Questions on Word Problems: Problems with consecutive odd even integers answered by real tutors!

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Find two integers whose product is 105 such that one of the integers is one more twice the other integer.
.
Let x = one of two integers
and y = second of two integers
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Since we have two unknowns, we'll need two equations.
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Equation 1 comes from:"whose product is 105 "
xy = 105
.
Equation 2 comes from:"one of the integers is one more twice the other integer"
x = 2y + 1
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Using the definition of 'x' from equation 2, we substitute it into equation 1 a nd solve for 'y':
xy = 105
(2y + 1)y = 105
2y^2 + y = 105
2y^2 + y - 105 = 0
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Since it can't easily be factored, we can use the "quadratic equation" to solve. Doing so yields (See reference below):
y = {7, -7.5}
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If y = 7, we can find 'x' by substituting it into equation 1 and solve for 'x':
xy = 105
x(7) = 105
x = 105/7
x = 15
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If y = -7.5
xy = 105
x(-7.5) = 105
x = 105/(-7.5)
x = -14
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Conclusion -- there are two possible answers:
(x,y) = (15,7)
(x,y) = (-14,-7.5)
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For reference, here is the details of the quadratic:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation ax^2+bx+c=0

(in our case 2x^2+1x+-105 = 0

) has the following solutons:

$x[12] = (b+-sqrt( b^2-4ac ))/2\a$

For these solutions to exist, the discriminant b^2-4ac

should not be a negative number.

First, we need to compute the discriminant b^2-4ac

.

Discriminant d=841 is greater than zero. That means that there are two solutions: $x[12] = (-1+-sqrt( 841 ))/2\a$ .

$x[1] = (-(1)+sqrt( 841 ))/2\2 = 7$
$x[2] = (-(1)-sqrt( 841 ))/2\2 = -7.5$

Quadratic expression 2x^2+1x+-105

can be factored:
2x+1x+-105 = 2(x-7)*(x--7.5)

Again, the answer is: 7, -7.5. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 2*x^2+1*x+-105 )