Find 2 numbers that the sum of 1/3 of the first plus 1/5 of the second will be equal to 13 and that if you multiply the first by 5 and the second by 7 you get 247 as the sum of the two products with replacement solution



Answer to a math question Find 2 numbers that the sum of 1/3 of the first plus 1/5 of the second will be equal to 13 and that if you multiply the first by 5 and the second by 7 you get 247 as the sum of the two products with replacement solution

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Let's set up a system of equations to find the two numbers. Let the first number be "x," and the second number be "y." From the first statement, we have: (1/3)x + (1/5)y = 13 From the second statement, we have: 5x + 7y = 247 Now, we can solve this system of equations. First, we can eliminate fractions by multiplying both sides of the first equation by the least common multiple (LCM) of 3 and 5, which is 15: (15 * (1/3))x + (15 * (1/5))y = 15 * 13 5x + 3y = 195 Now, we have the system of equations: 5x + 3y = 195 5x + 7y = 247 Subtract the first equation from the second equation to eliminate "x": (5x + 7y) - (5x + 3y) = 247 - 195 (5x - 5x) + (7y - 3y) = 52 4y = 52 Now, solve for "y": y = 52 / 4 y = 13 Now that we have the value of "y," we can substitute it back into one of the original equations. Let's use the second equation: 5x + 7y = 247 5x + 7(13) = 247 5x + 91 = 247 Subtract 91 from both sides: 5x = 247 - 91 5x = 156 Now, solve for "x": x = 156 / 5 x = 31.2 So, the two numbers that satisfy the conditions are x = 31.2 and y = 13.

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