8 Wyner Honors Algebra II Fall 2013

Transcription

1 8 Wyner Honors Algebra II Fall 2013 CHAPTER THREE: SOLVING EQUATIONS AND SYSTEMS Summary Terms Objectives The cornerstone of algebra is solving algebraic equations. This can be done with algebraic techniques, by graphing, or other methods. The same is true for systems of equations, which are sets of multiple equations. The solutions to a system of equations are the points where all of the equations are true, or cross together on a graph. This chapter focuses on linear equations and absolute value equations, but the concepts and procedures apply to other types as well. 3-A Graphs of Lines and Linear Inequalities Monday September 16 linear inequality ➊ Graph a linear equation. ➋ Graph a horizontal or vertical line. ➌ Graph a linear inequality. 3-B Linear Equations Tuesday September 17 solve ➊ Solve a linear equation. ➋ Solve a linear equation involving decimals. ➌ Solve a linear equation involving fractions. 3-C Systems Wednesday September 18 system consistent inconsistent ➊ Solve a system graphically. ➋ Solve a system of two equations using substitution. ➌ Solve a system of two equations using elimination. ➍ Solve a system of three equations. ➎ Identify algebraically whether a system is consistent or inconsistent. ➏ Graph a system of linear inequalities. 3-D Systems and Equations on the Calculator Friday September 20 ➊ Solve a system of two equations by graphically finding the points of intersection. ➋ Solve an equation on the calculator by finding the points of intersections of two graphs. 3-E Absolute Value Equations Monday September 23 absolute value extraneous solution ➊ Solve a simple absolute value equation. ➋ Check an equation for extraneous solutions. ➌ Solve a simple absolute value inequality. Review Tuesday September 24 Test Friday September 27

2 9 Wyner Honors Algebra II Fall A Graphs of Lines and Linear Inequalities A LINEAR Expression is a polynomial with no exponents. The graph of a linear equation is a line. ➊ Graph a linear equation. 1. Plug in x = 0 to find the y intercept, and plot this point. 2. Find and plot a second point by plugging in another number for x or plugging in y = Connect the two points. ➊ 4x + y = 1. 4(0) + y = y = () + y = y = -3 For any number c, y = c is a horizontal line and x = c is a vertical line. ➋ Graph a horizontal or vertical line. For y = c, start at (0, c) and draw a horizontal line. For x = c, start at (c, 0) and draw a vertical line. ➋ a) x = 4 b) y = -2 An INEQUALITY is a statement showing that two expressions are not necessarily equal, such as with <. ➌ Graph a linear inequality involving <, >,, or. 1. Graph it as if it were a line. 2. Make the line dotted if the inequality is < or >. 3. Choose a point not on the line. If it works in the inequality, shade that side; otherwise shade the other side. ➌ y > 2 / 3 x Graph y = 2 / 3 x / 3 (0) + 2 = 2 2 / 3 (6) + 2 = 6 2. It is > not, so make the line dotted. 3. For the point (0, 0), 0 > 2 / 3 (0) + 2 is not true, so shade the side not containing the point (0,0). 4x + y = y = 2 / 3 x + 2 (0, -2) y = -2 (0, 1) (4, 0) (6, 6) (0, 2) (0, 0) x = 4 (, -3)

3 10 Wyner Honors Algebra II Fall B Linear Equations SOLVING an Equation means finding all values of the variable that make the equation true. ➊ Solve a linear equation. 1. Distribute factors into expressions in parentheses, if any. 2. Isolate all the variable terms on one side. 3. Combine like terms. 4. Divide each side of the equation by the new coefficient. ➊ x + 2 = -3(2x 4) x + 2 = -6x x + 6x = x = x = 19 / 11 ➋ Solve a linear equation involving decimals. 1. Distribute factors into expressions in parentheses, if any. 2. Multiply each term by the same power of 10 to eliminate the decimals. 3. Do steps 2-4, above. ➋ 1.8(x 2) = 0.41x x 3.6 = 0.41x x 360 = 41x x 41x = x = 860 x = 860 / 139 ➌ Solve a linear equation involving fractions. 1. Distribute factors into expressions in parentheses, if any. 2. Reduce if possible. 3. Multiply each term by a common denominator of the fractions to eliminate the fractions. 4. Do steps 2-4, above. ➌ / 8 x + 8 = 6( 11 / 9 x ) 1. / 8 x + 8 = 66 / 9 x / 8 x + 8 = 22 / 3 x ( / 8 x + 8) = 24( 22 / 3 x 30) 1x = 176x = 176x 1x 912 = 161x 912 / 161 = x

4 11 Wyner Honors Algebra II Fall C Systems A SYSTEM of Equations is two or more equations connected by and. { means and. The solutions of a system are the points where all of the equations are true. This is where the graphs overlap each other. ➊ Solve a system graphically. 1. Graph the equations. 2. The solutions are where all of the graphs overlap each other. ➊ 2y = 6x 4 { 2x + 4y = 20 The lines cross at the point (2, 4). ➋ Solve a system of two equations using substitution. 1. Choose a variable and solve one of the equations for that variable. 2. Plug this expression into the other equation and solve. 3. Plug this solution into one of the original equations and solve for the other variable. ➋ 2y = 6x 4 { 2x + 4y = y = 3x x + 4(3x 2) = 20 2x + 12x 8 = 20 14x = 28 x = y = 6(2) 4 2y = 8 y = 4 ➌ Solve a system of two equations using elimination. 1. Align the like terms and equal signs of the equations. 2. Multiply one or both of the equations by a number so that one of the variables will cancel when the equations are added. 3. Add the equations and solve for the remaining variable. 4. Plug this solution into one of the original equations and solve for the other variable. ➌ 2y = 6x 4 { 2x + 4y = x + 2y = -4 { 2x + 4y = x 4y = 8 { 2x + 4y = x = 28 x = y = 6(2) 4 y = 4 (2, 4)

5 12 Wyner Honors Algebra II Fall 2013 ➍ Solve a system of three equations. 1. Use substitution and/or elimination until you have two equations that have the same two variables and no third variable. 2. Use substitution and/or elimination on this new system of two equations. 3. Plug the values of these two variables into one of the original equations and solve for the third variable. ➍ { x + 2y + 3z = 4 9x + z = 3y 6 y = 10x + 2z 1. x + 2(10x + 2z) + 3z = 4 { 9x + z = 3(10x + 2z) x + 7z = 4 {-21x z = -6 12x + 3z = 20 { -147x 3z = x = -22 x = 1 12(1) + 3z = 20 3z = -10 z = y = 10(1) + 2(-3) y = 4 A system is CONSISTENT if there is at least one solution, that is, all the graphs cross at one or more points. A system is INCONSISTENT if there is no solution, that is, the graphs never cross all at once (such as parallel lines). ➎ Identify algebraically whether a system is consistent or inconsistent. 1. Solve for one of the variables. a) A normal solution, such as x =, means the system is consistent. b) An impossible solution, such as 0 =, means the system is inconsistent. c) A tautological solution, such as =, means the graphs are the same and the system is consistent with infinitely many solutions. ➎ { y = 3x 1 6x 2y = 12 6x 2(3x 1) = 12 6x 6x + 2 = 12 2 = 12, so the system is inconsistent and has no solution. The solution set to a system inequalities is the region shaded by all of the inequalities in the system. ➏ Graph a system of linear inequalities. 1. Graph each inequality (see 3-A). 2. Shade darker the region that was shaded by each inequality. This darker region represents the solutions. ➏ { y 2x x < 1

6 13 Wyner Honors Algebra II Fall D Systems and Equations on the Calculator The calculator can give the exact points of intersection of two graphs, and thus the solution to a system of equations. ➊ Solve a system of two equations by graphically finding the points of intersection. 1. Graph each equation simultaneously. 2. If the points of intersection are off the screen, push [ZOOM] and try ZFit or use Zoom Out until they are visible. 3. Push [CALC] and choose intersect. 4. Push [ENTER] for First curve? and again for Second curve?.. For Guess? use the left or right arrow key to move the cursor to a point of intersection and push [ENTER]. 6. Repeat steps 2 for each additional point of intersection. ➊{ y = x2 6 3y = 6x Y 1 =X 2 6 Y 2 =2X 4. (-0.73,.46) or (2.73, 1.46) ➋ Solve an equation on the calculator by finding the points of intersections of two graphs. 1. For Y 1, enter the expression on the left side of the equation. 2. For Y 2, enter the expression on the right side of the equation. 3. Solve this system (see ➊). (Note that y is irrelevant since the original equation had no y terms.) ➋ x 4 8x = x 2 3x 1. Y 1 =X^4 8X 2. Y 2 =X 2 3X x = 0 or x (-0.73,.46) (2.73, 1.46) (0, 0) (1.90, -2.09)

7 14 Wyner Honors Algebra II Fall E Absolute Value Equations The ABSOLUTE VALUE of a number is the distance it is from zero. Since it is a distance, it cannot be negative. The graph of an absolute value function is a V shape (see right). ➊ Solve a simple absolute value equation. 1. Isolate the absolute value expression. 2. Remove the absolute value symbols by making the other side ±. 3. Solve the + version to get the first answer. 4. Solve the - version to get the second answer. ➊ 3 4x 10 = x 10 = x 10 = ±2 3. 4x 10 = 2 x = x 10 = -2 x = 2 An EXTRANEOUS Solution is a solution to the work that is not actually a solution to the equation. In a linear absolute value equation, when there is an x term outside the absolute value symbols whose coefficient has a higher absolute value than the coefficient of the x term inside the absolute value symbols, one or both of the solutions will be extraneous. ➋ Check an equation for extraneous solutions. 1. Solve the problem normally (see ➊). 2. Plug each solution into the original equation. 3. If a solution does not work but would have worked if a sign in the original equation were changed, that solution is extraneous. ➋ 1 / 3x + 6 = 2x x + 6 = 10x 4 (Note that 10 > 3, so there will be at least one extraneous solution.) 3x + 6 = ±(10x 4) 3x + 6 = 10x 4 3x + 6 = -10x = 7x 13x = 39 x = 1 / 7 x = / 3( 1 / 7 ) + 6 = 2( 1 / 7 ) 9 1 / 3(3) + 6 = 2(3) 9 1 / 13 / / 7 = 102 / 7 63 / 7 1 / = / 19 / 7 = 39 / 7 1 / 1 = / 7 = 39 / 7 is true, so x = 1 / 7 is a solution. 3 = -3 is not true, so x = 3 is extraneous. Note that x = 3 would work if the equation had -1 / instead of 1 /. (2, 6) (3, 6) ( 1 / 7, 39 / 7)

8 1 Wyner Honors Algebra II Fall 2013 ➌ Solve a simple absolute value inequality. 1. Isolate the absolute value expression. 2. Remove the absolute value symbols and solve to find the first boundary. 3. Repeat step 2, but this time reverse the direction of the inequality sign and multiply by -1 the side that did not include the absolute value. 4. If the absolute value expression was less than a constant, combine the two answers in a compound inequality: < x <. Or, if the absolute value expression was greater than a constant, combine the two answers using or: x < or x >. ➌ Solve. a) 3 4x 10 6 b) 3 4x x x x x 10 2 x 3 x x x 10-2 x 2 x x 3 x 2 or x 3