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# The length of a rectangle is 4 cmmore than 3 times its width. If the area of the rectangle is 15 cm, which equation could be used to find the width, w?

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The length of a rectangle is 4 cm more than 3 times its width. If the area of the rectangle is 15 cm, which equation could be used to find the width, w?

Guest Feb 23, 2017

#2
+10657
+10

w = width   and length=l = (3w+4)

area = w x l

area = 15

15 = w * (3w+4)

ElectricPavlov  Feb 23, 2017
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#1
+5

Area = L x W

Let the width of the rectangle =W

Then the length is =3W + 4

15 =W x [3W + 4], solve for W

Solve for W:
15 = W (3 W + 4)

15 = W (3 W + 4) is equivalent to W (3 W + 4) = 15:
W (3 W + 4) = 15

Expand out terms of the left hand side:
3 W^2 + 4 W = 15

Divide both sides by 3:
W^2 + (4 W)/3 = 5

Add 4/9 to both sides:
W^2 + (4 W)/3 + 4/9 = 49/9

Write the left hand side as a square:
(W + 2/3)^2 = 49/9

Take the square root of both sides:
W + 2/3 = 7/3 or W + 2/3 = -7/3

Subtract 2/3 from both sides:
W = 5/3 or W + 2/3 = -7/3

Subtract 2/3 from both sides:
Answer: |W = 5/3                                  or W = -3 Discard

Guest Feb 23, 2017
#2
+10657
+10

w = width   and length=l = (3w+4)

area = w x l

area = 15

15 = w * (3w+4)

ElectricPavlov  Feb 23, 2017
#3
+7157
0

The length of a rectangle is 4 cm more than 3 times its width. If the area of the rectangle is 15 cm^2, which equation could be used to find the width, w?

$$l=3w+4cm$$

$$15cm^2=(3w+4cm)\times w$$

$$15cm^2=3w^2+4cm\times w$$

$$3w^2+4cm\times w-15cm^2=0$$      $$[w = {-b + \sqrt{b^2-4ac} \over 2a}]$$

a           b                     c

$$\large w=\frac{-4cm+\sqrt{16cm^2+180cm^2}}{6}$$

$$w=\frac{-4cm+{14cm}}{6}=\frac{10cm}{6}$$

$$\large w=1\frac{2}{3}cm$$

$$l=3w+4cm=9cm$$

!

asinus  Feb 23, 2017

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