Question

find the two numbers such that the sum of twice the first and thrice the second is 92 and 4 times the first exceeds 7 times the second
by 2

Answer 1

Let the numbers be a and b

According to first condition,

2a + 3b = 92

=> 4a + 6b = 184 --------(1)

According to second condition,

4a - 7b = 2 -----------(2)


On subtracting equation 2 from 1, we get

4a + 6b - 4a + 7b = 184 - 2

=> 13b = 182

=> b = 14

Now,

On Substituting the value of b in equation 1, we get

4a + 6×14 = 184

=> 4a + 84 = 184

=> 4a = 100

=> a = 25


First number = 25

Second number = 14

Answer 2

Let the numbers be a and b

According to first condition,

2a + 3b = 92

=> 4a + 6b = 184 --------(1)

According to second condition,

4a - 7b = 2 -----------(2)

On subtracting equation 2 from 1, we get

4a + 6b - 4a + 7b = 184 - 2

=> 13b = 182

=> b = 14

Now,

On Substituting the value of b in equation 1, we get

4a + 6×14 = 184

=> 4a + 84 = 184

=> 4a = 100

=> a = 25

First number = 25

Second number = 14

hope it helped ❤️