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The larger of two numbers is twice the smaller increased by five. Find the two numbers if three times the larger exceeds double the smaller by 31

User Angordeyev
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1 Answer

5 votes

The larger number is 13 and smaller number is 4

Solution:

Let the larger number be "a"

Let the smaller number be "b"

Given that larger of two numbers is twice the smaller increased by five

larger number = twice the smaller increased by five

larger number = twice the smaller number + 5

a = 2b + 5 --- eqn 1

Also given that three times the larger exceeds double the smaller by 31

Here "exceeds" denotes addition and "times" denotes multiplication

"double" denotes being multiplied by 2

three times the larger number = double the smaller number + 31

3a = 2b + 31 ---- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "a" and "b"

Substitute eqn 1 in eqn 2

3(2b + 5) = 2b + 31

6b + 15 = 2b + 31

6b - 2b = 31 - 15

4b = 16

b = 4

From eqn 1,

a = 2b + 5

a = 2(4) + 5

a = 8 + 5

a = 13

Thus the larger number is 13 and smaller number is 4

User Kenny Wyland
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