Question 166487
let L = length of original rectangle.
let W = width of original rectangle.
formula for perimeter of the original rectangle:
2*L + 2*W = 46m (equation 1) *************************************
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let L+6 = length of modified rectangle.
let 2*W = width of modified rectangle.
formula for perimeter of modified rectangle:
2 * (L+6) + 2 * (2*W) = 70m (original equation 2)
expand this out more:
(2*L) + (2*6) + (2*2*W) = 70m
simplify:
2*L + 12 + 4*W = 70m 
subtract 12 from both sides:
2*L + 4*W = 58 (equation 2) **************************************
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you have 2 equations that you want to solve simultaneously so that the same solution is applicable to both.
they are:
2*L + 2*W = 46 (equation 1)
2*L + 4*W = 58 (equation 2)
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since they have a common term that can be eliminated (2*L), just subtract equation 1 from equation 2:
2*W = 12
W = 6
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you have W = 6.
substitute in equation 1:
2*L + 2*6 = 46
simplify:
2*L + 12 = 46
subtract 12 from both sides:
2*L = 34
divide both sides by 2:
L = 17
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you have:
W = 6
L = 17
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substitute in original equation 2:
2 * (17+6) + 2*(2*6) = 70
perform indicated operations:
2*23 + 2*12 = 70
46 + 24 = 70
70 = 70
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answer is good:
length of the original rectangle is 17 meters and width of the original rectangle is 6 meters.
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