Answer:
Larger number is 13 and Smaller number is 4.
Explanation:
Let the larger number be x.
Also let the smaller number be y.
We need to find the two numbers.
Given:
The larger of two numbers is twice the smaller increased by five.
framing in equation form we get;
⇒equation 1
Also Given:
Three times the larger exceeds double the smaller by 31
⇒equation 2
Now Substituting the value of x from equation 1 in equation 2 we get;
![3(2y+5) = 2y+31](https://img.qammunity.org/2020/formulas/mathematics/middle-school/r3dymsllobajjy0bfginzh3l0wscvbva0m.png)
Now Using Distributive property we get;
![6y+15 = 2y +31](https://img.qammunity.org/2020/formulas/mathematics/middle-school/idbfm9o5klpej0w9zx4z2gyjwpb0wpnxdw.png)
Combining the like terms we get;
![6y-2y = 31-15](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jkiv5nc0ihyp4gra6hinn0jozcxnhwv8zf.png)
Now Using Subtraction property we get;
![4y =16](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s7wjaqvw9y7v2m9g0mmadg29it3o11n0vq.png)
Now Using Division property we get;
![(4y)/(4)=(16)/(4)\\\\y =4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/65konqcr1h31wbpjalksi3hd0ks636tw7h.png)
Now Substituting value of y in equation 1 we will find the value of x.
![x=2y+5\\\\x = 2*4+5\\\\x=8+5\\\\x=13](https://img.qammunity.org/2020/formulas/mathematics/middle-school/udlki31f4j2f5f8vsxcbypm47i18gw9edm.png)
Hence Larger number is 13 and Smaller number is 4.