Solve using Cramer's Rule.
-The perimeter of a rectangle is 19 meters. If the width is doubled and the length is tripled, the perimeter is 50 meters. Determine the original width and length.
A) width $= 3.5 \mathrm {~m}$, length $= 6 \mathrm {~m}$
B) width $= 8 \mathrm {~m}$, length $= 17 \mathrm {~m}$
C) width $= 6 \mathrm {~m}$, length $= 10 \mathrm {~m}$
D) width $= 7 \mathrm {~m}$, length $= 12 \mathrm {~m}$

Question

Solve using Cramer's Rule.
-The perimeter of a rectangle is 19 meters. If the width is doubled and the length is tripled, the perimeter is 50 meters. Determine the original width and length. 
A) width $= 3.5 \mathrm {~m}$, length $= 6 \mathrm {~m}$
B) width $= 8 \mathrm {~m}$, length $= 17 \mathrm {~m}$
C) width $= 6 \mathrm {~m}$, length $= 10 \mathrm {~m}$
D) width $= 7 \mathrm {~m}$, length $= 12 \mathrm {~m}$

Quizplus · Accepted Answer

The original equations based on the problem are $2w + 2l = 19$ and $2(2w) + 2(3l) = 50$. Solving these equations using Cramer's Rule gives the width $= 3.5 \mathrm {~m}$ and length $= 6 \mathrm {~m}$.

Solve Using Cramer's Rule =3.5 m= 3.5 \mathrm {~m}=3.5 m , Length =6 m= 6 \mathrm {~m}=6 m

Solve Using Cramer's Rule $= 3.5 \mathrm {~m}$ , Length $= 6 \mathrm {~m}$