Question

Four times a number added to 3 times a larger number is 31. Seven subtracted from the larger number is equal to twice the smaller number. Let x represent the smaller number and y represent the larger number. Which equations represent this situation?
A y = negative four-thirds x + 31. y = 2 x + 7.
B y = negative four-thirds x + StartFraction 31 Over 3 EndFraction. Y = 2 x + 7.
C y = negative four-thirds x + 31. y = negative 2 x + 7.
D y = negative four-thirds x + StartFraction 31 Over 3 EndFraction. Y = negative 2 x + 7.

Answer 1

Required Answer:-

GiveN:-

• Considering x the smaller number and y the larger number.
• Four times a number added to 3 times a larger number is 31.
• Seven subtracted from the larger number is equal to twice the smaller number$.$

To FinD:-

According to Condition (1):

• 4x + 3y = 31

Then,

➝ 3y = 31 - 4x

➝ y = 31 - 4x / 3

➝ y = -4x / 3 + 31 / 3

According to Condition (2):

• y - 7 = 2x

Then,

➝ y = 2x + 7

Hence:-

The option that matches is

• y = Negative four-thirds x + fraction 31 over 3 (which means -4x / 3 + 31 / 3)
• and y = 2x + 7

The correct option is (Option B)

Answer 2

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GIVEN :

• Let the smaller number be x and larger number be y.
• Four times a number added to 3 times a larger number is 31.
• 7 subtracted from the larger number is equal to twice the smaller number.

TO FIND :

• According to the 1st condition,

$\leadsto \sf 4x \: + \: 3y \: = \: 31$

Now,

$\longrightarrow \sf 3y \: = \: 31 \: - \: 4x$

$\longrightarrow \sf y \: = \: 31 \: - \: \dfrac {4x}{3}$

$\longrightarrow \sf y \: = \: \dfrac {-4x}{3} \: + \: \dfrac {31}{3}$

• According to the 2nd condition,

$\leadsto \sf y \: = \: 2x \: + \: 7$

Therefore,

The correct answer is ( option B ).