math
posted by Abby .
could someone check my work please? I have to solve for 'x'
Questions ~
1. 3x – 19 = 2
Add ‘19’ to each side of the equation ~ 19+19+3x= 2+19
Combine like terms~ 19+19=0
0+3x= 2+19
3x= 2+19
Combine like terms~ 2+19=17
3x=17
Divide each side by‘3’~ x=5.666666667
X = 5.666666667
2. 4x  12x = 62 + 4(11)
Combine like terms ~ 4x+12x=8x
8x= 62+4(11)
Multiply ~ 4*11
8x=62+44
Combine like terms ~ 62+44=106
8x=106
Divide each side by ‘8’ X=13.25
X= 13.25
3. x/8 + 7 =  10
?? I can't figure this out
4. 4(5x – 9) = 2(3x + 7)
Add ‘6x’to each side of the equation ~ 36+20x+6x= 14+ 6x+6x
Combine like terms ~ 20x +6x=26x
36+26x=14+ 6x+6x
Combine like terms ~ 6x+6x=0
36+26x= 14+0
36+26=14
Add ‘36’ to each side of the equation ~ 36+36+26x=14+36
Combine like terms ~ 36+36=0
0+26x= 14+36
26x= 14+36
Combine like terms ~ 14+36=22
26x=22
Divide each side by ‘26’ ~ x= 0.8461538462
x= 0.8461538462
5. ½ x – 5 = (3/8)x + 2
?? I can't figure this one out
6. 3(x – 4) – (x + 5) = 2x + 11
Combine like terms ~ 12+ 5= 17
17+3x+ 1x=2x+11
Combine like terms ~ 3x+ 1x=2x
17+2x=2x+11
Add ‘2x’ to each side of the equation
17+2x+2x=11+2x+2x
Combine like terms~ 2x+ 2x=0
17+0=11+2x+2x
17=11+2x+2x
Combine like terms~ 2x+ 2x=0
17+11+0
17=11
I think I did something wrong with question #6 but I can’t figure out what it is, could someone help me figure it out?
Questions #3 and #5 I couldn’t figure out how to do them, I would really appreciate any help you could give.
Answer this Question

math 
Reiny
all your questions are correct, except the two you could not do
first of all, let's clear up that last one.
you ended up with 17 = 11
This of course is false even though you made no errors in your solution.
If that happens , and you end up with an absurd conclusion like that, simply say:
There is no solution to that equation
In general,
 if your variables drop out and you end up with a false statement, there is no solution
 if your variables drop out and you end up with a true statement, such as 17 = 17, any real value of the variable would be a solution.
Such an equation is called an identity,
e.g. 2x + 4 = 2x + 6  2
#3
x/8 + 7 = 10
x/8 = 17
multiply both sides by 8, cancelling the 8 on the left side
x = 136
#5
½ x – 5 = (3/8)x + 2
I see 4 terms, the LCM of my two denominators is 8
how about multiplying every term by 8 ?
4x  40 = 3x + 16
hey look ma, no more fractions !
4x + 3x = 16+40
7x = 56
x = 56/7 = 8
one more comment.
Unless your decimals come out exact, I would consider the corresponding fraction to be in better and proper form
e.g #1, leave it as 17/3 rather than 5.666667
and 22/26 as 11/13 instead of the repeating decimal 
math 
Abby
Alright! thank you so much! this really helps!
Respond to this Question
Similar Questions

Variable
SOLVING EQUATIONS WITH THE VARIABLE ON BOTH SIDES Solve the equation. Then check your solution. 4x9 = 7x+12 Just remember that one can add, subtract, multiply, divide, etc but whatever is done to one side must be done to the other … 
math
3(4u6)=2(4u3)(u8) what is u 12u18=8u6u+8 Can you take it from here? 
algebradistributive
how would you do this prob. negative4(5y+6) = negative 7y+3 thanks. happy halloween I'm going to assume your problem is this: 4  (5y + 6) = 7y + 3 If that's the case, distribute the negative over the terms in the parentheses first: … 
MATH PLEASE CHECK IT!!2
m2/m5*m+5/m2=1 If your problem is this: (m  2)/(m  5) * (m + 5)/(m  2) = 1 (m  2) cancels out in both the numerator and denominator on the lefthand side since you are multiplying. That leaves us with this: (m + 5)/(m  5) … 
mathpre alg
not sure i did right 3(f+2)+9=13+5f 3f+6+9 = 13+5f +9 3F+15 = 12+5F GOT MIXED UP SOMWERE HELP 3f +6 +9 = 13 + 5f 3f + 15 = 13 + 5f 2f = 2 f = 1 drwls is correct You are doing OK up to this point: 3f + 6 + 9 = 13 + 5f Now combine any … 
algebra
Can someone please help me with this one? 
Maths
you have to solve the following by finding the value of x and y. can u help please? 
math
2x+4(1x)=11+5x 2x+44x=11+5x 6x+4=11+5x 6x=7+5x x=7 i think this is it x = 7 is not correct. Here's how the problem should work out: Use the distributive property to get rid of the parentheses and you will end up with this: 2x + 4 … 
Math/Calculus
Please check my work and correct any errors/point out any errors. Thanks. Solve the initialvalue problem using the method of undetermined coefficients. y''4y=e^xcos(x), y(0)=1, y'(0)=2 r^24=0, r1=2, r2=2 yc(x)=c1*e^2x+c2*e^2x … 
math
could someone check my work please? I have to solve for 'x' Questions ~ 1. 3x – 19 = 2 Add ‘19’ to each side of the equation ~ 19+19+3x= 2+19 Combine like terms~ 19+19=0 0+3x= 2+19 3x= 2+19 Combine like terms~ 2+19=17