Mathematics
Precalculus. I am not that good at it

Question Description

I’m studying for my Calculus class and need an explanation.

  • Write a problem for a classmate to solve that can be translated to a system of two (2) or more equations in at least two (2) variables. Explain your answer.

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Final Answer

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If a system of linear equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent. If the system has an infinity number of solutions, it isdependent. Otherwise it is independent.


A linear equation in three variables describes a plane and is an equation equivalent to the equation

 

where A, B, C, and D are real numbers and A, B, C, and D are not all 0.


Example 2:


Let's create three equations from the given points. 


We are going to show you how to solve this system of equations three different ways:


1) Substitution, 2) Elimination 3) Matrices


SUBSTITUTION:


The process of substitution involves several steps:


Step 1: Solve for one of the variables in one of the equations. It makes no difference which equation and which variable you choose. Let's solve for [img width="17" height="15" align="BOTTOM" border="0" src="http://www.sosmath.com/soe/SE3001/img21.gif" alt="$C$"> in equation (1).

 

Step 2: Substitute this value for [img width="17" height="15" align="BOTTOM" border="0" src="http://www.sosmath.com/soe/SE3001/img21.gif" alt="$C$" > in equations (2) and (3). This will change equations (2) and (3) to equations in the two variables $A$ and . Call the changed equations (4) and (5), respectively. 

  
$\displaystyle (2)$
  
  
(4)
  
  
  
  
(5)
  
  

Step 3: Solve for [img width="16" height="14" align="BOTTOM" border="0" src="http://www.sosmath.com/soe/SE3001/img23.gif" alt="$A$" > in equation (4).

 

Step 4: Substitute this value of $A$ in equation (5). This will give you an equation in one variable.

 

Step 5:  Solve for $B$.
 

Step 6: Substitute this value of $B$ in equation (4) and solve for $A.$
 

Step 7: Substitute $-6$ for $A$ and $-8$ for $B$ in equation (1) and solve for $C$.
 

The solution: The equation of the circle that contains the points $\left( 3,-1\right) $$\left( -2,4\right) $, and $\left( 6,8\right) $ is
 

$\bigskip\bigskip\bigskip $ Step 8: Check the solutions:
 

ELIMINATION:


The process of elimination involves several steps: First you reduce three equations to two equations with two variables, and then to one equation with one variable.


Step 1: Decide which variable you will eliminate. It makes no difference which one you choose. Let us eliminate $C$ first..

 

Step 2: Add equations (1) and (2) to form equation (4), then add equations (2) and (3) to form equation (5). Equations (4) and (5) will contain the variables A and B.
 

Step 3: We now have two equations with two variables. Let's simplify these two equations.
 

Step 4: Add the simplified equations (4) and (5) to create equation (6) with just one variable.
Step 5: Solve for [img width="16" height="14" align="BOTTOM" border="0" src="http://www.sosmath.com/soe/SE3001/img23.gif" alt="$A$" > in equation (6).
 

Step 6: Substitute $-6$ for $A$ in equation (4) and solve for B.
 

Step 7: Substitute $-6$ for $A$ and $-8$ for $B$ in equation (1) and solve for $C$.
 

The equation of the circle is $x^{2}+y^{2}-6x-8y=0\medskip\medskip $ Check your answers as before. 

MATRICES:


Step 1: Create a three-row by four-column matrix using coefficients and the constant of each equation.

 


We want to convert the original matrix

 

to the equivalent matrix.
 


Step 2: We work with column 1 first. We want a 1 in Cell 11 [Row 1-Col 1]. To achieve this, multiply Row 1 by $\dfrac{1}{3}$ to form a new Row 1.

 

 

Step 3: Add -2 times Row 1 to Row 2 to form a new Row 2, and add -6 times Row 1 to Row 3 to form a new Row 3.
 

 

 

 

 

 

Step 6: Let's now manipulate the matrix so that there is a 1 in Cell 33. We do this by multiplying Row 3 by $-\dfrac{1}{6}.$
 

 

Step 7: Let's now manipulate the matrix so that there are zeros in Cells 13 and 12.

 

 

You can now read the answers off the matrix: $A=-6$$B=-8$, and $C=0$. check your answers by the method described above.$\bigskip\bigskip\bigskip $




Please let me know if you need any clarification. I'm always happy to answer your questions.

Iland A (747)
Boston College

Anonymous
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