x-5 = 2
x=2+5
x=7
is that right??
but i dont know how to do these. i can only get so far on them.
5x-2=4+3x
5x=4+3x+2
5x=3x+6
i dont know how to finish this????
and this one!!
1/3x-5=6
1/3x=6+5
1/3x=11
?? i dont know what else to do???
HELLLPPP. my math teacher cant explain things to me either because i live in the philippines and i only speak english so its confusing when she explains it so please help
2007-06-12 03:12:31 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Yes, the first one is correct.
Ok, for the second one, at this point (5x = 3x +6), we need to subtract 3x from each side of the equation. This give us the following equation:
2x = 6
Now, we need to divide each side by two (remember, it's ok to do something to one side of the equation as long as you do it to the other side also). Dividing by two give us:
x = 3
For the next problem, we need to multiply each side by 3. This will give us x = 33.
2007-06-12 03:20:41 · answer #1 · answered by Math Stud 3 · 0⤊ 0⤋
Your sooo close in finding the answers for both. you left off at:
5x=3x+6
So what do you need to do to get the 3x on the other side? You need to cross out the 3x, how do you cross it out? With a -3x.Now subtract the "x"s.
Now you need to cross out the 2. How do you do that? You divide both sides with a 2 to find out what x stands for:
This is how it should look like:
5x-2=4+3x
5x=4+3x+2
5x=3x+6
5x-3x=6
2x=6
x=3
Now the other one. How do you get rid of a fraction? You multiply, just remember, you always have to do the opposite to cross out the other number. And youv'e got your answer:
1/3x-5=6
1/3x=6+5
1/3x=11
(3/1)1/3x=11(3/1)
x=33
I hope this will help you.
2007-06-12 10:50:04 · answer #2 · answered by Caligirl82 2 · 0⤊ 0⤋
the first one is right
5x-2=4+3x
5x=4+3x+2
5x=3x+6
you're on the right track. Now you have to move the 3x to the left side of the equation
5x-3x=6
2x=6
and divide by 2
x=3
1/3x-5=6
1/3x=6+5
1/3x=11
Ok now divide by 1/3 (which is the same as multiplying by 3)
3(1/3x)=3(11)
x=33
2007-06-12 10:18:20 · answer #3 · answered by uʍop ǝpısdn ツ 3 · 0⤊ 0⤋
You've gotten really close on the first problem: 5x-2=4+3x. Starting where you left off...
You want to get all terms with "x" on one side of the equation and all terms without an "x" on the other side. So...
5x=3x+6
5x-3x=6
2x=6
x=6/2
x=3
For the second equation:
(1/3)x=11
x=11*3
x=33
To "undo" multiplication (5x, for instance), divide both sides of the equation by the number in front of x (in this case, 5).
To "undo" division (1/3 x, for instance), multiply both sides of the equation by the inverse (3, in this case).
2007-06-12 10:49:32 · answer #4 · answered by math_geek_in_the_pink 1 · 0⤊ 0⤋
For the first one: 5x=3x+6 =>
5x-3x=3x+6-3x => 2x=6 => x=3
For the second one:
multiply both sides by 3
x=33
2007-06-12 10:24:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋
for the first one: subtract the 3x on both sides.
That leaves you with 2x=6
Divide by 2 and the answer is 3
The second one just multiply by 3 on both sides and the answer is 33. Look at it like this: If one third of x is 11 then x must be three times x. 3 times 11 is 33.
2007-06-12 10:17:38 · answer #6 · answered by Joe L 2 · 2⤊ 0⤋
5x= 3x+6
5x - 3x =6
2x = 6
divide by 2, x=3
1/3x =11,
multiply by 3,
x = 33
2007-06-12 10:18:15 · answer #7 · answered by raja 2 · 2⤊ 0⤋
I will state instructions after each line:-
Question 1
5x - 2 = 4 + 3x
ADD 2
5x = 6 + 2x
SUBTRACT 2x
3x = 6
DIVIDE by 3
x = 2
Question 2
(1/3).x - 5 = 6
ADD 5
(1/3).x = 11
MULTIPLY by 3
x = 33
2007-06-12 10:23:01 · answer #8 · answered by Como 7 · 1⤊ 0⤋
(1) Yes 7 is correct
(2) x= 8
(3) x = 33
2007-06-12 10:34:24 · answer #9 · answered by miles p 2 · 0⤊ 0⤋
5x-2=4+3x
5x=4+3x+2
5x=3x+6
5x-3x=6
2x=6
x=3
1/3x-5=6
1/3x=5+6
1/3x=11
1/3x*3=11*3
x=33
2007-06-12 10:25:03 · answer #10 · answered by WJVV 4 · 0⤊ 0⤋