Systems of Equalities

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SRC="0d181600.gif" border=0 width="129" height="96" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></b></div> <div> <B>Keep the equations balanced.</b></div> <div> <B>Remember:  What you do to one side of the equation, you must do to the other.</b></div> <bR> <div> <B>Be sure to sign my guest book to let me know you were here.</b></div> </td> <td valign="top" colspan=2 height=94 BGCOLOR="#FF3300" TEXT="#080000" bgproperties="fixed" background="1693c5e1dvmwlb.gif"> <div> <FONT SIZE="5" COLOR="#000000" FACE="Times New Roman" ><B>GMHS: Mrs. Bossie's  Algebra I Tutorial</b></FONT><B>     |    </b><B> </b><A HREF="index.htm" TARGET="_top" TITLE="GMHS: Mrs. Bossies Algebra I Tutorial"><U><B>home</b></U></A></div> <div> </div> </td> </tr><tr> <td valign="top" height=386 colspan=2 BGCOLOR="#FFFFFF" TEXT="#080000"> <div> <I>Systems of Equalities</I></div> <bR> <bR> <div> Objective 1:  Solve each inequality</div> <div> Objective 2--Graph the system of equalities to solve</div> <bR> <div> Sample problem 1. </div> <div> y = x - 2</div> <div> y = -x + 4</div> <bR> <div> First draw your x and y axis.  Then put a dot on the y-intercept indicated in the first equation.  You will put a dot on -2.  </div> <div> Second look at the slope, 1.  The fraction is 1/1; this means starting at -2 go up one to (0, -1) and right one to (1,-1).  Put another dot at (1, -1).  Draw a line through (0, -2) and (1, -1)  </div> <div> Then look at the second equation and follow the steps above to draw the line indicated by the equation.</div> <bR> <div> The solution to the system of equalties is the point where the two lines cross.</div> <bR> <bR> <div> Objective 2:  Solving a system of equalities using the <FONT SIZE="2" COLOR="#0000BF" FACE="Arial" ><B>substitution</b></FONT> method.</div> <bR> <div> -x + y = 1</div> <div> 2x + y = -2</div> <bR> <bR> <div> First look at the two statements.  See which statement would be easier to solve.  In this case the first one is easier.  </div> <div> -x + y = 1</div> <div> +x              +x          Add x to both sides of the equation.</div> <div> y = 1+ x</div> <bR> <div> Second substitute (1 + x) for the y in the second equation.</div> <bR> <div> 2x + (1 + x) = -2 Remove the parenthesis</div> <div> 2x + 1 + x = -2  Combine like terms</div> <div> 3x + 1 = -2          Subtract one from both sides of the equation</div> <div>        -1    -1</div> <div> 3x + -3                  Divide by 3</div> <div> x = -1</div> <bR> <div> Last go to the first equation and use the solution y = x + 1.</div> <div> Substitute -1 for x</div> <div> y = -1 + 1</div> <div> y = 0</div> <bR> <div> The solution is (-1, 0)</div> <bR> <div> <B>Since question #7 has both a positive and a negative y, </b><FONT SIZE="2" COLOR="#BF0000" FACE="Arial" ><B>linear combinations</b></FONT> would be easier to use to solve the system of equations.</div> <bR> <div> 3x + y = 5</div> <div> <U>2x - y = 10</U></div> <div> 5x = 15                       Add the two rows</div> <div> 5x/5 = 15/5                 Divide both sides of the sum by 5</div> <div> x = 3</div> <bR> <div> 3(3) + y = 5                Substitute 3 for x in one of the original statements</div> <div> 9 + y = 5                     Simplify the equation by multiplying 3 x 3.</div> <div> -9         -9                      Subtract 9 from both sides of the equation.</div> <div> y = -4</div> <bR> <div> The solution is (3, -4)</div> <bR> <div> To check you answers, substitute 3 for x and -4 for y in both original statements.</div> <div> 3(3) + (-4) = 5</div> <div> 9 - 4 = 5         This is true.</div> <bR> <div> 2(3) - (-4) = 10</div> <div> 6 + 4 = 10       This is true.</div> <bR> <div> (3, -4) is the correct solution to the system of linear equations.</div> <bR> <div> Question 8:  Use substitution because x is given in the second statement.</div> <div> 2x + 2y = 15</div> <div> x = 2y</div> <bR> <div> 2(2y) + 2y = 15         Using the first statement, substitute 2y for x.</div> <div> 4y + 2y = 15</div> <div> 6y + 15</div> <div> y = 15/6 which reduces to 5/2</div> <bR> <div> Use the second statement, x = 2y.  Substitute 5/2 for y</div> <div> x = 2(5/2)</div> <div> x = 10/2</div> <div> x = 5</div> <bR> <div> The solution is (5, 5/2)</div> <bR> <div> To check the answer, substitute x with 5 and y with 5/2 in both original statements.</div> <bR> <div> Problem 9:  I would use linear combinations because there is a negative x in the second statement and a positive x in the first statement.</div> <bR> <div> x + 2y = 4</div> <div> <U>-x + y = -7</U>          Add the two rows.</div> <div>       3y = -3          Divide both sides of the equation by 3</div> <div>         y = -1</div> <bR> <div> Substitute -1 for y in one of the original statements.</div> <div> -x + (-1) = -7</div> <div> -x -1 = -7</div> <div>     +1   +1</div> <div> -x = -6</div> <div>   x =  6</div> <bR> <div> The solution is (6, -1).  Check by replacing x with 6 and y with -1 in both of the original statements.</div> <bR> <div> Problem 10.  Use substitution.</div> <div> 2x + 2y = 3</div> <div> x - 4y = -1         -----------------> x = 4y -1</div> <bR> <div> 2(4y -1) + 2y = 3</div> <div> 8y -2 + 2y = 3   Distribute 2 throughout the quantity (4y - 1)</div> <div> 10 y - 2 = 3        Combine terms with a y variable.</div> <div>         +2   +2       Add 2 to both sides of the equation.</div> <div> 10y =5               Divide both sides of the equation by 10</div> <div> y = 1/2</div> <bR> <div> Go to the revised equation x = 4y - 1; substitute 1/2 for y.</div> <div> x = 4(1/2) - 1</div> <div> x = 2 -1</div> <div> x = 1</div> <bR> <div> The solutionis (1, 1/2)</div> <bR> <div> Objective 3:  Use linear combinations to solve the systems:</div> <bR> <div> Problem 11.</div> <div> 4x + 3y = 16         The y variable has coefficients of 3 and -3; therefore, we can add the rows to</div> <div> <U>2x - 3y = 8   </U>          eliminate y.         </div> <div>    6x = 24               First add the rows, then divide both sides by 6.</div> <div>      x = 4                   </div> <bR> <div> Substitute 4 for x in one of the original statements.</div> <div> 2(4) - 3y = 8</div> <div> 8 - 3y = 8</div> <div> -8          -8</div> <div> <U>-3</U>y = <U>0</U></div> <div> -3      -3</div> <div> y = 0</div> <bR> <div> The solution is (4, 0).  To check your solution, substitute 4 for x and 0 for y in both of the original statements.</div> <bR> <div> Problem 12.</div> <div> 5x - 4y = 3</div> <div> 2x + 8 y = -2    Notice that if we divide row 2 by 2, we will have y coefficients of -4 and 4.</div> <bR> <div> <U>2x + 8y = -2  </U>   'Remember that all three terms in the equation is divided by 2.</div> <div>           2</div> <bR> <div> x + 4y = -1       Now rewrite the system of equations using the revised statement for statement 2.</div> <bR> <div> 5x - 4y = 3</div> <div> <U>1</U><U>x + 4y = -1</U><U> </U>     Add the two rows</div> <div> <U>6</U>x = <U>2 </U>               Divide both sides of the equation by 6</div> <div> 6       6</div> <div> x = 1/3</div> <bR> <div> Substitute 1/3 for x in one of the original statements.</div> <bR> <div> 2(1/3) + 8y = -2</div> <div> 2/3 + 8y = -2             Multiply the entire statement by 3 to eliminate the fraction.</div> <div> 2/3 (3) + 8y (3) = (-2)(3)</div> <div> 2 + 24y = -6</div> <div> -2              -2</div> <div> <U>24</U>y = <U>-8</U></div> <div> 24       24</div> <div> y = -1/3</div> <bR> <div> The solution is (1/3, -1/3)</div> <bR> <div> Problem 13.  The key to this problem is to align the variables in columns.</div> <div> 2u = 4v + 8</div> <div> 3v = 5u - 13</div> <bR> <div> 2u - 4v = 8         Move the 4v to the left side (be sure to change the positive sign to a negative sign)</div> <div> -5u + 3v = -13     Move the 5u to the left side (be sure to change the positive sign to a negative.)</div> <div>                          There are no common coefficients or variables.  We have to create them.</div> <div> (5)(2u - 4v) = 5(8)  Multiply the top row by 5.  (Look at the coefficient with the u)</div> <div> (2)(-5u + 3v) = -13 (2)   Multiply the bottom by 2 -- the coefficient of u in the top row.</div> <bR> <div> 10 u - 20v = 40</div> <div> <U>-10u +  6v = -26  </U>     Add the rows</div> <div>        -<U>1</U><U>4</U> v = <U>14</U></div> <div>        -14       -14</div> <div>        v = -1</div> <bR> <div> Substitute -1 for v in one of the original statements.</div> <div> 2u = 4v + 8                                    3v = 5u -13</div> <div> 2u = 4(-1) + 8                                3(-1) = 5u - 13  </div> <div> 2u = -4 +8                                       -3 = 5u - 13</div> <div> 2u = 4                                             10 = 5u</div> <div> u = 2                                                   2 = u</div> <bR> <div> <B>Since u comes before v in the alphabet, we write the solution as (u, v) or (2, -1)</b></div> <bR> <div> You should be able to figure out problems 14 and 15.</div> <div> Objective 4:  Word problems</div> <bR> <div> First make a three by three table.</div> <div>  Add across                               Total</div> <div> <TABLE BORDER="2" CELLPADDING="2" CELLSPACING="1" WIDTH="239" BORDERCOLORLIGHT="#C0C0C0" BORDERCOLORDARK="#808080" FRAME="BOX" RULES="ALL" ALIGN="BOTTOM" HSPACE="0" VSPACE="0" > <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> X</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> Y</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 22</div> </tD> </tR> <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> 150</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> 225</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><IMG BORDER="0" SRC="1x1.gif" HEIGHT="16" ALIGN="bottom" WIDTH="71" HSPACE="0" VSPACE="0"></tD> </tR> <tR> <tD VALIGN=TOP BGCOLOR=#C0DCC0 HEIGHT= 16 WIDTH="71"><div> <B>150x</b></div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 225y</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 3900</div> </tD> </tR> </TABLE></div> <bR> <div> Then read the problem.  Look for the variables that need to be solved.</div> <div> An office supply company sells two types of fax machines.  They charge $150 for one machine and $225 for the other.  <FONT SIZE="2" COLOR="#0000BF" FACE="Arial" >Let x be the fax machine that sells for $150 and y be the machine that sells for $225. </FONT>If the company sold 22 fax machines for a total of $3900 last month, how many of each did they sell.  </div> <bR> <div> In the top row put the variables and the total number of sold fax machine.  </div> <div> The variables will be x and y and the total sold is 22.  </div> <div> In the second row put the cost of the machines. </div> <div>  Under x put the cost of machine x and under y put the cost of machine y.  </div> <div> Multiply down the first two columns.  </div> <div> If the total is given, put it in the last cell of column three.  </div> <div> If the total is not given (that is, if  words like "per" or "each" are given, put the unit cost in the second cell of the last column and multiply to find the total.  Put that total in the last cell of column three.  </div> <div> <B>In this problem, the total is given.</b></div> <bR> <div> Now set up your system of equations using the information in the top and bottom rows.</div> <bR> <div> x + y = 22</div> <div> 150x + 225 y = 3900</div> <bR> <div> Use substitution to solve the system of equations.  </div> <div> Solve for x + y = 22  ------------>   x = 22 - y</div> <bR> <div> Substitute (22-y) for x in the second statement.</div> <bR> <div> (150)(22-y) + 225y = 3900</div> <div> 3300 - 150 y + 225 y = 3900</div> <div> -3300                             -3300</div> <div>          -150 + 225 y = 600</div> <div>                  75 y = 600</div> <div>                      y = 14</div> <bR> <div> Substitute your answer in the top row to see how many x machines were sold.</div> <div> x + y = 22</div> <div> x + 14 = 22</div> <div>      - 14    -14</div> <div> x = 8</div> <bR> <div> The company sold 8 machines for $225 each and 14 machines for $150 each.</div> <bR> <div> To check your answers multiply 8 (225) and 14 (150), then combine your answers to see if the result is 3900.  8 (225) + 14 ( 150) = 1800 + 2100 = 3900.  The answers are correct.</div> <bR> <div> Follow the same steps for the problem 17.</div> <bR> <div> A toy maker produces wooden planes and trains.  Trains require 3 ounces of paint each while planes requie 5 ounces each.  If the toy maker has 64 ounces of paint and wants to paint 14 toys, how many of each can he paint?</div> <bR> <div> Let t = trains and p = planes.  The total number of toys is <FONT SIZE="2" COLOR="#0000CC" FACE="Arial" >14.</FONT></div> <div> He has a <FONT SIZE="2" COLOR="#0000CC" FACE="Arial" >total of 64</FONT> ounces of paint.</div> <bR> <bR> <div> <TABLE BORDER="2" CELLPADDING="2" CELLSPACING="1" WIDTH="239" BORDERCOLORLIGHT="#C0C0C0" BORDERCOLORDARK="#808080" FRAME="BOX" RULES="ALL" ALIGN="BOTTOM" HSPACE="0" VSPACE="0" > <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> t</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> p</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 14</div> </tD> </tR> <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> 3</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> 5</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><IMG BORDER="0" SRC="1x1.gif" HEIGHT="16" ALIGN="bottom" WIDTH="71" HSPACE="0" VSPACE="0"></tD> </tR> <tR> <tD VALIGN=TOP BGCOLOR=#C0DCC0 HEIGHT= 16 WIDTH="71"><div> <B>3t</b></div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 5p</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> 64</div> </tD> </tR> </TABLE></div> <bR> <div> t + p = 14  --------------> t = 14 - p</div> <div> 3t + 5p = 64</div> <bR> <div> 3(14-p) + 5p = 64</div> <div> 42  - 3p + 5p = 64</div> <div> 42 + 2p = 64</div> <div> -42           -42</div> <div> 2p = 22</div> <div> p = 11</div> <bR> <div> t + p = 14</div> <div> t + 11 = 14</div> <div>      -11    -11</div> <div> t = 3                                                                     He can paint 3 trains and 11 planes.</div> <bR> <div> Problem 18 is a little different because we are not given the total for the last cell in column 3.</div> <bR> <div> A chemist needs to make<FONT SIZE="2" COLOR="#0000CC" FACE="Arial" > 30 ounces</FONT> of a <FONT SIZE="2" COLOR="#0000FF" FACE="Arial" >25%</FONT> alcohol solution by mixing a 15% alcohol solution with a 40% solution.  How much of each can he use?  Let x be the 15% solution and y be the    40% solution.</div> <bR> <div> <TABLE BORDER="2" CELLPADDING="2" CELLSPACING="1" WIDTH="265" BORDERCOLORLIGHT="#C0C0C0" BORDERCOLORDARK="#808080" FRAME="BOX" RULES="ALL" ALIGN="BOTTOM" HSPACE="0" VSPACE="0" > <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> X</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> Y</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="97"><div> 30 ounces</div> </tD> </tR> <tR> <tD VALIGN=TOP HEIGHT= 16 WIDTH="71"><div> 0.15</div> </tD> <tD VALIGN=TOP WIDTH="71"><div> 0.40</div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="97"><div> 25%</div> </tD> </tR> <tR> <tD VALIGN=TOP BGCOLOR=#C0DCC0 HEIGHT= 16 WIDTH="71"><div> <B>0.25x</b></div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0 WIDTH="71"><div> <B>0.40y</b></div> </tD> <tD VALIGN=TOP BGCOLOR=#C0DCC0><NOBR><div> 30 (0.25)= 7.5</div> </NOBR></tD> </tR> </TABLE></div> <bR> <bR> <div> x + y = 30 ----------------- x = 30 - y</div> <div> 0.15x  + 0.40y = 7.5         (Multiply by 100 to remove the decimal points)</div> <div> 15 x + 40 y - 750              Substitute (30 - y) for x</div> <div> 15(30 - y) + 40y = 750</div> <div> 450 - 15y + 40y = 750         Combine like terms</div> <div> 450 + 25 y = 750</div> <div> -450              -450</div> <div> <U>25 </U>y = <U>300</U>                            divide both sides by 25</div> <div> 25         25</div> <bR> <div> y = 12</div> <bR> <div> Go to the top row.  x + y = 30  Substitute 12 for y</div> <div> x + 12 = 30</div> <div>       -12   -12</div> <div> x = 18</div> <bR> <div> He needs to mix 18 ounces of 15% alcohol solution with 12 ounces of 40% solution to get 30 ounces of 25% solution.</div> <bR> <div> Problem 19 requires a little geometryknowledge.</div> <bR> <div> <IMG SRC="0ce4e4f0.gif" border=0 width="78" height="79" ALT="Bitmap Image" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> This is an isosceles triangle with height 4 and sides x. The base is y.  The area of an isosceles triangle is 1/2bh.  The perimeter is y + y + x or 2y + x.</div> <bR> <div> An iscoseles triangle has a perimeter of 16 inches and an area of <FONT SIZE="2" COLOR="#BF0000" FACE="Arial" >12</FONT> inches with a perpendicular height of <FONT SIZE="2" COLOR="#BF0000" FACE="Arial" >4</FONT> inches.  What are the measurements of the sides?<FONT SIZE="2" COLOR="#0000CC" FACE="Arial" >  Let b = x and h = 4 </FONT></div> <bR> <div> Area                                                         Perimeter</div> <div> <TABLE BORDER="2" CELLPADDING="2" CELLSPACING="1" WIDTH="440" BORDERCOLORLIGHT="#C0C0C0" BORDERCOLORDARK="#808080" FRAME="BOX" RULES="ALL" ALIGN="BOTTOM" HSPACE="0" VSPACE="0" > <tR> <tD VALIGN=TOP HEIGHT= 144 ><NOBR><div> A = 1/2 bh </div> <div> 12 = 1/2(4)x</div> <div> 12 = 2x</div> <div> 6 = x</div> <div> (use this to find the perimeter</div> </NOBR></tD> <tD VALIGN=TOP><NOBR><div>  P = 2y + x                                        </div> <div> 16 = 2y + x</div> <div> 16 = 2y + x</div> <div> 16 = 2y + 6</div> <div>  -6            -6</div> <div> <U>10</U> = <U>2</U>y</div> <div>  2      2</div> <div> 5 = y</div> <bR> </NOBR></tD> </tR> </TABLE></div> <div>            </div> <div> The base is 6 and the sides are 5.  To check this substitute 5 for y and 6 for x in the formulas for area and perimeter.</div> <div> A = 1/2bh                                                                          P = y + y + x</div> <div> A = (1/2)(6)(4)                                                                   P = 5 + 5 + 6</div> <div> A = 24/2                                                                             P = 16</div> <div> A = 12  </div> <div> This was the given area in the problem.                         This was the given perimeter in the problem</div> <bR> <bR> <div>  For problem 20, simply substitute 1 for x and 2 for y.</div> <bR> <div> Determine if (1,2) is a solution to the system of equalities.</div> <div> x + 2y = 5                      1 + (2)(2) = 5 ---------> 1 + 4 = 5  (This is true)</div> <div> 5x - y = 3                       5(1) - 2 = 3 -------------> 5 - 2 + 3  (This is true)</div> <bR> <div> Therefore, (1,2) is a solution to the system.  The answer is "yes."</div> <bR> <div> For Objective 5: See Review for Inequalities, Objective 6 test, and tips and hints on this web site.</div> <bR> <bR> <div> <IMG SRC="0d237370.gif" border=0 width="55" height="55" ALT="Bitmap Image" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> Good luck!</div> <bR> <bR> </td> </tr></table></body>