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I personally understand what you are saying... everyone has a teacher like that at least once in their schooling.
Doing two-step equations can sometimes be tricky, even if you understand the topic pretty well. All two-step equations are: equations that require two objectives to solve for the variable (That's why it's called a <em>two-step </em>equation).
There are multi-step equations...but let's take one bite at a time.
First, let's establish one simple rule:
Anything you do to one side of the algebraic equation, you must do to the other side for the equation to remain true.
For example:
So, we have this just chillin' out with the
. When trying to solve for a variable, this is usually not what we want. The
needs to be alone for us to know what
equals.
For the first step of getting alone, we can add
on each side of the equation to cancel out the
on the variable's side. Remember to do it on both sides of the equation!
becomes
Did you see what I did there? All I did was basically add to
and
. The equation remains equal, but using that simple rule, we were able to get one step closer to isolating the
in the equation.
To continue...
turns into
<em>(This is because the
and the
cancel out.)</em>
Now we have
This next step should look familiar to you (assuming you already know how to solve for one-step equations, I'll keep it short and sweet).
Now, since is the same as saying
times
, we can do the opposite of multiplication...which is <em>division, </em>like so:
The 's will cancel each other out and
simplifies to
.
Awesome! Looks like we solved our two-step equation!
In the equation: ,
Some things to remember:
- What ever you do to one side of the equation, you must do to the other.
- You want to apply the opposite sign of the number you are trying to cancel out of the equation while isolating the
.
- Never make it harder than it needs to be. It is called a two-step equation for a reason, multi-step requires more than two...but wait until you have to work those before you start using fifty different ways to solve the equation.
Hope I was able to help you out! Have a good one, and God bless!
<em>Also, if you are confused about something I answered, or I may be incorrect, feel free to let me know!</em>
Let us list out the given information:
Cost per foot = $5
Total fencing cost = $1200
Let L be the length of the barn that will maximize the area of the pasture and W be the width of the rectangular cow pasture.
From the given information, we need to do the fencing for three sides covering 1 L and 2 W's because other L will be the fourth side is part of the side of a barn. We need to write the equation based on the cost information.
If cost/foot is $5, then cost of length ' L ' will be 5L dollars and cost of '2W' width equals 10W dollars. Total cost is $1200. We can set up second equation as
5L + 10W = 1200. This can be simplified by dividing throughout the equation by 5 to get L+ 2W= 240. Solving for L we get L = 240 - 2W
Area of a rectangle is given by A = L ×W.
Substituting 240 - 2W for L we get the area function as..
A=(240 - 2W)×W= 240W-2W²
A = -2W²+240W.
When we have a function of the form f(x) = ax^2 + bx + c, then maximum value of the function can be obtained for x = -b/2a. Using that idea we can find the maximum area.
Here in the area function a = -2 and b = 240
For W = -b/2a = -240/2(-2) = -240/-4 = 60 feet, we get maximum area.
The question is asking us to find Length. We can substitute W = 60 in the equation L = 240 - 2W to find the Length.
L = 240 - 2(60) = 240 - 120 = 120 feet.
Conclusion: The length of the side parallel to the barn that will maximize the area of the pasture is 120 feet.
Note: Here the information 400 feet in the sentence "the fourth side is part of the side of a barn that is 400 feet long" given to distract us, we might not require that information to find the Length.
