Question

write and solve a system of equations to solve each problem. explain why you chose the method you used. 13. geometry the length of a rectangle is 3 times the width. the perimeter is 44 cm. what are the dimensions of the rectangle? 14. farming a farmer grows only pumpkins and corn on her 420 - acre farm. this year she wants to plant 250 more acres of corn than pumpkins. how many acres of each crop should the farmer plant? 15. coins you have a total of 21 coins, all nickels and dimes. the total value is $1.70. how many nickels and how many dimes do you have? 16. business suppose you start an ice cream business.

Explanation:

Problem 13

Step1: Define variables

Let $w$ = width (cm), $l$ = length (cm)

Step2: Set up equations

Length is 3× width: $l = 3w$
Perimeter formula: $2l + 2w = 44$

Step3: Substitute $l$ into perimeter

Substitute $l=3w$: $2(3w) + 2w = 44$

Step4: Solve for $w$

Simplify: $6w + 2w = 44 \implies 8w = 44 \implies w = \frac{44}{8} = 5.5$

Step5: Solve for $l$

$l = 3(5.5) = 16.5$
Chosen substitution method: Simple linear relation between variables makes substitution fast.

Step1: Define variables

Let $p$ = pumpkin acres, $c$ = corn acres

Step2: Set up equations

Total farm: $p + c = 420$
Corn = pumpkins +250: $c = p + 250$

Step3: Substitute $c$ into total

Substitute $c=p+250$: $p + (p + 250) = 420$

Step4: Solve for $p$

Simplify: $2p + 250 = 420 \implies 2p = 170 \implies p = 85$

Step5: Solve for $c$

$c = 85 + 250 = 335$
Chosen substitution method: Direct relation between $c$ and $p$ simplifies substitution.

Step1: Define variables

Let $n$ = number of nickels, $d$ = number of dimes

Step2: Set up equations

Total coins: $n + d = 21$
Total value (in cents): $5n + 10d = 170$

Step3: Simplify value equation

Divide by 5: $n + 2d = 34$

Step4: Subtract coin equation

$(n + 2d) - (n + d) = 34 - 21 \implies d = 13$

Step5: Solve for $n$

$n = 21 - 13 = 8$
Chosen elimination method: Simplified value equation makes elimination efficient.

Answer:

Width = 5.5 cm, Length = 16.5 cm

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Problem 14