SOLUTION: 1)If 3 is subtracted from the numerator of a certain fraction,the value of the fraction becomes 3/5.If 1 is subtracted from the denominator of the same fraction,it becomes 2/3.Find
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Question 122228: 1)If 3 is subtracted from the numerator of a certain fraction,the value of the fraction becomes 3/5.If 1 is subtracted from the denominator of the same fraction,it becomes 2/3.Find the original fraction.
2)Find the product of 2 numbers such that twice the first added to the second equals 19 and three times the first is 21 more than the second.
3)The second of the four numbers is three less than the first,the third is four more than the first and the fourth is two more than the third.Find the fourth number if the sum is 35.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1)If 3 is subtracted from the numerator of a certain fraction,the value of the fraction becomes 3/5.If 1 is subtracted from the denominator of the same fraction,it becomes 2/3.Find the original fraction.
Let the "certain fraction" =
If 3 is subtracted from the numerator, the value is 3/5.
=
Cross multiply:
5(x-3) = 3y
5x - 15 = 3y
5x - 3y = 15; eq1
:
If 1 is subtracted from the denominator of the same fraction,it becomes 2/3.
=
Cross multiply:
3x = 2(y-1)
3x = 2y - 2
3x - 2y = -2;eq2
:
We're going to use elimination: mult eq1 by 2 and eq2 by 3:
10x - 6y = 30
9x - 6y = -6
------------------subtracting eliminates y; find x
1x = 36 is the numerator
:
Find y using: 5x - 3y = 15
5(36) - 3y = 15
180 - 3y = 15
-3y = 15 - 180
-3y = -165
y =
y = 55 is the denominator
is our certain fraction
:
You can check the solutions by substitution
:
:
2)Find the product of 2 numbers such that twice the first added to the second equals 19 and three times the first is 21 more than the second.
Let the two numbers = x & y; find the two numbers, then worry about the product
:
twice the first added to the second equals 19
2x + y = 19
:
3 times the first is 21 more than the second
3x = y + 21
3x - y = 21
:
add the two equations
2x + y = 19
3x - y = 21
-------------adding eliminates y, find x
5x = 40
x =
x = 8
:
Find y using 2x + y = 19
2(8) + y = 19
y = 19 - 16
y = 3
Obviously the product is 8 * 3 = 24
:
Check solutions in 3x - y = 21
;
:
3)The second of the four numbers is three less than the first,the third is four more than the first and the fourth is two more than the third.Find the fourth number if the sum is 35.
:
Let our 4 numbers be = to 35 as it says.
w + x + y + z = 35
:
The second of the four numbers is 3 less than the first:
x = (w-3)
:
the third is 4 more than the first
y = (w+4)
:
the fourth is two more than the third
z = (y+2)
we want all the variables in terms of w, substitute (w+4) for y
z = (w+4) + 2
z = (w+6)
:
Substitute for x, y & z in the total equation
w + (w-3) + (w+4) + (w+6) = 35
4w - 3 + 4 + 6 = 35
4w + 7 = 35
4w = 35 - 7
w =
w = 7
:
The want the 4th number (z)
z = 7 + 6
z = 13
:
Check solutions by substitution, see if I'm right
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