}^2}=\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2}\\ =\sqrt{\frac{34}{4}}=\frac{\sqrt{34}}{2}\end{array}$

$\begin{array}{c}D=\sqrt{{\left(1-\frac{7}{2}\right)}^2+{\left(5-\frac{7}{2}\right)}^2}=\sqrt{{\left(\frac{5}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2}\\ =\sqrt{\frac{34}{4}}=\frac{\sqrt{34}}{2}\end{array}$

1. Find the distance between (2, 1) and (7, 10).
2. • (a) Find the midpoint of the line segment between (2, 1) and (7, 10).
   • (b) Use the distance formula to find the distance between (2, 1) and the midpoint from part (a), as well as the distance between the midpoint and (7, 10). What is true of these two distances?
3. • (a) Find the distance between ($-2,-7$) and (3, 4).
   • (b) Determine the midpoint of the line segment between ($-2,-7$) and (3, 4).
4. • (a) Find the distance between ($-6,-7$) and (1, $-3$).
   • (b) Determine the midpoint of the line segment between ($-6,-7$) and (1, $-3$).

$\begin{array}{c}3\left(4\right)+2(-3)\stackrel{?}{=}6\\ 12+(-6)\stackrel{?}{=}6\\ 6=6\text{True}\end{array}$

$\begin{array}{c}3\left(1\right)+2\left(1\right)\stackrel{?}{=}6\\ 3+\left(2\right)\stackrel{?}{=}6\\ 5=6\text{True}\end{array}$

$\begin{array}{c}3\left(0\right)+2y=6\\ 0+2y=6\\ 2y=6\\ y=3\end{array}$

$\begin{array}{c}3x+2\left(0\right)=6\\ 3x+0=6\\ 3x=6\\ x=2\end{array}$

$\begin{array}{ll}4x+6y=4(3+x)\hfill & \begin{array}{c}\text{Remove}\text{the}\text{parentheses}\\ \text{by}\text{multiplying}\text{.}(\text{Use}\text{the}\\ \text{distributive}\text{law}\text{.})\end{array}\hfill \\ 4x+6y=12+4x\hfill & \begin{array}{c}\text{Subtract}4x\text{from}\text{both}\text{sides}\text{of}\text{the}\\ \text{equation}.\end{array}\hfill \\ 6y=12\hfill & \begin{array}{c}\text{Divide}\text{both}\text{sides}\text{of}\text{the}\text{equation}\\ \text{by}6,\text{the}\text{coefficient}\text{of}y.\end{array}\hfill \\ y=2\hfill & \hfill \end{array}$

1. Is (3, 4) a solution to $4x+3y=10$?
2. Is (1, 2) a solution to $4x+3y=10$?
3. Find the $x$- and $y$-intercepts for $4x+3y=10$. Draw a line that represents the solution set.
4. Express $4x-5y=2(3+x)$ in the form $ax+by=c$.
5. Is $\left(2,-\frac{2}{5}\right)$ a solution to the linear equation in Practice Exercise 4?
6. Find the $x$- and $y$-intercepts for the linear equation in Practice Exercise 4. Draw a line that represents the solution set.
7. Graph the solution set of $6x+4y=4(2+y)$.
8. Graph the solution set of $14x+2y=2(5x-3)+4x$.

• 1. (5, 1) and (20, 6)
• 2. (4, $-2$) and (8, $-6$)
• 3. (5, $-2$) and (5, 2)
• 4. ($-3,8$) and ($-5,8$)
• 5-8. Determine whether the slope of each line in the graph below is positive, negative, zero, or undefined.
  
• 9. Determine the slope of the line given the information in the graph below.

• Press $\begin{array}{||}\hline \text{Y}\\ \hline\end{array}=$. Clear out any stored functions.
• Enter the function, $mx+b$, as Y1 and then press $\begin{array}{||}\hline \text{ZOOM}\\ \hline\end{array}\begin{array}{||}\hline 6\\ \hline\end{array}$ to graph the line in the standard viewing window ($\text{X}min=\&-10,\text{X}max=10,\text{Y}min=-10,\text{Y}max=10$]).
• If needed, press WINDOW and adjust the scaling.

1. A line crosses the $y$-axis at 4 and has slope $\frac{4}{5}$. Write an equation that describes the line.
2. A line has $y$-intercept of (0, $-3$) and slope of 2. Identify a second point on the line.
3. A line is described by the equation $y=\frac{5}{4}x+2$. Starting at the $y$-intercept, use the slope to plot a second point that lies on the line that has an $x$-coordinate equal to 4. Draw a line through the $y$-intercept and second point.
4. Plot two points on the line described by $y=-0.25x+5$. Draw a graph of the line. (You may have to extend your horizontal axis in order to see where your graph crosses the $x$-axis.)
5. • (a) The points (2, 5) and (4, 9) lie on a line. What is the slope of the line?
   • (b) Substitute the slope into the equation $y=mx+b$. Substitute one of the points and solve for $b$.
   • (c) Finally, write an equation that describes the line through the points (2, 5) and (4, 9).
6. Repeat Practice Exercise 5 using the points (1, 7) and (3, 2).

7. $y=2.8x+3$
8. $y=-\frac{3}{2}x+\frac{1}{2}$. (Adjust the window settings: $\text{X}min=-5,\text{X}max=5,\text{Y}min=-5,\text{and}\text{Y}max=5$.)

• We draw a solid line in cases where the points on the line are included in the solution set (the inequality is inclusive, $\le$ or $\ge$) and a dashed line when the points on the line are not part of the solution set (the inequality is not inclusive, $<$ or $>$).
• To determine which side of the line to shade, we substitute a test point. A test point can be any point in the plane that is not on the line. In most cases, choosing the origin as a test point allows for easy calculation. If substituting the test point results in an inequality that is true, shade the region on the side of the line containing the test point; otherwise, shade the region on the opposite side of the line to the test point.

• Adding or subtracting the same quantity to both sides of the inequality.
• Multiplying or dividing both sides of the inequality by the same positive quantity.
• Multiplying or dividing both sides of the inequality by the same negative number and reversing the direction of the inequality sign.

• STEP 1. Solve the inequality for $y$.

  $\begin{array}{ll}3x+2y\le 12\hfill & \begin{array}{l}\text{Subtract}3x\text{from}\text{both}\text{sides}\text{of}\text{the}\\ \text{inequality}.\end{array}\hfill \\ 2y\le -3x+12\hfill & \begin{array}{l}\text{Divide}\text{both}\text{sides}\text{of}\text{the}\text{inequality}\text{by}\\ 2,\text{the}\text{coefficient}\text{of}y.\end{array}\hfill \\ y\le -\frac{3}{2}x+6\hfill & \hfill \end{array}$

• STEP 2. Using the test point (0, 0), we find that $0\le -\frac{3}{2}\left(0\right)+6\text{or}0\le 6$, which is a true statement. The solution set is the region on the side of the line containing (0, 0).

• STEP 3. Enter the linear function $y=\frac{3}{2}x+6$ into your calculator’s function list.

  • Press $\begin{array}{||}\hline \text{Y}=\\ \hline\end{array}$ and clear any stored functions.
  • Enter the function as Y1. Press $\begin{array}{||}\hline (\\ \hline\end{array}\begin{array}{||}\hline (-)\\ \hline\end{array}\begin{array}{||}\hline 3\\ \hline\end{array}\begin{array}{||}\hline ÷\\ \hline\end{array}\begin{array}{||}\hline 2\\ \hline\end{array}\begin{array}{||}\hline )\\ \hline\end{array}\begin{array}{||}\hline \text{X},\text{T},\text{θ},\text{n}\\ \hline\end{array}\begin{array}{||}\hline 6\\ \hline\end{array}$ and then $\begin{array}{||}\hline \text{ZOOM}\\ \hline\end{array}\begin{array}{||}\hline 6\\ \hline\end{array}$ to graph the line in the standard viewing window.
  • The test point (0, 0) lies below the line. So next we shade that region. Press $\begin{array}{||}\hline \text{Y}=\\ \hline\end{array}$. The cursor should be to the right ofY1. Use the left arrow key to move the cursor to the left of Y1. Press $\begin{array}{||}\hline \text{ENTER}\\ \hline\end{array}$ repeatedly until the “shade below” symbol is displayed. Press $\begin{array}{||}\hline \text{GRAPH}\\ \hline\end{array}$. The result appears below.
  

• STEP 1. Solve the inequality for $y$.

  $\begin{array}{ll}5x-10y\le 5\hfill & \begin{array}{l}\text{Subtract}5x\text{from}\text{both}\text{sides}\text{of}\text{the}\\ \text{inequality}.\end{array}\hfill \\ -10y\le 5x+5\hfill & \begin{array}{l}\text{Divide}\text{both}\text{sides}\text{of}\text{the}\text{inequality}\text{by}\\ -10\text{and}\text{reverse}\text{the}\text{direction}\text{of}\\ \text{the}\text{inequality}.\end{array}\hfill \\ y\ge \frac{-5}{-10}x+|\frac{5}{-10}\hfill & \text{Simplify}.\hfill \\ y\ge \frac{1}{2}x-\frac{1}{2}\hfill & \hfill \end{array}$

  Page AR-17
• STEP 2. Using the test point (0, 0), we find $0\ge \frac{1}{2}\left(0\right)-\frac{1}{2}=-\frac{1}{2}$, which is true. So, we will shade the side of the line that contains the test point.

• STEP 3. Graph $y=\frac{1}{2}x-\frac{1}{2}$, determine which side of the line to shade and then graph the solution set. The result appears below. The viewing window is set so that the values for $x$ and $y$ are between $-5$ and 5. The region above the line is shaded because the test point (0, 0) lies above the line.

  

1. Does (1, 3) belong to the solution set of $4x-2y>1$?
2. Does (4, 2) belong to the solution set of $2x-4y\le 0$?

3. Graph the solution set of $5y-10x>20$ as follows:

   • (a) Graph $5y-10x=20$. (Should you draw this line solid or dashed?)
   • (b) Select a test point. Does your test point make the inequality true or false?
   • (c) Shade the region corresponding to the solution set.
4. Graph the solution set of $4x-2y>1$. Use the test point from Practice Exercise 1 to determine which region should be shaded. (If this test point does not help, select a different test point.)
5. 5. Graph the solution set of $2x-4y\le 0$. Use the test point from Practice Exercise 2 to determine which region should be shaded. (If this test point does not help, select a different test point.)
6. Solve $3x-4y\ge 6$ for $y$. [In other words, express in the form $y(\ge \text{or}\le )mx+b$.]
7. Use a graphing calculator to graph the solution set of $y\le -3x+6$. Make a sketch of the graph.
8. Use a graphing calculator to graph the solution set of $3x-4y\ge 6$. Make a sketch of the graph.

• STEP 1. Solve one of the equations for one of the variables. Substitute this expression into the second equation to obtain one equation in one variable.
• STEP 2. Solve this new equation using the techniques from Part IV, item A, Linear Equations in One Variable (page AR-10).
• STEP 3. Use the solution from Step 2 and either of the original equations to compute the corresponding value for the second variable.

$\begin{array}{ll}x+y=5\hfill & \left(1\right)\hfill \\ x+2y=7\hfill & \left(2\right)\hfill \end{array}$

• STEP 1. Solve equation (1) for $y$ and then substitute the expression for $y$ into equation (2).

  $\begin{array}{ll}x+y=5\hfill & \text{Add}-x\text{to both}\text{sides}\text{of}\text{the}\text{equation}.\hfill \\ y=-x+5\hfill & \text{Substitute}-x+5\text{for}y\text{in}\text{equation}\left(2\right).\hfill \\ x+2(-x+5)=7\hfill & \hfill \end{array}$

• STEP 2. Solve the equation from Step 1 for $x$.

  $\begin{array}{ll}x+2(-x+5)=7\hfill & \begin{array}{l}\text{Use}\text{the}\text{distributive}\text{law}\text{to}\text{remove}\\ \text{the}\text{parentheses}.\end{array}\hfill \\ x-2x+10=7\hfill & \begin{array}{l}\text{Subtract}10\text{from}\text{both}\text{sides}\text{of}\text{the}\\ \text{equation}.\end{array}\hfill \\ \begin{array}{l}x-2x+10-10\\ =7-10\end{array}\hfill & \text{Simplify}\text{by}\text{combining}\text{terms}.\hfill \\ -x=-3\hfill & \begin{array}{l}\text{Multiply}\text{both}\text{sides}\text{of}\text{the}\text{equation}\\ \text{by}-1.\end{array}\hfill \\ x=3\hfill & \hfill \end{array}$

• STEP 3. Find the corresponding value for $y$ by substituting 3 for $x$ in either equation (1) or (2).

  $\begin{array}{ll}x+y=5\hfill & \text{Substitute}3\text{for}x\text{in}\text{equation}\left(1\right).\hfill \\ 3+y=5\hfill & \begin{array}{l}\text{Subtract}3\text{from}\text{both}\text{sides}\text{of}\text{the}\\ \text{equation}.\end{array}\hfill \\ y=5-3=2\hfill & \hfill \end{array}$

$\begin{array}{ll}\text{Check}(3,2)\text{in}\hfill & \text{Check}(3,2)\text{in}\hfill \\ x+y=5\hfill & x+2y=7\hfill \\ 3+2=5\hfill & 3+2\left(2\right)\stackrel{?}{=}7\hfill \\ 5=5\text{True}\hfill & 7=7\text{True}\hfill \end{array}$

$\begin{array}{c}y=-x+5(1{)}^{\prime }\\ y=\frac{1}{2}x+\frac{7}{2}(2{)}^{\prime }\end{array}$

• Press $\begin{array}{||}\hline \text{Y}=\\ \hline\end{array}$, clear any previously stored functions, and enter the expressions for $(1{)}^{\prime }$ and $(2{)}^{\prime }$ as functions Y1 and Y2, respectively.
  
• Press $\begin{array}{||}\hline \text{WINDOW}\\ \hline\end{array}$ and adjust the window settings to $\text{Xmin}=-1,\text{Xmax}=8,\text{Xscl}=1;\text{Ymin}=-1,\text{Ymax}=6,\text{YScl}=1.\text{Press}\begin{array}{||}\hline \text{GRAPH}\\ \hline\end{array}$.
  

• Press $\begin{array}{||}\hline \text{2nd}\\ \hline\end{array}\begin{array}{||}\hline \text{TRACE}\\ \hline\end{array}$ (for CALC) $\begin{array}{||}\hline \text{5}\\ \hline\end{array}$ (for intersect).
• Press $\begin{array}{||}\hline \text{ENTER}\\ \hline\end{array}$ twice in response to “First curve?” and “Second curve?”.
• Use the arrow keys to move the cursor close to the point of intersection and press $\begin{array}{||}\hline \text{ENTER}\\ \hline\end{array}$.
• Read off the solution $x=3\text{and}y=2$ or (3, 2) as shown in the screenshot.
  

1. Check to see if (4, 1) is a solution to the system of equations

   $\begin{array}{c}2y-x=2\\ y+2x=9\end{array}$

2. $y+3x=6$$2y-x=4$
3. $2y+3x=12$$4y+7x=12$

4. • (a) Use algebraic procedures to solve the following system of equations:

     $\begin{array}{c}3y+2x=7\\ 6y+4x=10\end{array}$

   • (b) What problem did you encounter?
   • (c) Graph the system of equations, and explain why this system has no solution.

5. Use a graphing calculator to solve the following system of equations:

   $\begin{array}{l}y=3x+9\\ y=-2x+1\end{array}$

6. Check whether (1, 3) is in the solution set to $x\ge 0,y+2x\le 4,\text{and}y\ge 3x$.
7. Graph the solution set of $x\le 6,y\ge 4,\text{and}y\ge 2x+3$.
8. Graph the solution set of $x>1,2y-x\ge 2,y+2x\le 9$.