One number is 5 more than another number. Four times the larger number plus 3 times the smaller is 97. find the numbers. Show your work.

So you need to make two equations, and then use substitution.

I used A for the larger number and B for the smaller:

A=B+5
4A+3B=97

Then you plug in (B+5) for A in the second equation...

4(B+5)+3B=97
4B+20+3B=97
7B+20=97
7B=77
B=11

A=5+11
A=16

To solve this problem, we can use a system of equations. Let's assume that the smaller number is represented by 'x', and the larger number is represented by 'y'.

From the problem, we know that one number is 5 more than the other. So, we can write the first equation as:

y = x + 5 ----(Equation 1)

The problem also states that four times the larger number plus three times the smaller number is 97. We can express this as the second equation:

4y + 3x = 97 ----(Equation 2)

Now, we have a system of two equations (Equation 1 and Equation 2). To find the values of x and y, we can solve this system.

Let's solve this system using the method of substitution:

First, substitute the value of y from Equation 1 into Equation 2:

4(x + 5) + 3x = 97 (replace 'y' with 'x + 5')

Now, simplify and solve for x:

4x + 20 + 3x = 97 (distribute 4 to x + 5)

7x + 20 = 97 (combine like terms)

7x = 97 - 20 (subtract 20 from both sides)

7x = 77 (simplify the right side)

x = 11 (divide both sides by 7)

Now that we have found the value of x, substitute it back into Equation 1 to find y:

y = x + 5

y = 11 + 5

y = 16

Therefore, the smaller number is 11, and the larger number is 16.