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Question 817230: I really have trouble with word problems! I'd love some help. Thanks! :)
One number is 3 less than twice another. If their product is 405, find the numbers.
I tried: x=3-2x, 3-2x=x ??
I'm just not sure how to do this.
Found 3 solutions by mananth, richwmiller, rothauserc:
Answer by mananth(16946)   (Show Source):
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! you have two numbers
one number is x
is 3 less than -3
twice another
and another is y
2y twice another
x=2y-3
their product means multiply them
xy=405
now we have
x=2y-3
xy=405
x=405/y
both equations =x so they equal each other
2y-3=405/y
2y^2-3y=405
2y^2-3y-405=0
find factors of -810 (2*-405) that add up to -3
2y^2-
(y-15) (2 y+27) = 0
y = 15 y = -27/2
x=27
| Solved by pluggable solver: Factoring using the AC method (Factor by Grouping) |
Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last term is  .
Now multiply the first coefficient by the last term to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient ?
To find these two numbers, we need to list all of the factors of (the previous product).
Factors of :
1,2,3,5,6,9,10,15,18,27,30,45,54,81,90,135,162,270,405,810
-1,-2,-3,-5,-6,-9,-10,-15,-18,-27,-30,-45,-54,-81,-90,-135,-162,-270,-405,-810
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*(-810) = -810
2*(-405) = -810
3*(-270) = -810
5*(-162) = -810
6*(-135) = -810
9*(-90) = -810
10*(-81) = -810
15*(-54) = -810
18*(-45) = -810
27*(-30) = -810
(-1)*(810) = -810
(-2)*(405) = -810
(-3)*(270) = -810
(-5)*(162) = -810
(-6)*(135) = -810
(-9)*(90) = -810
(-10)*(81) = -810
(-15)*(54) = -810
(-18)*(45) = -810
(-27)*(30) = -810
Now let's add up each pair of factors to see if one pair adds to the middle coefficient :
| First Number | Second Number | Sum | | 1 | -810 | 1+(-810)=-809 | | 2 | -405 | 2+(-405)=-403 | | 3 | -270 | 3+(-270)=-267 | | 5 | -162 | 5+(-162)=-157 | | 6 | -135 | 6+(-135)=-129 | | 9 | -90 | 9+(-90)=-81 | | 10 | -81 | 10+(-81)=-71 | | 15 | -54 | 15+(-54)=-39 | | 18 | -45 | 18+(-45)=-27 | | 27 | -30 | 27+(-30)=-3 | | -1 | 810 | -1+810=809 | | -2 | 405 | -2+405=403 | | -3 | 270 | -3+270=267 | | -5 | 162 | -5+162=157 | | -6 | 135 | -6+135=129 | | -9 | 90 | -9+90=81 | | -10 | 81 | -10+81=71 | | -15 | 54 | -15+54=39 | | -18 | 45 | -18+45=27 | | -27 | 30 | -27+30=3 |
From the table, we can see that the two numbers  and  add to (the middle coefficient).
So the two numbers and both multiply to and add to
Now replace the middle term  with  . Remember, and add to . So this shows us that  .
 Replace the second term with .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out  from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
 Combine like terms. Or factor out the common term 
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Answer:
So  factors to .
In other words,  .
Note: you can check the answer by expanding to get or by graphing the original expression and the answer (the two graphs should be identical).
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Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! start with what you are given and then write the related equations
let the numbers be x and y
x = 2y - 3
xy = 405
substitute for x in second equation
(2y - 3)y = 405
2y^2 -3y = 405
put equation in standard form
2y^2 -3y -405 = 0
use quadratic formula to solve for y
y = (3 + square root(3^2 -4*2*(-405)) / 2*2
y = (3 - square root(3^2 -4*2*(-405)) / 2*2
y = 15 or -13.5
solve for x in first equation
x = 27 or -30
check the answers
15*27 = 405
(-30)*(-13.5) = 405
our answers check, therefore there are two solution sets
x = 27 and y = 15
x = -30 and y = -13.5
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