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

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

From eqn 1,

a = 2b + 5

a = 2(4) + 5

a = 8 + 5

Thus the larger number is 13 and smaller number is 4

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.