Solve the system using elimination. -2x+6y=18 -4x-6y=18
2 answers:
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<span>1.1 What are the different ways you can solve a system of linear
equations in two variables?
</span><span>A linear equation can be written in many ways, so the way we can use to solve a system of
linear equations depends on the form the system is written. There are
several methods to do this, the more common are: Method of Equalization,
Substitution Method, Elimination Method.
<span>1.2. Method of Equalization
</span>a. Write the two equations in the
style [variable = other terms] either variable x or y.<span>
</span>b. Equalize the two equations
c. Solve for the other variable and then for the
first variable.
</span>
<span>
1.3. Substitution Method
a. Write one of the equations (the one that looks the simplest equation) in
the style [variable = other terms] either variable x or y.
b. Substitute that variable in the other equation and solve using the
usual algebra methods.
c. Solve the other equation for the other variable.<span>
1.4. Elimination Method
Eliminate</span> means to remove, so this method works by removing
variables until there is just one left. The process is:
<span>
</span>a. Multiply an equation by a constant "a" such that there is a term in one equation like [a*variable] and there is a term in the other equation like [-a*variable]
b. Add (or subtract) an equation on to another equation so the aim is to
eliminate the term (a*variable)
c. Solving for one variable and the for the other.
</span>
2.1 Benefits of Method of Equalization<span>
</span>
It's useful for equations that are in the form [variable = other terms] (both equations). In this way, it is fast to solve the system of linear equation by using this method. The limitations come up as neither of the equations are written in that way, so it would be necessary to rewrite the two equations to achieve our goal
2.2. Benefits and limitations of Substitution Method
This method is useful for equations that at least one of them is in the form [variable = other terms]. So unlike the previous method, you only need one equation to be expressed in this way. Hence, it is fast to solve the system of linear equation by using this method. The limitations are the same that happens with the previous method, if neither of the equations is written in that way, we would need to rewrite one equation to achieve our goal.
2.3. Benefits and limitations of Elimination Method
It's useful for equations in which the term [a*variable] appears in one equation and the term [-a*variable] appears in the other one, so adding (or subtracting).
3.1. What are the different types of solutions
When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.
One solution: It's also called a consistent system. It happens as each equation gives new information, also called Linear Independence.
An infinite number of solutions: It's also called a consistent system, but happens when the two equations are really the same, also called Linear Dependence.
No solutions: It happens when they are actually parallel lines.
3.2. Graph of one solution system
See figure 1, so there must be two straight lines with different slopes.
3.3. Graph of infinite number of solutions system
See figure 2. So there must be two straight lines that are really the same.
3.4. Graph of no solution system
See figure 3. So there must be two straight lines that are parallel.
4. Explain how using systems of equations might help you find a better deal on renting a car?
Q. A rental car agency charges $30 per day plus 10 cents per mile to rent a certain car. Another agency charges $20 per day plus 15 cents per mile to rent the same car. If the car is rented for one day in how many miles will the charge from both agencies be equal?
A. Recall: 10 cents = 0.1$, 15 cents = 0.15$
If you rent a car from the first car agency your cost for the rental will be:
(1)
<span>If you rent a car from the second car agency the total amount of money to pay is:
(2) </span>
Given that the problem says you want to rent the car just for one day, then D = 1, therefore:
First agency:
Second agency:
<span>At some number of miles driven the two costs will be the same:
Solving for M:
There is a representation of this problem in figure 4. Up until you drive 200
miles you would save money by going with the second company.</span>
<span>
5. Describe different situations in the real world that could be modeled and
solved by a system of equations in two variables </span>
<span>
</span>
For example, if you want to choose between two phone plans. The plan with the first company costs a certain price per month with calls costing an additional charge in cents per minute. The second company offers another plan at a certain price per month with calls costing an additional charge in cents per minute. So depending on the minutes used you should one or another plan.
If the diameter of the cone is 5, then the radius of it will be half that, or 2.5, thus
Two ways to solve the problem:
A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)
B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1 [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation
is given by
= 
- h() / h'()....(2)
Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000
Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal
Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748
So the solutions are S={-3.99989997696619,0.6692468328777476} both accurate to 14 places after the decimal
A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)
B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1 [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation
Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000
Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal
Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748
So the solutions are S={-3.99989997696619,0.6692468328777476} both accurate to 14 places after the decimal