Answer:
The length can be 5 and the width 10.
Explanation:
perimeter is equal to 2L + 2W
area is equal to L * W
you get 2 equations:
30 = 2L + 2W
50 = L*W
we'll solve by substitution.
you can use either equation to solve for either variable.
we'll use the second equation to solve for L.
second equation is:
50 = L*W
divide both sides of this equation by W to get:
L = 50/W
we now have a value of L that we can substitute in the first equation which we can then use to solve for W.
the first equation is:
30 = 2L + 2W
replace L with 50/W to get:
30 = 2*(50/W) + 2W
simplify to get:
30 = 100/W + 2W
multiply both sides of this equation by W to get:
30W = 100 + 2W^2
subtract 30W from both sides of this equation to get:
2W^2 - 30W + 100 = 0
factor this equation to get:
(2W - 20) * (W - 5) = 0
solve for W to get:
W = 10 or W = 5
if W = 10, then use the equation of 50 = L*W to get L = 5
if W = 5, then use the equation of 50 = L*W to get L = 10
you have 2 possible solutions.
L = 5 and W = 10 or:
L = 10 and W = 5
both are plausible so we'll test them both out.
when L = 5 and W = 10:
the first equation becomes 2*5 + 2*10 = 10 + 20 = 30
the second equation becomes 5*10 = 50
both equations confirm that the solution of L = 5 and W = 10 is good.
when L = 10 andW = 5:
the first equation becomes 2*10 + 2*5 = 20 + 10 = 30
the second equation becomes 10*5 = 5
both equations confirm that the solution of L = 10 and W = 5 is good.
the ticket can have either:
a length of 5 and a width of 10 or:
a length of 10 and a width of 5.
since the length is usually longer than the width, you are free to choose that the ticket most likely has a length of 10 and a width of 5.
alternatively, you can simply say that the dimensions are 5 by 10 and let the recipient of that information determine which is the length and which is the width.
mathematically, the length can be 10 and the width 5 or the length can be 5 and the width 10.