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.