SOLUTION: Please help show me where i am going wrong; On the two days before a swim meet, jesse swan a total of 2050 meters. The second day she swam 250 meeters more than the first day. Ho

Question 100197This question is from textbook middle grades math
: Please help show me where i am going wrong; On the two days before a swim meet, jesse swan a total of 2050 meters. The second day she swam 250 meeters more than the first day. How many meters did she swim on each day?
m+250=2050 m=1800
1800-250=500
500+250=750 1st day
2050-750=1600 2nd day
I know this to be wrong, book says 900 and 1150 This question is from textbook middle grades math

Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
On the two days before a swim meet, jesse swan a total of 2050 meters. The second day she swam 250 meeters more than the first day. How many meters did she swim on each day?
----------------
1st day she swam "x" miles
2nd day she sham "x+250" miles
EQUATION:
x + x+250 = 2050
2x = 1800
x = 900 (distance she swam on the 1st day)
x+250 = 1150 ( distance she swam on the 2nd day)
================
Cheers,
Stan H.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
On the first day Jesse swam fewer meters than the second day, so let's call that x. The total is 2050, so on the second day Jesse swam 2050-x meters.
We also know that on the second day Jesse swam 250 meters more than the first day.
Recall that the first day was x and that the second day is 2050-x. So to make them equal, you have to add 250 meters to the first day, so it becomes x+250.
x + 250 = 2050 - x
Adding an x to both sides of the equation, we obtain:
2x + 250 = 2050
Subtracting 250 from both sides, we obtain:
2x = 1800
Dividing both sides of the equation by 2:
x = 900
So, Jesse swam 900 meters on the first day.
Adding 250 we get 1150, which is the number of meters swam on the second day.
Checking these answers, we find that 900 + 1150 = 2050, which is the correct total.
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