The sum of two numbers is 13. Twice the larger increased by three times the smaller is 31. Find the numbers.

Answer 1

Let's set up , let's choose A = Larger #

B = Smaller #

1st Step) Set up the equation, this is a system of equations type of problem.

A + B = 13 (Sum means add)

2A + 3B = 31 (Twice the larger (A) means double, increase means add, is means equal)

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2nd step) Let's switch the order of the equations so we can solve it easier first.

2A + 3B = 31

1A + B = 13

3rd step) Now you want to get rid of one of the variable by multiplication, find the LCM of any of the coefficient of the A variable or B. Let's get rid of the A first. The LCM of "2" and 1" in front of the Variable A would be 2. So to get there, we multiply the the whole 2nd equation by 2 in order to get 1A becomes 2A . We don't touch the first equation (2A + 3B = 31) since the co-efficient there is already 2.

2 (1A + B) = 13 (2) ( Note: whatever you do on the left side you do on the right side)

2A + 2B = 26 (Distributive Property)

4th Step) Put both equations together and subtract to get rid of the A's. We want to find out what one variable is equal to first, before we use its answer to find the 2nd variable.

2A + 3B = 31

- - -

2A +2B = 26

5th) 1B = 5 (After subtract we get B = 5)

6th) Look at the original equation, one of them is A + B = 13 , we just found B = 5 so whenever we see B we're going to plug in 5 for B.

A + B = 13

so A + 5 = 13

7th) Solve for A by subtract 5 on both sides to get A by itself.

A + 5 = 13

-5 -5

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A = 8

8) That's the answer, A (the larger #) = 8 , and B (the smaller # ) = 5

Answer 2

Larger number = 8

Smaller number = 5

Let x = the larger number

Let y = the smaller number

We are given two equations and two unknowns

Equation 1

x + y = 13

Equation 2

2x + 3y = 31

We can use elimination or substitution. I will multiply Equation 1 by negative 2 and use elimination below.

-2(x + y = 13)

-2x - 2y = -26

Now we can eliminate x when we combine the equation above with Equation 2

-2x - 2y = -26

2x + 3y = 31

Leaves

y = 5

We can use this to solve for x in the original Equation 1

x + y = 13

x + 5 = 13

Subtract 5 from both sides of the equation

x = 8

Checking in both equations

x + y = 13

8 + 5 = 13

2x + 3y = 31

2(8) + 3(5) = 31

16 + 15 = 31

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