Question 1037961: The product of two consecutive even integers is four more than twice their sum. Find the integers (will have two sets of answers). Set up and solve using an algebraic equation.
Thank you so much
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Same problem, different numbers.
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Polynomials-and-rational-expressions/1037960: One number is one less than twice the other. Their product is 120.
Find the numbers (should have two sets of answers.) Set up using an Algebraic Equation and use a complete sentence.
Am I on the right path?
x= 2x+1
y= the other number
y=2x-1
(y)(x)=120
y(2x-1)=120
2x^2-x=120
(2x+15)(x-8)=0
x=8
y=15
The two numbers that are one numberless than twice the other are 8 and 15.
Thank you
1 solutions
Answer 652635 by Alan3354(51277) About Me on 2016-06-12 17:07:37 (Show Source):
You can put this solution on YOUR website!
One number is one less than twice the other. Their product is 120.
Find the numbers (should have two sets of answers.) Set up using an Algebraic Equation and use a complete sentence.
Am I on the right path?
x= 2x+1 ******************** That's not right.
y= the other number
y=2x-1
(y)(x)=120
y(2x-1)=120
2x^2-x=120
(2x+15)(x-8)=0
x=8
y=15
The two numbers that are one numberless than twice the other are 8 and 15.
Thank you
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Check your answers.
15 is 2*8 - 1, so that's OK.
The product is 120. --> 8 & 15 is a solution.
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do it like this:
y = 2x - 1
x*y = 120
Sub for y
x*(2x - 1) = 120
2x^2 - x - 120 = 0
(x - 8)*(2x + 15) = 0
x = 8, y = 15
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x = -7.5, y = -14
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-14 = 2*(-7.5) - 1 (the 2nd solution)
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