Determine two positive integers such as 5 less than two times the first number is the second number.
The sum of the second number and the square of the first is 115.
s = 2 f - 5
s + f^2 = 115
f^2 + 2 f - 5 = 115
f^2 + 2 f - 120 = 0
(f + 12)(f - 10) = 0
f - 10 = 0 ... f = 10
a = first number
b = second number
b = 2 a - 5
b + a ^ 2 = 115
2 a - 5 + a ^ 2 = 115
- 5 + a ^ 2 + 2 a = 115 Subtract 115 to both sides
- 5 + a ^ 2 + 2 a - 115 = 115 - 115
- 5 + a ^ 2 + 2 a - 115 = 0
Try to slove this equation.
The solutions are :
a = - 12 and a = 10
- 12 isn't positive integer so a = 10
b = b = 2 a - 5 = 2 * 10 - 5 = 20 -5 = 15
To solve this problem, we need to set up a system of equations based on the given information.
Let's assume the first number is x, and the second number is y.
According to the first statement, "5 less than two times the first number is the second number," we can write the equation:
2x - 5 = y (Equation 1)
According to the second statement, "the sum of the second number and the square of the first is 115," we can write the equation:
y + x^2 = 115 (Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) that we can solve simultaneously to find the values of x and y.
To solve the system, we can use substitution or elimination method. In this case, let's solve it using the substitution method.
Step 1: Solve Equation 1 for y in terms of x:
y = 2x - 5
Step 2: Substitute this value of y in Equation 2:
2x - 5 + x^2 = 115
Rearranging the equation:
x^2 + 2x - 120 = 0
Step 3: Factor or use the quadratic formula to solve for x:
(x + 12)(x - 10) = 0
This gives us two possible solutions for x: x = -12 or x = 10.
However, we are looking for positive integers, so we discard the negative value.
Therefore, x = 10.
Step 4: Substitute the value of x back into Equation 1 to find y:
2(10) - 5 = y
y = 15
So the two positive integers that satisfy both conditions are x = 10 and y = 15.