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Here's the first problem I can't solve...
(x – 7)/5 – 5/2 = (x + 9)/10

Any help would be greatly appreciated :)

2007-04-10 05:42:45 · 8 answers · asked by chaoswoman_007 2 in Science & Mathematics Mathematics

8 answers

[ (x-7)/5 - 5/2 ] * 10 = [ (x+9)/10 ] * 10
2(x-7) - 25 = x + 9
2x - 14 - 25 - x - 9 = 0
x - 48 = 0
x= 48

2007-04-10 05:46:48 · answer #1 · answered by ........ 5 · 0 0

(x-7)/5 - 5/2 = (x+9)/10 First multiply each side by 10

2(x-7) - 25 = x + 9 Clear up that parenthesis

2x - 14 - 25 = x + 9 Get x by itself on one side by adding and subtracting

x = 48

2007-04-10 12:48:41 · answer #2 · answered by ajfmf90 2 · 0 0

You need to multiply each expression you have by the LCD. In this case (which is relatively simple), you multiply by the LCD, 10. Then you will have,
2(x-7) - 5(5) = x+9 (distribute)
2x-14-25 = x+9 (combine like terms)
2x - 39 = x+ 9 (isolate expressions with variable x)
x= 30
That's your answer- 30. Good luck!

2007-04-10 12:48:03 · answer #3 · answered by kash 3 · 0 0

x should equal 48...

just multiply the ten off of the right side to the left to get 10(x-7)/5 - 10(5)/2= x+9

from there just simplify and solve for x!

hope that helps you :o)

2007-04-10 12:47:24 · answer #4 · answered by The_Economy_of_Mercy 2 · 0 0

LCM(2,5,10)=10

2(x-7)-25=x+9

2x-14-25=x+9

x=48

2007-04-10 12:47:01 · answer #5 · answered by iyiogrenci 6 · 0 0

LHS becomes [ {x - 7}/5 - 5/2]
RHS becomes {x + 9}/10

Multiply both sides by 10

LHS becomes 10 x[{x-7}/5 -5/2]
RHS becomes {x +9}

Simplify LHS [2 x {x - 7} - 25]
becomes 2x - 14 - 25
becomes 2x -39

So IF LHS = 2x - 39
and RHS = x + 9

Collect terms

x = 48

2007-04-10 12:54:54 · answer #6 · answered by Rod Mac 5 · 0 0

get a common denominator on the left hand side of the equation, after that move the right hand side to the left and make it equal to zero, then solve. 5x-7/10-25/10-(x+9)/10=0

2007-04-10 12:49:50 · answer #7 · answered by lilo 2 · 0 0

x=4445 i believe. i am only 8th grade, but i think this is the answer

2007-04-10 12:49:16 · answer #8 · answered by Anonymous · 1 0

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