Question

a number exceeds another number by3 and their product is40​

Answer 1

Let the smaller number be x.

Let the smaller number be x.The bigger number will thus be x + 3.

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0(x + 11)(x - 8) = 0

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0(x + 11)(x - 8) = 0x + 11 = 0 or x - 8 = 0

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0(x + 11)(x - 8) = 0x + 11 = 0 or x - 8 = 0x = -11 or x = 8

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0(x + 11)(x - 8) = 0x + 11 = 0 or x - 8 = 0x = -11 or x = 8Since the numbers are positive, we reject x = -11.

Let the smaller number be x.The bigger number will thus be x + 3.The product of the two numbers is 88, so we have:x(x + 3) = 88x^2 + 3x = 88x^2 + 3x - 88 = 0(x + 11)(x - 8) = 0x + 11 = 0 or x - 8 = 0x = -11 or x = 8Since the numbers are positive, we reject x = -11.Hence, the answer is x = 8.

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