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find two numbers such that the sum of twice the first and thrice the second is 92 ,and four times the first exceeds seven times the second by 2​

User Glutinous
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1 Answer

3 votes

Answer:

First number = 25

Second number = 14

Explanation:

Let the numbers be x & y

The sum of twice the first and thrice the second is 92

2x + 3y = 92 -------------------(I)

Four times the first exceeds seven times the second by 2​

4x - 7y = 2 --------------------(II)

Multiply equation (I) by -2 and then add the equations

(I)*-2 -4x - 6y = -184

(II) 4x - 7y = 2 { Now add, thus x will be eliminated}

- 13y = -182 {Divide both sides by -13}

y = -182/-13

y = 14

Plugin the value of y in equation (I)

2x + 3*14 = 92

2x + 42 = 92

2x = 92 - 42

2x = 50

x = 50/2

x = 25

First number = 25

Second number = 14

User Rashane
by
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