SOLUTION: I just wanted to make sure that I solved these correctly
5/8X + 1/16X= 5/16 + X The solution is X= 1/32
also,
Use the distributive property to help you solve the equatio
Algebra.Com
Question 206626: I just wanted to make sure that I solved these correctly
5/8X + 1/16X= 5/16 + X The solution is X= 1/32
also,
Use the distributive property to help you solve the equation
3(w-5)=6 W=___ ( I came up with W= 7)
Solve using the multiplication principle.
11X = -99 the solutionis X=-9 (Is this correct)
Thank you
Found 2 solutions by jim_thompson5910, Theo:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1
I'm assuming that the equation is right? If so, then...
Start with the given equation.
Multiply both sides by the LCD to clear any fractions.
Distribute and multiply.
Combine like terms on the left side.
Subtract from both sides.
Combine like terms on the left side.
Divide both sides by to isolate .
Reduce.
----------------------------------------------------------------------
Answer:
So the solution is
# 2
Start with the given equation.
Distribute.
Add to both sides.
Combine like terms on the right side.
Divide both sides by to isolate .
Reduce.
----------------------------------------------------------------------
Answer:
So the solution is
So you are correct. Good Job.
# 3
Start with the given equation.
Divide both sides by to isolate .
Reduce.
----------------------------------------------------------------------
Answer:
So the solution is
Once again, you're correct. Congrats.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
PROBLEM 1
-----
5/8X + 1/16X= 5/16 + X The solution is X= 1/32
-----
if i understand the problem correctly, your equation is:
-----
i could not verify your answer as correct.
-----
i would have solved as follows:
-----
multiply both sides of the equation by 16x to get:
which becomes:
which becomes:
-----
i used the quadratic formula to get a solution of:
x = -1 or x = .6875
-----
substituting in the original equation of should confirm the answer as correct.
-----
if we let x = -1, then becomes
-----
This becomes -(5/8) - (1/16) = (5/16) - 1
which becomes -11/16 = -11/16 confirming x = -1 is correct.
-----
if we let x = .6875, then becomes
-----
This becomes .90909090909... + .0909090909091... = .3125 + .6875
which becomes 1 = 1 confirming x = .6875 is also correct.
-----
a graph of this equation would look like this:
-----
an alternative interpretation of this problem would be:
-----
i would have solved as follows:
multiply both sides of this equation by 16 to get:
which becomes:
11x = 5 + 16x
subtract 11x from both sides and subtract 5 from both sides to get:
5x = -5
divide both sides by 5 to get:
x = -1
-----
substitute in the original equation of 11x = 5 + 16x gets -11 = -11 confirming the answer is correct.
-----
PROBLEM 2
-----
3(w-5)=6 W=___ ( I came up with W= 7)
-----
this looks good.
substituting in your original equation we get:
3*(7-5) = 6
which becomes 3*2 = 6 which becomes 6 = 6 which means w = 7 is good.
-----
PROBLEM 3
-----
11X = -99 the solution is X=-9 (Is this correct)
-----
this also looks good.
substituting in your original equation we get:
11*(-9) = -99 which becomes -99 = -99 which means x = -9 is good.
-----
RELATED QUESTIONS
I think I solved this correctly, but just wanted to check:
9y+3x=18; -3y-x=-6... (answered by Alan3354,rothauserc)
With the equation {{{ x^2/16 + y^2/4 = 1 }}}, solve for y. I got that {{{ y=2-x/2 }}}, I... (answered by solver91311)
Ok, I just don't get this one at all would I have to find the LCM first? Or would I... (answered by solver91311)
please help me solve: 5/8x+1/16x=9/16+x
the solution is x=?
*these are fractions is the (answered by rfer)
SOLVE.
5/8x + 1/16x = 3/16 + x
(these are all fractions)
HELP, how do I do this?... (answered by stanbon)
Tutor,
I was wondering if I solved these problems correctly would you be able to help (answered by bucky)
Good Evening tutor,
Could you tell me if I have these correct?
Solve by... (answered by checkley75)
Hello, I have to use 'completing the square' to put this equation into standard form,... (answered by ewatrrr)