1. First of all, let us acknowledge the fact that the lengths on either side of a rectangle are equivalent (this would mean the top length is equal to the bottom length, and the left length is equal to the right length). Knowing this, we can use the dimensions in terms of x and y that are given and equate them for either side. Thus:
Equation 1 (top and bottom): x + y + 2 = 2x + 1
Equation 2 (left and right): x + 2 = 2y
2. Now that we have our equations, we need to solve them. As you have requested the substitution and elimination methods, I will show these as method a) and b), respectively. Before we do this however, I would like to simplify equation 1 first as there are like terms that could be brought together (this will also help us with solving the equations later on):
Equation 1: x + y + 2 = 2x + 1
y + 2 = x + 1 (Subtract x from both sides)
y = x - 1 (Subtract 2 from both sides)
3. a) Substitution method
The substitution method is dependent on our rearranging one of the equations so that it has either x or y as the subject, and then substituting this into the other equation. Since we have already effectively done this in equation 1 by collecting like terms, we have no need to do any further rearranging. Thus our equations are:
1) y = x - 1
2) x + 2 = 2y
Now, we can substitute y = x - 1 into x + 2 = 2y as such:
x + 2 = 2(x - 1) (Replace y with (x - 1) )
x + 2 = 2x - 2 (Expand 2(x - 1) )
2 = x - 2 (Subtract x from both sides)
4 = x (Add 2 to both sides)
Now that we know that x = 4, we can substitute this back into equation 1 to get:
y = x - 1
if x = 4: y = 4 - 1 (Replace x with 4)
y = 3
Therefor, x = 4 and y = 3.
b) Elimination method
The elimination method relies on our subtracting (or adding) the equations from (or to) each other in order to cancel out or 'eliminate' either x or y and then solve for the remaining variable. Let us remind ourselves of our two equations:
1) y = x - 1
2) x + 2 = 2y
Looking at them, we can see that both have x in them, therefor this will be easier to cancel out than the y, where one equation has y and the other 2y (note that this could always be overcome by simply multiplying both sides of the equation with y by 2, however this would require more working than choosing to eliminate x).
In this case, we will be subtracting the two equations, since both x's are positive (thus we will have x - x = 0), however note that if one were negative you would add the two equations since -x + x = 0.
Before we start we could also write the equations out as such:
1) x - 1 = y (I have simply switched the two sides of the equation here)
2) x + 2 = 2y
(This is not compulsory but it is a little easier to visualise how each side will be subtracted)
Thus, if we subtract equation 1 from equation 2 (note that we could very well do this the other way around and get the same answer), we get:
x + 2 - (x - 1) = 2y - y (Subtract the left & right sides of the equations from each other)
x + 2 - x + 1 = y (Expand -(x - 1) ; 2y - y = y)
3 = y (x - x = 0 ; 2 + 1 = 3)
Now, given that y = 3, we can substitute this back into either equation 1 or 2. Let's choose equation 1:
x - 1 = y
if y = 3: x - 1 = 3
x = 4 (Add 1 to both sides)
Thus, x = 4 and y = 3 (note that this is the same answer we had in a) - this is always a good way of checking that you're on track).
4. Now that we have x = 4 and y = 3, we simply need to substitute these values back into one of the expressions for the length of the rectangle and one of the expressions for the width of the rectangle.
Let us choose the following:
L (length) = 2x + 1 (bottom line)
W (width) = 2y (left side)
1) if x = 4: L = 2(4) + 1
L = 8 + 1
L = 9 cm
2) if y = 3: W = 2(3)
W = 6 cm
5. We are at the final step now! All that remains is to find the perimeter. The formula for the perimeter of a rectangle is given by:
P = 2L + 2W
(Note that with long problems it is often easy to forget what you were originally solving for, so always check to make sure you haven't forgotten something.)
Now, given that L = 9 and W = 6, we can substitute these into the formula for the perimeter above to get:
P = 2(9) + 2(6)
P = 18 + 12
P = 30 cm
Thus, the perimeter of the rectangle is 30cm.
Note that the other way of solving for the perimeter is to add all the expressions for the lengths of the sides of the rectangle given (since the perimeter is effectively the sum of the lengths of the sides) and then substitute x = 4 and y = 3 into the resulting expression:
P = x + y + 2 + (x + 2) + (2x + 1) + 2y
= x + x + 2x + y + 2y + 2 + 2 + 1
= 4x + 3y + 5
Given that x = 4 and y = 3:
P = 4(4) + 3(3) + 5
P = 16 + 9 + 5
P = 30 cm
Yet again, we find that the perimeter of the rectangle is 30cm. One of the times where the first method will be better than this one is when you actually have to find the width and length of the rectangle as well as the perimeter, otherwise since you only need the perimeter, either method will work fine.
