document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197204>Question 1197204</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a grain dealer sold to one customer 9th bushels of wheat 3 of corn and 5 of rye for $40.90 to another 3 of wheat 5 of corn and 9 of rye for $47.30 and to a third 5 of wheat 9 of corn and 3 of rye for $46.10 what was the price per bushel of corn </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #830373 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=830373\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Answer: <font color=red>$3.10 per bushel</font> for the price of corn\r<BR>\n" );
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document.write( "Explanation:\r<BR>\n" );
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document.write( "Given information<BR>\n" );
document.write( "9 bushels of wheat 3 of corn and 5 of rye for $40.90<BR>\n" );
document.write( "3 bushels of wheat 5 of corn and 9 of rye for $47.30<BR>\n" );
document.write( "5 bushels of wheat 9 of corn and 3 of rye for $46.10\r<BR>\n" );
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document.write( "x = price for 1 bushel of wheat<BR>\n" );
document.write( "y = price for 1 bushel of corn<BR>\n" );
document.write( "z = price for 1 bushel of rye\r<BR>\n" );
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document.write( "The three facts in the given information allow us to form these three equations in this order<BR>\n" );
document.write( "9x+3y+5z = 40.90<BR>\n" );
document.write( "3x+5y+9z = 47.30<BR>\n" );
document.write( "5x+9y+3z = 46.10\r<BR>\n" );
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document.write( "The matrix form looks like this<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28matrix%283%2C4%2C9%2C3%2C5%2C40.90%2C3%2C5%2C9%2C47.30%2C5%2C9%2C3%2C46.10%29%29 ALIGN=MIDDLE  ALT=\"%28matrix%283%2C4%2C9%2C3%2C5%2C40.90%2C3%2C5%2C9%2C47.30%2C5%2C9%2C3%2C46.10%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Each coefficient makes up the 3x3 matrix on the left hand side; the right hand side values are along the far right column. Ultimately we have a 3x4 matrix.\r<BR>\n" );
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document.write( "I'll represent this matrix as a table having 3 rows and 4 columns like this<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >9</TD><TD >3</TD><TD >5</TD><TD >40.9</TD></tr><tr><TD >3</TD><TD >5</TD><TD >9</TD><TD >47.3</TD></tr><tr><TD >5</TD><TD >9</TD><TD >3</TD><TD >46.1</TD></tr>\n" );
document.write( "</TABLE>We'll do matrix operations to get this matrix into row echelon form (abbreviated as REF).\r<BR>\n" );
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document.write( "Let's swap row1 (denoted R1) with row3 (aka R3)<BR>\n" );
document.write( "The notation is R1 <--> R3<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >5</TD><TD >9</TD><TD >3</TD><TD >46.1</TD><TD >R1 <--> R3</TD></tr><tr><TD >3</TD><TD >5</TD><TD >9</TD><TD >47.3</TD><TD ></TD></tr><tr><TD >9</TD><TD >3</TD><TD >5</TD><TD >40.9</TD><TD ></TD></tr>\n" );
document.write( "</TABLE><BR>\n" );
document.write( "To get R1 to have 1 as the first item, we'll multiply everything by 1/5<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD >(1/5)*R1 --> R1</TD></tr><tr><TD >3</TD><TD >5</TD><TD >9</TD><TD >47.3</TD><TD ></TD></tr><tr><TD >9</TD><TD >3</TD><TD >5</TD><TD >40.9</TD><TD ></TD></tr>\n" );
document.write( "</TABLE>The goal is to zero out everything below this '1' in the top left corner.\r<BR>\n" );
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document.write( "Compute the operation R2 - 3*R1 which says<BR>\n" );
document.write( "\"triple an item in R1, then subtract from R2\". We subtract straight down. This is why I think the grid lines or boxes from a table are handy to keep things lined up. Use of scratch paper is recommended. A more stronger recommendation is to use a spreadsheet if possible. \r<BR>\n" );
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document.write( "The result of R2-3*R1 replaces R2. This is what the notation R2-3R1 --> R2 is referring to.<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD ></TD></tr><tr><TD >0</TD><TD >0.4</TD><TD >-7.2</TD><TD >-19.64</TD><TD >R2  -  3*R1 --> R2</TD></tr><tr><TD >9</TD><TD >3</TD><TD >5</TD><TD >40.9</TD><TD ></TD></tr>\n" );
document.write( "</TABLE>Now compute R3 - 9*R1 --> R3\r<BR>\n" );
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document.write( "Multiply 9 times an item in R1, subtract from R3. The result replaces the value in R3<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD ></TD></tr><tr><TD >0</TD><TD >0.4</TD><TD >-7.2</TD><TD >-19.64</TD><TD ></TD></tr><tr><TD >0</TD><TD >-13.2</TD><TD >-0.4</TD><TD >-42.08</TD><TD >R3  -  9*R1 --> R3</TD></tr>\n" );
document.write( "</TABLE>-----------------------------------<BR>\n" );
document.write( "The first nonzero value in R2 is the 0.4 in the second entry.\r<BR>\n" );
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document.write( "Multiply this and all of R2 by 1/0.4 so that the first nonzero entry turns into a 1. This is the pivot for this row.<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD ></TD></tr><tr><TD >0</TD><TD >1</TD><TD >-18</TD><TD >-49.1</TD><TD >(1/0.4)*R2 --> R2</TD></tr><tr><TD >0</TD><TD >-13.2</TD><TD >-0.4</TD><TD >-42.08</TD><TD ></TD></tr>\n" );
document.write( "</TABLE>\r<BR>\n" );
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document.write( "Now compute R3 + 13.2*R2 to have it replace R3<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD ></TD></tr><tr><TD >0</TD><TD >1</TD><TD >-18</TD><TD >-49.1</TD><TD ></TD></tr><tr><TD >0</TD><TD >0</TD><TD >-238</TD><TD >-690.2</TD><TD >R3 + 13.2*R2 --> R3</TD></tr>\n" );
document.write( "</TABLE>\r<BR>\n" );
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document.write( "The last step to get it to REF is to multiply every item in R3 by (-1/238)<TABLE  border = \"1\" cellpadding = \"5\">\n" );
document.write( "<tr><TD >1</TD><TD >1.8</TD><TD >0.6</TD><TD >9.22</TD><TD ></TD></tr><tr><TD >0</TD><TD >1</TD><TD >-18</TD><TD >-49.1</TD><TD ></TD></tr><tr><TD >0</TD><TD >0</TD><TD >1</TD><TD >2.9</TD><TD >(-1/238)*R3 --> R3</TD></tr>\n" );
document.write( "</TABLE>The matrix is now in row echelon form\r<BR>\n" );
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document.write( "We have gone from this augmented matrix<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28matrix%283%2C4%2C9%2C3%2C5%2C40.90%2C3%2C5%2C9%2C47.30%2C5%2C9%2C3%2C46.10%29%29 ALIGN=MIDDLE  ALT=\"%28matrix%283%2C4%2C9%2C3%2C5%2C40.90%2C3%2C5%2C9%2C47.30%2C5%2C9%2C3%2C46.10%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "to this matrix in row echelon form<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28matrix%283%2C4%2C1%2C1.8%2C0.6%2C9.22%2C0%2C1%2C-18%2C-49.1%2C0%2C0%2C1%2C2.9%29%29 ALIGN=MIDDLE  ALT=\"%28matrix%283%2C4%2C1%2C1.8%2C0.6%2C9.22%2C0%2C1%2C-18%2C-49.1%2C0%2C0%2C1%2C2.9%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The bottom row has <BR>\n" );
document.write( "0, 0, 1, 2.90<BR>\n" );
document.write( "which translates back to this equation<BR>\n" );
document.write( "0x + 0y + 1z = 2.90<BR>\n" );
document.write( "or in short<BR>\n" );
document.write( "z = 2.90\r<BR>\n" );
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document.write( "The price of one bushel of rye is $2.90<BR>\n" );
document.write( "In other words, rye costs $2.90 per bushel.\r<BR>\n" );
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document.write( "Then use back substitution to find y<BR>\n" );
document.write( "The second row has<BR>\n" );
document.write( "0, 1, -18, -49.1<BR>\n" );
document.write( "which gives this equation<BR>\n" );
document.write( "0x+1y-18z = -49.1<BR>\n" );
document.write( "aka<BR>\n" );
document.write( "y - 18z = -49.1\r<BR>\n" );
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document.write( "We'll use the recently found z value to find y<BR>\n" );
document.write( "y - 18z = -49.1<BR>\n" );
document.write( "y - 18(2.90) = -49.1<BR>\n" );
document.write( "y - 52.2 = -49.1<BR>\n" );
document.write( "y = -49.1 + 52.2<BR>\n" );
document.write( "y = <font color=red>3.10</font><BR>\n" );
document.write( "The price of a bushel of corn is <font color=red>$3.10</font>\r<BR>\n" );
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document.write( "Use further back substitution to find x<BR>\n" );
document.write( "The first row is <BR>\n" );
document.write( "1, 1.8, 0.6, 9.22<BR>\n" );
document.write( "giving us<BR>\n" );
document.write( "1x + 1.8y + 0.6z = 9.22\r<BR>\n" );
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document.write( "Plug in y = 3.10 and z = 2.90 and solve for x.<BR>\n" );
document.write( "1x + 1.8y + 0.6z = 9.22<BR>\n" );
document.write( "1x + 1.8*3.10 + 0.6*2.90 = 9.22<BR>\n" );
document.write( "1x + 5.58 + 1.74 = 9.22<BR>\n" );
document.write( "x + 7.32 = 9.22<BR>\n" );
document.write( "x = 9.22 - 7.32<BR>\n" );
document.write( "x = 1.90<BR>\n" );
document.write( "The price of wheat is $1.90 per bushel\r<BR>\n" );
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document.write( "Summary:<BR>\n" );
document.write( "x = 1.90<BR>\n" );
document.write( "y = <font color=red>3.10</font><BR>\n" );
document.write( "z = 2.90<BR>\n" );
document.write( "which represents the per bushel cost of wheat, corn, and rye in that order.<BR>\n" );
document.write( "All costs are in dollars.\r<BR>\n" );
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document.write( "--------------------------------------------------\r<BR>\n" );
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document.write( "You can use computer software or a graphing calculator to verify the answers. <BR>\n" );
document.write( "The <a  rel=nofollow HREF= \"http://www.math.odu.edu/~bogacki/lat/\">linear algebra toolkit</a> is a very valuable resource for situations like this. Let me know if you have any questions using it. <BR>\n" );
document.write( "Tools like WolframAlpha and GeoGebra are useful as well.\r<BR>\n" );
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document.write( "Or you can plug each x,y,z value into the original equations mentioned.\r<BR>\n" );
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document.write( "If we plugged x = 1.90, y = 3.10, and z = 2.90 into the first equation, then,<BR>\n" );
document.write( "9x+3y+5z = 40.90<BR>\n" );
document.write( "9*1.90+3*3.10+5*2.90 = 40.90<BR>\n" );
document.write( "17.10+9.30+14.50 = 40.90<BR>\n" );
document.write( "40.90 = 40.90<BR>\n" );
document.write( "We get the same thing on both sides, so this is a true equation.<BR>\n" );
document.write( "This confirms those x,y,z values work for the first equation.\r<BR>\n" );
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document.write( "I'll let you check the other equations. You should get the same number on both sides after replacing those x,y,z values with the numbers mentioned. <BR>\n" );
document.write( "Once you discover the three equations are true for those x,y,z values, you will have fully confirmed the solution.<BR>\n" );
document.write( "</font> </SPAN>\n" );
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