the sum of the squares of two numbers is 34.the first number is one less than twice the second number find the numbers
Answer (2)
Answer:Here's how to solve this problem: 1. Set up the equations: - Let 'x' be the first number.- Let 'y' be the second number. We are given the following information: - "The sum of the squares of two numbers is 34": x² + y² = 34- "The first number is one less than twice the second number": x = 2y - 1 2. Substitute to solve for one variable: Since we have an expression for 'x' (x = 2y - 1), we can substitute it into the first equation: (2y - 1)² + y² = 34 3. Expand and simplify: - 4y² - 4y + 1 + y² = 34- 5y² - 4y - 33 = 0 4. Solve the quadratic equation: This quadratic equation can be factored: - (5y + 11)(y - 3) = 0 This gives us two possible solutions for 'y': - y = -11/5- y = 3 5. Find the corresponding values of 'x': - If y = -11/5: x = 2(-11/5) - 1 = -27/5- If y = 3: x = 2(3) - 1 = 5 6. The solutions: Therefore, the two possible pairs of numbers are: - x = -27/5, y = -11/5- x = 5, y = 3
Answer:Let's denote the two numbers as and . According to the given information: 1. The sum of the squares of the two numbers is 34: 2. The first number is one less than twice the second number: We can now solve this system of equations to find the values of and . Substitute the value of from the second equation into the first equation: Now, we can solve this quadratic equation to find the values of and then use that value to find . Let's proceed with solving this equation:The solutions to the quadratic equation are and . Now, we need to find the corresponding values of using the equation . For : For : Therefore, the two numbers are: 1. and 2. and
