tag:blogger.com,1999:blog-19247466430696873992024-02-18T17:42:01.907-08:00Linear Equation in Two VariablesIn this website, we will be looking at systems that have two equations and two unknowns. We will look at solving them in three different ways: graphing, substitution method and elimination method. This will lead us into solving word problems with systems. That is where we get to answer the infamous question, when will we use this? But first, we have to learn how to work with systems in general. That is why we use generic variables like x and y at this point.Kei Rebolledo and Carlos Mendozahttp://www.blogger.com/profile/07327032589872868366noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-1924746643069687399.post-33936492433499586212007-12-08T21:12:00.001-08:002007-12-13T04:16:38.566-08:00<span style="color:#33cc00;"><span style="font-size:180%;"><span style="color:#009900;"><span style="FONT-WEIGHT: bold">History On The Use Of Variables</span> </span><br /></span></span><br /><div align="center">The use of letters to represent general numbers goes back to the Greeks. Aristotle frequently used capital letters <span class="blsp-spelling-error" id="SPELLING_ERROR_0">of</span> two letters for the designation of magnitude or number.<br /><br />The use of z,y,x to represent unknowns is due to Rene Descartes in his La <span class="blsp-spelling-error" id="SPELLING_ERROR_1">geometrie</span>. He introduces the use of the first letters to signify known quantities while he used last letters to signify unknown quantities. He used the last letter z for the first unknown and proceeded backwards to y and x for the second and third, respectively.<br /><br />The predominant use of the letter x to represent an unknown value came in an interesting way. During the printing of Descartes’ La <span class="blsp-spelling-error" id="SPELLING_ERROR_2">Geometrie</span> which introduced coordinate geometry, the printer reached a dilemma. While the text was being typeset, the printer began to run short of the last letters of the alphabet. The publisher asked Descartes if it mattered whether x, y, or z was used in each of the book’s many equations. He replied that it made no difference. The printer selected x for the most unknowns, since the letters y and z are used in the French language more frequently than is x.</div><div align="center"> </div><div align="center"></div><p align="center"><img style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 216px; CURSOR: hand; HEIGHT: 197px; TEXT-ALIGN: center" height="190" alt="" src="http://www.gla.ac.uk/departments/philosophy/Personnel/susan/Webpages0506/JadeBrian/descartes.jpg" border="0" /><span style="color:#663300;"><strong></strong></span></p><p align="center"><span style="color:#663300;"><strong>Rene Descartes</strong></span></p>Kei Rebolledo and Carlos Mendozahttp://www.blogger.com/profile/07327032589872868366noreply@blogger.com0tag:blogger.com,1999:blog-1924746643069687399.post-89881495195385210172007-12-08T21:10:00.000-08:002007-12-13T03:16:00.011-08:00<div align="left"><span style="color:#009900;"><span style="FONT-WEIGHT: bold;font-size:180%;" >Discussion with Examples</span><br /></span><br /><span style="COLOR: rgb(255,102,0);font-family:verdana;font-size:130%;" ><span style="FONT-WEIGHT: bold">* What is a linear equation in two variables?</span></span> <span style="font-family:verdana;"></span><br /><br /></div><div style="TEXT-ALIGN: center" align="center"><span style="font-family:verdana;">A linear equation in two variables is a first-degree equation that can be put in the form:</span><span style="font-family:verdana;"><br /><br /></span><span style="FONT-STYLE: italic;font-family:verdana;" >Ax + By = C</span> <span style="font-family:verdana;"></span><span style="font-family:verdana;">Where A, B, and C are real numbers, A and B are not 0, and x and y are the two variables<br /></div></span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)"></span><br /><div style="TEXT-ALIGN: center" align="center"><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">e.g. x + y = 10, 2x + 5y = 15</span> </div><span style="COLOR: rgb(255,102,0);font-size:130%;" ><span style="FONT-WEIGHT: bold;font-family:verdana;" ><br /><div align="left">* What is the solution of a linear equation in two variables?</span></span></div><div align="left"><span style="font-family:verdana;"><br /></div></span><div style="TEXT-ALIGN: center" align="center"><span style="font-family:verdana;">A solution of linear equation in two variables requires two values, one for each variable, which will satisfy both equations. These two variables can be represented by an ordered pair (a,b).<br /></div></span><img id="BLOGGER_PHOTO_ID_5142298040487282514" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-9Ef4lVWnqy62jRSqGQSeZEb8IB3xGWDp9l3bkbGfP-QTaUVIOwH9coFc9j_3pg8E9hgCWDJH_wEZ4om-frKSIrEgCf6IGXU0xj1y4BSkps1e7rxNS7VeH4j_MNwDvg6eNO6oqmzpMX_4/s320/x+and+y+value.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold"><span style="font-family:verdana;"><span style="font-size:130%;color:#ff6600;">* How do you decide whether an ordered pair satisfies an equation in two variables?</span><br /></span></span><br /><div style="TEXT-ALIGN: center"><span style="font-family:verdana;"></span></div><div style="TEXT-ALIGN: center"><span style="font-family:verdana;">An ordered pair (a,b) can belongs to an equation in two variables if a true statement results when a and b are substituted for x and y.<br /><br /></span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">e.g. (3,2); 2x + 3y = 12<br /><br />To see whether (3,2) is a solution of the equation 2x + 3y = 12, we substitute 3 for x and 2 for y in the given equation..</span> </div><br /><br /><div style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic; TEXT-ALIGN: center"><span style="font-family:verdana;">2x + 3y = 12</span> <span style="font-family:verdana;">2(3) + 3(2) = 12 ? </span></div><br /><div style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic; TEXT-ALIGN: center"><span style="font-family:verdana;">Let x = 3; let y = 2</span> </div><br /><div style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic; TEXT-ALIGN: center"><span style="font-family:verdana;">6 + 6 = 12 ?</span> <span style="font-family:verdana;">12 = 12 True</span> </div><br /><div style="TEXT-ALIGN: center"><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)"></span></div><br /><div style="TEXT-ALIGN: center"><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">The result is true, so (3,2) satisfies 2x + 3y = 12<br /><br /></div></span><span style="color:#ff6600;"><span style="font-size:130%;"><span style="FONT-WEIGHT: bold"><span style="font-family:verdana;">* What are the three kinds of the systems of equations?</span></span> </span></span><span style="font-family:verdana;"><br /></span><br /><div style="TEXT-ALIGN: center"><span style="font-family:verdana;">To illustrate what the three kinds of the systems are, let us observe these graphs of a linear system:<br /><br /></span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">1) x – y = 4, x + y = 2<br /></span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)"></div></span><img id="BLOGGER_PHOTO_ID_5142300780676417410" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiyjX1AltgxouF7qB2jK80vGz7heBoArM2yMv7sBYbBqxxus-dX3aiVmXM0zQolJEVQ0NXszzAGUVz2QbALSVwkJs-oKkC3rp5hFrCN06m13q7srp1fMYJpaKscXNsEZ0gdGK5ixCFAMY9C/s320/1st.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)"><p align="center"></span><span style="color:#663333;"><strong>2) 5x – 2y = 10, 5x – 2y = 6 </strong></span></p><br /><div align="center"><img id="BLOGGER_PHOTO_ID_5142301351907067794" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhrOCvM8etujFC3qV4irC_BLKjp86mjyptynbfMpbM21oi3xw341q_s3GwHdKlSA0wzDadK0G0C33JfyllvyZvGJFwqzoRwT8ecOEOX2IdB-nuv9tAD4bCK3iVRiFKCYkXQKzZHv4kQOtRG/s320/2nd.bmp" border="0" /><span style="color:#663333;"><strong> 3) x + y = 2, 2x + 2y = 41</strong></span><span style="font-family:verdana;"><img id="BLOGGER_PHOTO_ID_5142299165768714098" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwcbhbZh-RtKYjhz-Jl3ayYonaorDfeipFDG9kjdix3Xfbhztw_h4raUUz01ao_P4u6CA0WY6fNtxgTkXHlB8asegzCBjgkxapLX-G45RybBDkyCyiomVzaxQFpc0zeh_TmevKhlpGztp4/s320/3rd.bmp" border="0" /><br />1) In the first graph, the lines intersect at a single point. In this case, it has only one solution. This is a consistent system.<br /><br /></span><span style="font-family:verdana;">2) In the second graph, the lines are parallel to each other. Therefore, the lines do not intersect at any point and sp the system has no solution. This is an inconsistent system.<br /><br /></span><span style="font-family:verdana;">3) In the third graph, the lines are identical; thus, they overlap and end up being the same line. In this case, the system has an infinite number of solutions since all the solutions of one equation are also solutions of the other equation.</span></div><div align="center"><span style="font-family:Verdana;"></span> </div><div align="center"><span style="font-family:verdana;"></span></div><div align="center"></div><div align="left"><span style="COLOR: rgb(255,102,0);font-size:130%;" ><span style="font-family:verdana;"><span style="FONT-WEIGHT: bold">* How do you determine whether a system is dependent, consistent, or inconsistent?<br /></span></span></span><br /></div><div style="TEXT-ALIGN: center" align="center"><span style="font-family:verdana;"></span></div><div align="center"><span style="font-family:verdana;">In general, the system of equations Ax + By = C and ax + by = c is:<br /></div></span><div align="center"><span style="font-family:verdana;">1) consistent if A/a is not equal to B/b and B/b is not equal to C/c</span> <span style="font-family:verdana;">e.g. Consider the consistent system (x – y = 4, x + y = 2) given in the sample graph. Substituting the coefficients as stated in the formula above, 1/1 is not equal to -1/1 and -1/1 is not equal to 4/2, proving that it is indeed a consistent system.<br /><br /></span><span style="font-family:verdana;">2) inconsistent if A/a is equal to B/b and B/b is not equal to C/c</span> <span style="font-family:verdana;">e.g. Consider the inconsistent system (5x – 2y = 10, 5x – 2y = 6) given in the sample graph. Substituting the coefficients as stated in the formula above, 5/5 is equal to -2/-2 and -2/-2 is not equal to 10/6, proving that it is indeed an inconsistent system<br /><br /></span><span style="font-family:verdana;">3) dependent if A/a is equal to B/b and B/b is equal to C/c</span> <span style="font-family:verdana;">e.g. Consider the inconsistent system (x + y = 2, 2x + 2y = 4) given in the sample graph.<br />Substituting the coefficients as stated in the formula above, 1/2 is equal to 1/2 and 1/2 is equal to 1/2, proving that it is indeed a dependent system</span> </div><div align="center"><br /></div><span style="FONT-WEIGHT: bold;font-size:130%;" ><span style="COLOR: rgb(255,102,0);font-family:verdana;" ></span></span><div align="center"><span style="FONT-WEIGHT: bold;font-size:130%;" ><span style="COLOR: rgb(255,102,0);font-family:verdana;" >* How do you solve for the solution of systems of linear equations?<br /></span></span><br /><span style="font-family:verdana;">There are three ways to solve systems of linear equations in two variables:<br /><br />1) graphing<br /><br />2) elimination method<br /><br />3) substitution method </span></div><div align="center"><span style="font-family:Verdana;"></span></div><div style="TEXT-ALIGN: center" align="left"></div><div style="TEXT-ALIGN: center" align="right"><span style="font-family:Verdana;font-size:130%;color:#ff6600;"><strong>* </strong></span><span style="FONT-WEIGHT: bold"><span style="font-family:verdana;"><span style="font-size:130%;color:#ff6600;">How do you use graphing to solve for the solution of systems on linear equations?</span><br /><br /></span></span><span style="font-family:verdana;">To solve for the solution of systems on linear equations using graphing, find the x and y intercepts of each solution. The x-intercept is found by taking y = 0 and then solving for the value of x. The y-intercept is found by taking x = 0 and solving for the value of y. The x-intercept (a,0) and y-intercept (0,b) will serve as the two points to create the line.</span> </div><span style="font-family:verdana;"><div align="center"><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic">e.g. Solve the system of equation by graphing:</span></span> <span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)"><br /><br />In x + y = 3 and x- y = 1, you will get (respectively): </span></div><div align="center"><strong><span style="color:#663333;"></span></strong></div><div align="center"><strong><span style="color:#663333;">x - intercept - (3,0); y - intercept - (0,3)</span></strong></div><div align="center"><strong><span style="color:#663333;">x - intercept - (1,0); y - intercept - (0,-1)</span></strong></div><div align="center"><strong><span style="color:#663333;"></span></strong> </div><div align="center"><strong><span style="color:#663333;"></span></strong> </div><div align="center"><strong><span style="color:#663333;"></span></strong></div><div align="center"><strong><span style="color:#663333;"></span></strong></div><div align="center"><strong><span style="color:#663333;"></span></strong></div><div align="center"><strong><span style="color:#663333;"></span></strong></div><div align="center"><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Therefore, (3, 0) and (0, 3) are on the line x + y = 3 and (1, 0) and (0, -1) are on the line x – y = 1. Plotting these points and drawing the line that contains each pair we obtain this graph:</span> </div><div align="center"></div><div align="center"></div><div align="center"> </div><div align="center"><br /></div><span style="font-family:verdana;"><img id="BLOGGER_PHOTO_ID_5142302760656340898" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" height="196" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkB0v20Bd4aj6YWQYEzDH17Ghrb_VpWAI4HNi_foecR-6XyeDOb-iRHXtcJIEBqHwtpdz83sYaliZR7KBii78C_avROkCle6zr2FCAxUtW0oX79K1SJRsyNu3zHWCjM3OseOyt__I9-tih/s320/4th.bmp" width="331" border="0" /> <p align="center"><span style="font-family:verdana;"><strong><span style="color:#663333;">Conclusion: Since the lines intersect at (2,1), the solution is (2,1)</span></strong></span></p><p align="center"><span style="FONT-WEIGHT: bold;font-size:130%;" ><span style="COLOR: rgb(255,102,0)">* How do you use the substitution method to solve for the solution of systems on linear equations?</span></span></span> <span style="font-family:verdana;"><br /><br /></p></span><div style="TEXT-ALIGN: center"><span style="font-family:verdana;">To solve for the solution of systems on linear equations using the substitution method, follow these steps:</span><br /><br />1) <span style="font-family:verdana;">Simplify if needed. This involves removing parentheses and fractions</span><br /><br />2) <span style="font-family:verdana;">Solve one equation for either variable.</span><br /><br /><span style="font-family:verdana;">3) Substitute what you get for step 2 into the other equation. </span><br /><br /><span style="font-family:verdana;">4) Solve for the remaining variable.</span><br /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">e.g. Solve the system of graphing by the substitution method:</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0); FONT-STYLE: italic"> </div><img id="BLOGGER_PHOTO_ID_5142303147203397554" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfOFwYZpKk6X5rXAMfqXlp5wtmsBMDRNhGUwQPNipL4PTdGsEA5wQuiPUCyDkRATGFwLUIoi9jr8a5xJgNdkITB2Hk4swB4obxsGKpE7T8W9ZxmNGkVMM1h_aMUUz1otgCKLsQUE_Mkbm_/s320/1st+graphing.bmp" border="0" /></span><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">Step 1: It is already simplified</span><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0); FONT-STYLE: italic"></span><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">Step 2: Solving for the second equation</span><br /><img id="BLOGGER_PHOTO_ID_5142303860167968706" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgX5mAiRIso6dVMvhFnnVI7rKrExI484a34OXrKCZxn0VYGUGMAmq56NZIphmyRSqq9wKbj-wyevUGcwQaZNomgR6Zynl8qy5xEKSUX9IE9RnVhHEtY_hyphenhypheni82MQ1tfdaymOoNXrnqteur6c/s320/2ng+graphing.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">Step 3 and 4: Substitute the expression y + 1 for x into the first equation and solve for y:</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0); FONT-STYLE: italic"> </span><br /><img id="BLOGGER_PHOTO_ID_5142304186585483218" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiA9TDi4gzUOgd0294wz0NZPyOfuO92eUT206WFz3cUG3PaaVB1NrwGGJ_q8Z1Kqgkr2c_1HdAf0ZuAZlPhGBtUZHo4rIMof8U74gbR68Ycfcsry4i9pIcQMEZiQR8_zL4TJ6nbM1GmObaV/s320/3rd+graph.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">Step 5: Plug in 3 for y into the equation in step 2 to find the value of x.</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0); FONT-STYLE: italic"> </span><br /><img id="BLOGGER_PHOTO_ID_5142304534477834210" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjkPGSGZx26gDui-JLGfuo29O92aeF0K_92mCkRaW6mmRRMGw3OWeNw979QDBR7fPC-y6lkYR0HEXs3VWhbWE3mBU7ffDeiiJAMwrmaD0X_b5evcWZWkU1HnoVLIH9GUT-YYtyZKkttzmTm/s320/4th+graph.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0)">Conclusion: (4, 3) is the solution</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,0); FONT-STYLE: italic"> </span><br /><span style="font-family:verdana;"><br /><span style="font-size:130%;"><span style="FONT-WEIGHT: bold;color:#ff6600;" >* How do you use the elimination method to solve for the solution of systems on linear equations?</span></span></span><br /><span style="font-family:verdana;"><br /></span><div style="TEXT-ALIGN: center"><span style="font-family:verdana;">To solve for the solution of systems on linear equations using the elimination method, follow these steps:</span><br /><br /><span style="font-family:verdana;">1) Simplify and put both equations in the form Ax + By = C if needed.</span><br /><br /><span style="font-family:verdana;">2) Multiply one or both equations by a number that will create opposite coefficients for either x or y if needed.</span><br /><br /><span style="font-family:verdana;">3) Add equations.</span><br /><br /><span style="font-family:verdana;">4) Solve for second variable</span><br /><br /><span style="font-family:verdana;">5) Solve for remaining variable.</span><br /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic">e</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">.g. Solve the system of graphing by the elimination method:</span><br /><img id="BLOGGER_PHOTO_ID_5142305470780704754" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjz_fP3bBwcsAEuGjtBuFTuE4qlwjkL_LhBu-CCHp-MI31ZKws-_oXmvOj18SbPWbvL8U2QRxh-IRYClIU2-xyObBzYwO6KoyRgWo3XAj_5UzfUe9puVyNJgN90itWy7pyLQvfFR5iM8Mx_/s320/elim1st.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Step 1: Multiplying each equation by it's respective LCD we get:</span><br /><img id="BLOGGER_PHOTO_ID_5142305801493186562" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8TfWHU0BY4pWDAtB1OSY0q-E3KfDTmuRDNpu2BouVlENbErTzJf1qJP9LDrr24ITAJzepQvdsO5IkyjZ-KuEpHM34ZjHTVhXEqGG_pSV25hSHOfEjQMKOBV9FbyOa0wqEEy2kaSSTiPXm/s320/elim2nd.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Step 2: Multiplying the first equation by 5 and the second equation by -2 we get:</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic"> </span><br /><img id="BLOGGER_PHOTO_ID_5142306235284883474" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMfaCa1OULa3duaJ7jTWxhSs4OvbIrUCPRYRpSCp-Ficl00JycH4-IcQULCMih4uQ0WBjWpUYz-GiGAbKxgrj2kTo_BV-J8b3yKZpm9mjM0_bGLPL4iNOZyqgbLw82rNvprvPhAbuP9nCG/s320/elim4th.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Step 3:</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic"> </span><br /><img id="BLOGGER_PHOTO_ID_5142306707731286050" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgajQCAkTPKnhNG1UmhPvFh4TviQFFMaYG-012NjZMhTZTb55TTxDfZhGaxEQxYRLHIYqRYwKu9Xx5OClJ1tEVz_bDtHR7xf6Z98LyJa5TXfqT2EVUwon2KM_CWeIbF8o-8oumvOAR8479u/s320/elim3rd.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Step 4:</span><br /><img id="BLOGGER_PHOTO_ID_5142307029853833266" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-020OWZ-1tHVWdTPCQpsNCB_hv5nejRjMcdAf69umqxEZO1AN5KhK-IyDIyiYeLTVv6Sq4EUXGmx_3gxpAiG4Hw0tJwu3445iFE1aedMeVUwfaQFPefwxvnTxhzic11MeDmywmWq0DEzs/s320/elim5th.bmp" border="0" /><br /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Step 5: I choose to plug in 10 for x into the first simplified equation (found in step 1) to find the value of y.</span><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51); FONT-STYLE: italic"> </span><img id="BLOGGER_PHOTO_ID_5142307437875726402" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3G4BC4hIoI7xcG8NfFccRK4cVvNT0SF0RCNgQYJW_jwodWI8JF8qz_N4k5kcLpoSlJs0HP0f04RrycWdM38aRaJdCDchkdPKQA050kMop-JT_SvlC7wYvpTMHSn8GZIVzVN4h9gkQNPHz/s320/elim6th.bmp" border="0" /><span style="FONT-WEIGHT: bold; COLOR: rgb(102,51,51)">Conclusion: The solution is (10, 24)</span></div>Kei Rebolledo and Carlos Mendozahttp://www.blogger.com/profile/07327032589872868366noreply@blogger.com0tag:blogger.com,1999:blog-1924746643069687399.post-12526771779079658282007-12-08T21:01:00.000-08:002007-12-13T04:28:12.995-08:00<div align="center"><span style="font-size:180%;"><span style="FONT-WEIGHT: bold; COLOR: rgb(0,153,0)">Exercises</span><br /></span><br /><span style="color:#ff6600;"><span style="FONT-WEIGHT: bold;font-size:130%;" >A. Determine if the given system of equation is consistent, dependent, or inconsistent</span><br /></span><br />1. x + 3y = 5<br />3x + 9y = 45<br /><br />2. 3x – 6y =5<br />2x – 4y = 11<br /><br />3. 6x + 4y = 3<br />6x + 4y = 5<br /><br /><br /><span style="font-size:130%;"><span style="FONT-WEIGHT: bold; COLOR: rgb(255,102,0)">B. Solve the following using the method of elimination</span></span><br /><br />1. 4x – y = 9<br />x – 3y = 16<br /><br />2. 0.7x – 0.3y = 0.5<br />-0.4x + 0.7y = 1.3<br /><span style="COLOR: rgb(255,102,0);font-size:130%;" ><br /><span style="FONT-WEIGHT: bold">C. Use substitution to solve these equations</span><br /></span><br />1. 5x + y = 8<br />3x – 4y =14<br /><br />2. 4x + 12y = 4<br />5x – y = -11<br /><span style="COLOR: rgb(255,102,0);font-size:130%;" ><span style="FONT-WEIGHT: bold"><br />D. Use comparison method to solve these equations</span></span><br /><br />1. x – 3y = 5<br />5x – y = -14<br /><br />2. 3x – 8y = 11<br />x + 6y = 8<br /><br /><span style="font-size:130%;color:#ff6600;"><strong>Answers </strong></span><br /><br /><span style="color:#ff6600;"><strong><span style="font-size:130%;">A</span> </strong></span><br /><strong><span style="color:#ff6600;"></span></strong><br /><strong><em><span style="color:#996633;">1. dependent<br />2. inconsistent<br />3. inconsistent</span></em><br /></strong><br /><strong><span style="font-size:130%;color:#ff6600;">B</span></strong><br /><br /><strong><em><span style="color:#996633;">1. (-1 , -17/3)<br />2. (2 , 3)</span></em></strong><br /><br /><span style="color:#ff6600;"><strong><span style="font-size:130%;">C</span></strong> </span><br /><span style="color:#ff6600;"><br /></span><strong><em><span style="color:#996633;">1. (2 , -2)<br />2. (-2 , 1)</span></em></strong><br /><br /><span style="font-size:130%;"><strong><span style="color:#ff6600;">D</span></strong> </span><br /></span><span style="font-size:130%;"></span><br /><strong><em><span style="color:#996633;">1. (-3 , -1)<br />2. (5 , ½)</span></em></strong></div>Kei Rebolledo and Carlos Mendozahttp://www.blogger.com/profile/07327032589872868366noreply@blogger.com0tag:blogger.com,1999:blog-1924746643069687399.post-76532548950500570742007-12-08T20:49:00.000-08:002007-12-13T04:35:20.334-08:00<div align="center"><span style="FONT-WEIGHT: bold"><span style="COLOR: rgb(51,204,0)"><span style="font-size:130%;"><span style="font-family:verdana;"><span style="font-size:180%;"><span style="COLOR: rgb(0,153,0)">Applications</span> </span></span><br /><br /></div></span></span></span><span style="FONT-WEIGHT: bold"><span style="COLOR: rgb(51,204,0)"><span style="font-size:130%;"></span></span></span><div align="center"></div><div align="center">The systems of linear equations can be used for a variety of problems such as coin, geometry, age, work, mixture, investment, motion, and digits. That is why we should exert effort in understanding them because they can be helpful to real life situations like other math concepts.</div><div align="center"><br /></div><div align="center"><span style="color:#ff6600;"><strong><span style="font-size:130%;">Age</span> <span style="font-size:130%;">Word</span> <span style="font-size:130%;">Problem</span></strong></span></div><div align="center"><br /></div><div align="center"><strong><span style="color:#996633;"><em>In January of the year 2000, my husband John was eleven times as old as my son William. In January of 2012, he will be three times as old as my son. How old was my son in January of 2000?</em></span></strong> </div><div align="center"> </div><div align="center"></div><div align="center"></div><div align="center">Let "J " stand for my husband John's age, and let "W " stand for William's age. Then J = 11W in the year 2000. In the year 2012, John and William will be twelve years older, so their ages will be J + 12 and W + 12. Also, John will be three times as old as William, so J + 12 = 3(W + 12). So now you've got two equations, each with two variables:</div><div align="center"><br /></div><div align="center">J = 11W J + 12 = 3W + 36<br /></div><div align="center">If you know how to solve systems of equations, you can proceed with those techniques. Otherwise, use the first equation to simplify the second; since J = 11W, plug in "11W " for "J " in the second equation:</div><div align="center"><br /></div><div align="center">J + 12 = 3W + 36 (11W) + 12 = 3W + 36 11W – 3W = 36 – 12 8W = 24 W = 3 </div><div align="center"><br /></div><div align="center">Remember that the problem did not ask for the value of the variable W; it asked for the age of a person. So the answer is: William was three years old at the beginning of the year 2000.</div><div align="center"><br /></div><div align="center"><span style="font-size:130%;color:#ff6600;"><strong>Coin Word Problem</strong></span></div><div align="center"><br /></div><div align="center"><em><strong><span style="color:#996633;">Your uncle walks in, jingling the coins in his pocket. He grins at you and tells you that you can have the coins if you can figure out how many of each kind of coin he is carrying. You're not too interested until he tells you that he's been collecting those gold-tone dollar coins. The twenty-six coins in his pocket are dollars and quarters, and they add up to seventeen dollars. How many of each coin does he have? Here's how you figure it out:</span></strong></em> </div><div align="center"><br /></div><div align="center">There are twenty-six coins in total. Some of them are dollar coins; let "d" stand for the number of dollar coins. The rest of the coins are quarters; let "q" stand for the number of quarters. Then d + q = 26.<br />If your uncle has only one quarter, then 25×1 = 25 cents comes from quarters. If he has two quarters, then 25×2 = 50 cents comes from quarters. Since he has q quarters, then 25×q = 25q cents comes from quarters.<br />For the dollar part to work, you'll have to convert to cents. That is, one dollar is one hundred cents. Since he has d dollars, then he has 100d cents from the dollar coins.<br />He has seventeen dollars in total, or 1700 cents, part of which is from quarters and part of which is from dollars. To help keep things straight, I'll set up a table:</div><div align="center"><br /></div><div align="center"></div><div align="center"><br /></div><div style="TEXT-ALIGN: center" align="center"></div><img id="BLOGGER_PHOTO_ID_5143422223456710850" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxsNK8S51EbBgqqG8SDlHoe6dM8nHQpp4Ky8128_wOqVqOasy3-M1DS15pN3FB05QSky6RXu8D93l0NH8lRaAg9IkgoZeXTeMbk5wislDuwjc2a_fXMIQJDJXP-ayEdpj7EJgmHjX3T7da/s320/wa.bmp" border="0" /> <p align="center">If you know how to solve systems of linear equations, you can see that you have the following system:<br />d + q = 26 100d + 25q = 1700<br />If you don't know about systems yet, don't worry: we're nearly done. Since we have two equations related to the same situation, I'll solve one of them and plug the result into the other one. For instance, starting with the two equations::<br />d + q = 26 100d + 25q = 1700<br />...I'll solve the first equation above to get:<br />q = 26 – d<br />(I could have solved the first equation for d instead of for q; the end result would have been the same, though the computations would have looked different in the middle. Choosing to solve for q in this case was purely a matter of taste.)<br />Then I'll plug this into the second equation in place of q to get:<br />100d + 25(26 – d) = 1700<br />Then I'll solve:<br />100d + 25(26 – d) = 1700 100d + 650 – 25d = 1700 75d + 650 = 1700 75d = 1050 d = 14<br />In other words, fourteen of the coins are dollar coins. Since the remainder of the twenty-six coins are quarters, there are 12 quarters.<br /></p><br /><div align="center"><strong><em><span style="color:#996633;">The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended? </span></em></strong></div><div align="center"><strong><em><span style="color:#996633;"></span></em></strong> </div><div align="center"><strong><em><span style="color:#996633;"></span></em></strong></div><div align="center">Now, though, you can use systems of equations, which means you can use two variables.<br />a + c = 2200 4a + 1.5c = 5050<br />Then solve the system for the number of adults and the number of children:<br />a = 2200 – c<br />4(2200 – c) + 1.5c = 5050 </div><div align="center">8800 – 4c + 1.5c = 5050 </div><div align="center">8800 – 2.5c = 5050 </div><div align="center">–2.5c = –3750 </div><div align="center">c = 1500<br />a = 2200 – (1500) = 700<br />There were 1500 children and 700 adults.</div><strong><em><span style="color:#996633;"><p align="center"></span></em></strong><span style="font-size:130%;color:#ff6600;"><strong>Distance Problem</strong></span><br /></span><br /><strong><em><span style="color:#996633;">A 555-mile, 5-hour plane trip was flown at two speeds. For the first part of thtrip, the average speed was 105 mph. Then the tailwind picked up, and the remainder of the trip was flown at an average speed of 115 mph. For how long did the plane fly at each speed? </span></em></strong></p><div style="TEXT-ALIGN: center"></div><div style="TEXT-ALIGN: center">Let's set up a grid: </div><br /><br /><img id="BLOGGER_PHOTO_ID_5143423030910562514" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwp322v9waa9KkUU6BSDP1KEhJyooA4TyzigwiggFxz8DLUE4b7Fdkbt6ofhCjOC5uV9bTNx7ITYw1pMbR2Bdy497wr-VPOyyz-Aoj-C1IqU2Przwgn8097Q2KUEKUAlDVO4Y5tAuUmK0Z/s320/wa+2.bmp" border="0" /> <p align="center">Using "d = rt", the first row gives us d = 105t and the second row gives us:<br />555 – d = 115(5 – t)<br />Since the two distances add up to 555, add the two distances:<br />555 = 105t + 115(5 – t)<br />Then we get:<br />555 = 105t + 575 – 115t 555 = 575 – 10t –20 = –10t 2 = t<br />According to our grid, "t" stands for the time spent on the first part of the trip, so the answer is "The plane flew for two hours at 105 mph and three hours at 115 mph."<br />You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?</p><div align="center"></div><div align="center"><strong><span style="font-size:130%;color:#ff6600;">Digit Problem</span></strong></div><div align="center"><br /><strong><em><span style="color:#cc9933;"><span style="color:#996633;">The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number</span>.</span></em></strong> </div><div align="center"><br />I'll use "t" for the "tens" digit of the original number and "u" for the "units" (or "ones") digit. Keeping in mind that the tens digit stands for "ten times this value", I then have:<br />t + u = 7 10u + t = 10t + u + 27<br />(The "10t + u" on the right-hand side of the second equation represents the original number.) First I'll simplify the second equation:<br />10u + t = 10t + u + 27 9u – 9t = 27 u – t = 3<br />This gives me:<br />u + t = 7 u – t = 3<br />Adding down, I get 2u = 10, so u = 5. Then t = 2. Checking, this means that the original number was 25 and the new number (gotten by switching the digits) is 52. Since 52 – 25 = 27, this solution checks out.</div><div align="center">The number is 25.</div><div align="center"> </div><div align="center"></div><div align="center"></div><div align="center"><span style="font-size:130%;color:#ff6600;"><strong>Business Problem</strong></span></div><div align="center"> </div><div align="center"><strong><span style="font-size:130%;color:#ff6600;"></span></strong></div><div align="center"><span style="color:#996633;"><strong><em>A landscaping company placed two orders with a nursery. The first order was for 13 bushes and 4 trees, and totalled $487. The second order was for 6 bushes and 2 trees, and totalled $232. The bill does not list the per-item price. What is the cost of one bush and of one tree?</em></strong> </span></div><div align="center"><span style="color:#996633;"><br /></span>I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this wouldn't get me anywhere, because I don't have subtotals for the bushes and trees. So I'll pick variables ("b" for the number of bushes and "t" for the number of trees) and set up a system of equations:<br />13b + 4t = 487 6b + 2t = 232<br />Multiplying the second row, I get:<br />13b + 4t = 487 –12b – 4t = –464<br />This says that b = 23. Back-solving, I get that t = 47.<br />Bushes cost $23 each; trees cost $47 each.</div><div align="center"></div>Kei Rebolledo and Carlos Mendozahttp://www.blogger.com/profile/07327032589872868366noreply@blogger.com0