Math word problem!!!! help?
2008-10-07 1:58 pm
The difference of two numbers is 33. If the bigger number is one less than thrice the smaller number, what is the sum of the two numbers?
回答 (6)
2008-10-07 2:04 pm
✔ 最佳答案
Let x = smaller number3x - 1 = bigger number
-----
*(3x - 1) - x = 33
*2x = 33 + 1
*2x/2 = 34/2
*x = 17 (smaller number)
To get the bigger number:
3 (17) - 1 = 50 (bigger number)
The sum of the two numbers:
ANSWER: 17 + 50 = 67 (sum of the two numbers)
Checking:
50 - 17 = 33 (Correct!)
2008-10-07 2:40 pm
x - y = 33 (solve by using substitution)
x = 3y - 1
x - y = 33
(3y - 1) - y = 33
3y - 1 - y = 33
3y - y = 33 + 1
2y = 34
y = 34/2
y = 17
x - y = 33
x - 17 = 33
x = 33 + 17
x = 50
â´ x = 50 , y = 17
x = 3y - 1
x - y = 33
(3y - 1) - y = 33
3y - 1 - y = 33
3y - y = 33 + 1
2y = 34
y = 34/2
y = 17
x - y = 33
x - 17 = 33
x = 33 + 17
x = 50
â´ x = 50 , y = 17
2008-10-07 2:19 pm
let no. be x,y
acc. to ques. we have
x-y=33
second case:
x=3y-1
x-3y= -1
solving simultaneouly by eliminating method we have
x-y=33 ---------------------- (1)
x-3y= -1 --------------------------------- (2)
(subtracting both we get)
2y=34
y=17
putting the value of "y" in equation 1 we get
x-y=33
x=33+y
x=33+17
x=50
ans. x=50,y=17
acc. to ques. we have
x-y=33
second case:
x=3y-1
x-3y= -1
solving simultaneouly by eliminating method we have
x-y=33 ---------------------- (1)
x-3y= -1 --------------------------------- (2)
(subtracting both we get)
2y=34
y=17
putting the value of "y" in equation 1 we get
x-y=33
x=33+y
x=33+17
x=50
ans. x=50,y=17
參考: myself
2008-10-07 2:08 pm
Let
bigger number = a,
smaller number = b
Difference of numbers is 33.
Therefore,
a - b = 33--------------------(1)
Bigger number is one less than thrice the smaller number.
Therefore,
a = 3b - 1--------------------(2)
Using the value of a from equation (2) in equation (1),
3b - 1 - b = 33
Or 2b - 1 = 33
Or 2b = 33 + 1
Or 2b = 34
Or b = 34/2
Or b = 17
Putting this value of b in (2),
a = 3*17 - 1 = 51 - 1 = 50
Sum of numbers = a + b = 50 + 17 = 67
Ans: 67
bigger number = a,
smaller number = b
Difference of numbers is 33.
Therefore,
a - b = 33--------------------(1)
Bigger number is one less than thrice the smaller number.
Therefore,
a = 3b - 1--------------------(2)
Using the value of a from equation (2) in equation (1),
3b - 1 - b = 33
Or 2b - 1 = 33
Or 2b = 33 + 1
Or 2b = 34
Or b = 34/2
Or b = 17
Putting this value of b in (2),
a = 3*17 - 1 = 51 - 1 = 50
Sum of numbers = a + b = 50 + 17 = 67
Ans: 67
2008-10-07 2:07 pm
x - y = 33
and
x = 3y -1
now substitute the 2nd equation into the first:
3y -1 - y = 33
2y = 34
y =17,
and since x - y = 33,
x - 17 = 33
x = 50
and
x = 3y -1
now substitute the 2nd equation into the first:
3y -1 - y = 33
2y = 34
y =17,
and since x - y = 33,
x - 17 = 33
x = 50
2008-10-07 2:06 pm
let x = the larger #
let y = the smaller #
"The difference of two numbers is 33"
x - y = 33
x = 33+y [eq 1]
"the bigger number is one less than thrice the smaller number"
x = 3y - 1 [eq 2]
substituting eq 1 into eq 2
x = 33+y [eq 1]
x = 3y - 1 [eq 2]
33+y = 3y-1
33+1 = 3y - y
34 = 2y
y = 17
substitute this into eq 1:
x = 33+y [eq 1]
x = 33+17
x = 50
The sum of the two is 17+50 = 67
.
let y = the smaller #
"The difference of two numbers is 33"
x - y = 33
x = 33+y [eq 1]
"the bigger number is one less than thrice the smaller number"
x = 3y - 1 [eq 2]
substituting eq 1 into eq 2
x = 33+y [eq 1]
x = 3y - 1 [eq 2]
33+y = 3y-1
33+1 = 3y - y
34 = 2y
y = 17
substitute this into eq 1:
x = 33+y [eq 1]
x = 33+17
x = 50
The sum of the two is 17+50 = 67
.
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