Finding pair of linear equations

width is one of two short sides, so you need 15 times 2 for the equation. then you need the long side length which is 2 times the width plus 7. the long side is 37, now you need to add those. 30 plus 74 to get the perimeter/

3600

Be sure to show your work.

We are given the following data:

length = 5/2 * width
length = 10 inWe know through the transitive

property of equality that the following must be true:

5/2 * width = 10 in

We then solve this equation algebraically for width:

width = (10 in) * 2/5
width = 4 in

The perimeter of the rectangle is given by substituting our known values of length and width into the following general equation:

Perimeter = 2 * length + 2 * width
Perimeter = 2(10 in) + 2(4 in) = 28 in

In a similar fashion, we substitute these known values into the general equation for the area of the rectangle to solve for the same:

Area = length * width
Area = (10 in)(4 in)=40 in^2

Ok
2L + 2W = 50 (i)
W = L/4 + 5 (ii)

So 4W = L + 20 from (ii) and
4W = 100 - 4L from (i)
L + 20 = 100 - 4L
5L = 80
L = 16
W = 9 from (ii)

check
(i) 2*16 + 2*9 = 50
(ii) 16/4 + 5 = 9 = W

160 by 140

The perimeter is 600, so 2w+2l=600 where w is the width and l is the length.
Divide both sides by 2: w+l=300
The length is 20 more than the width: l=w+20
Substituting in the previous equation: w+(w+20)=300
Collecting terms: 2w+20=300
Subtract 20 from both sides: 2w=280
Divide by 2: w=140
Thus the width is 140. Substituting into the equation for length: l=140+20
Simplifying: l=160
The width is 140 and the length is 160

Solve the resulting quadratic equation
2X^2+3X-90=0
Discriminant: (3)^2-4(2)(-90)=729=(27)^2
Two roots
X_1=(1/4)*(-3+27)=6
X-2=(1/4)*(-3-27) =-(15/2), negative
Since the width must be positive, reject the negative root and keep X_1=6
Width =6
Length=2(6)+3=15
Check 6*(15)=90. Checks OK

Rectangle has two pairs of equal sides.
L=length=3 units
w=width =1 unit
Perimeter = sum of the measures of all sides.
2 sides with measure L: sum of lengths=2*L
2 sides with measure w: sum of widths=2*w
Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

Rectangle has two pairs of equal sides.
L=length=3 units
w=width =1 unit
Perimeter = sum of the measures of all sides.
2 sides with measure L: sum of lengths=2*L
2 sides with measure w: sum of widths=2*w
Perimeter = 2L+2w=2(L+w)=2(1+3)=8 units.

The rectangle is 11cm by 19cm.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

Translate the English into Mathematics.
W=L-5 (in inches)
P=2(L+W)=2(L+L-5)=2(2L-5)
Use distirbutive property of multiplication with respect to addition to open up the parentheses (brackets)
P=4L-10.
Set P= 50 (inches), to get 4L-10=50
Solve for L: Do it'!
Find W= L (the one you just found) -5 =

Now, with W the value you just calculated
the new length is L'=-4+3W
and the new perimeter is P'=2(L'+W)

Now your turn to do some work.