The sum of two numbers is 52. One of the numbers is 14 less than two times the other number. What is the smaller of the two numbers

Answer 1

Hi Amyah R

You can solve this several ways by keeping track of the available substitutions within the information

Given

x + y = 52

And one of the numbers is 14 less than two times the other number, you can either one just keep track. I let

y = 2x -14

Solution 1 you can obviously put everything in terms of x with the following substitution in

x + y = 52

x + (2x - 14) = 52

Solve that for x

You can also set up an equality as follows:

Since

x + y = 52

y = 52 - x

We also choose above to have

y = 2x - 14

So

2x - 14 = 52 - x

This can also be used to solve for x

You can also set up a system of equations from the information

Equation 1

x + y = 52

Equation 2 comes from

y = 2x - 14

Move the variables to one side of the equation leave the constant on the other

-2x + y = -14

x + y = 52

-2x + y = -14

If you eliminate y by multiplying Equation 2 by negative 1 you can solve for x

You should get the same value for x in all the situations above, give them a try, compare your number notice the situations check each other.

If you have any questions send me a message

By the way I got x = 22

Answer 2

First thing you need to do in a problem like this is decide what variables you would like to use. I'll help you set it up by using x and y. In this problem you will have two equations and will need to use substitution to solve it out.

First equation:

The sum (+) of two numbers (x and y) is (=) 52.

x + y = 52

Second equation:

One of the numbers (you can choose either x or y, i'll choose y) is (=) 14 less (-14) than two times the other number (2x)

y = 2x - 14

So your system would be

x + y = 14

y = 2x - 14

Since your second equation is y equals, you will substitute that equation in for y in the first equation, so the first equation would become:

x + (2x - 14) = 14

Once you solve that equation for x, you will plug your x value into the second equation to get your y value.

Hope this helped!