SOLUTION: I have 2 questions and if someone could help i would really appreciate it!! 1) 1/7(2x-5)=3/8x-5/7 2) 5/6(x-2)=x-3 If it helps i am a college student using the textbook "

Algebra ->  Linear-equations -> SOLUTION: I have 2 questions and if someone could help i would really appreciate it!! 1) 1/7(2x-5)=3/8x-5/7 2) 5/6(x-2)=x-3 If it helps i am a college student using the textbook "      Log On


   

Question 502886: I have 2 questions and if someone could help i would really appreciate it!!
1) 1/7(2x-5)=3/8x-5/7

2) 5/6(x-2)=x-3
If it helps i am a college student using the textbook "Elementary Algebra For College Students" (8th edition) and i am learning about Solving Linear Equations and Inequalities i believe.
Thank you so much!!
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1) (1/7)(2x-5)=(3/8)x-(5/7)
Multiply both sides by 56 to get:
8(2x-5) = 7*3 - 8*5
------
16x-40 = 21-40
16x = 21
x = 21/16
---------------------
2) (5/6)(x-2)=x-3
Multiply both sides by 6 to get:
5(x-2) = 6x-18
5x-10 = 6x-18
x = 8
--------------
Cheers,
Stan H.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you started.

1)

%281%2F7%29%282x-5%29=%283%2F8%29x-5%2F7 Start with the given equation.

%281%2F7%29%282x%29%2B%281%2F7%29%28-5%29=%283%2F8%29x-5%2F7 Distribute.

%282x%29%2F7-5%2F7=%283%2F8%29x-5%2F7 Multiply.

8%282x%29-8%285%29=7%283x%29-8%285%29 Now multiply EVERY term by the LCD 56 to clear out the fractions.

16x-40=21x-40 Multiply.

16x=21x-40%2B40 Add 40 to both sides.

16x-21x=-40%2B40 Subtract 21x from both sides.

-5x=-40%2B40 Combine like terms on the left side.

-5x=0 Combine like terms on the right side.

x=%280%29%2F%28-5%29 Divide both sides by -5 to isolate x.

x=0 Reduce.

So the solution is x=0