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.