The sum of two numbers is 15. Three times the second number minus four times the first number equals 4. What are the missing variables? Please show your work.

x+y = 15

3y-4x = 4

now just crank it out.

TO BE WORKED

To solve this problem, we will use a system of equations. Let's assume the two numbers are x and y.

According to the problem, the sum of the two numbers is 15. This can be represented as:

Equation 1: x + y = 15

The second condition states that three times the second number minus four times the first number equals 4. This can be represented as:

Equation 2: 3y - 4x = 4

We now have a system of two equations with two variables. To solve it, we can use either substitution or elimination. Let's use the substitution method.

First, solve Equation 1 for x:

From Equation 1: x = 15 - y

Now substitute this value of x into Equation 2:

3y - 4(15 - y) = 4

Simplify the equation:

3y - 60 + 4y = 4

Combine like terms:

7y - 60 = 4

Add 60 to both sides:

7y = 64

Divide both sides by 7:

y = 64/7

So the second number, y, is equal to 64/7 or approximately 9.14.

We can now substitute the value of y back into Equation 1 to find x:

x + 9.14 = 15

Subtract 9.14 from both sides:

x = 15 - 9.14

x ≈ 5.86

Therefore, the missing variables are x ≈ 5.86 and y ≈ 9.14.