Question

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

Answer 1

Answer:

Larger number is 13 and Smaller number is 4.

Step-by-step 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;

$x=2y+5$ ⇒equation 1

Also Given:

Three times the larger exceeds double the smaller by 31

$3x=2y+31$ ⇒equation 2

Now Substituting the value of x from equation 1 in equation 2 we get;

$3(2y+5) = 2y+31$

Now Using Distributive property we get;

$6y+15 = 2y +31$

Combining the like terms we get;

$6y-2y = 31-15$

Now Using Subtraction property we get;

$4y =16$

Now Using Division property we get;

$\frac{4y}{4}=\frac{16}{4}\\\\y =4$

Now Substituting value of y in equation 1 we will find the value of x.

$x=2y+5\\\\x = 2\times4+5\\\\x=8+5\\\\x=13$

Hence Larger number is 13 and Smaller number is 4.