can anyone solve this math problems??

Question

8r+4=10+2r
28-0.3y=0.7y-12
3c-4c+1=5c+2+3
5-(t+3)=5c+2(t-3)


---------PLEASE SHOW STEPS THANKS!-----------

Answer 1

8r+4=10+2r
6r+4=10 (subtract 2r)
6r=6 (subtract 4)
r=1 (divide by 6)

28-0.3y=0.7y-12
40-0.3y=0.7y (add 12)
40=1.0y (add 0.3y)
40=y (divide by 1)

3c-4c+1=5c+2+3
-1c+1=5c+2+3 (subtract 3c and 4c)
-1c+1=5c+5 (add 2 and 3)
-1c=5c+4 (subtract 1)
-6c=4 (subtract 5c)
c=-4/6 (divide by -6)
c=-2/3 (reduce)

5-(t+3)=5c+2(t-3)
5-1(t+3)=5c+2(t-3) (act as if there is a negative one in front of the parenthesis.
5-t-3=5c+2(t-3) (distribute)
5-t-3=5c+2t-6 (distribute)
2-t=5c+2t-6 (subtract 3 from 5)
8-t=5c+2t (add 6)
8-3t=5c (subtract 2t)
5c=-3t+8 (rearrange *optional*)
c=-3/5t+1.6

Hope this helps!

~queenpan1295

Answer 2

1.) 8r + 4 = 10 + 2r
- 4 -4
8r = 6+ 2r
- 2r -2r
6 r = 6
r = 1

2.) 28 - 0.3 y = 0.7y - 12
+12 +12
40 - 0.3y = .7y
+.3y +.3y
40= 1y

3.) 3c - 4c + 1 = 5c +2 + 3
1c + 1 = 5c + 5
1c = 5c + 4
-4c = 4
c = -1

Answer 3

8r + 4 = 10 + 2r
Subtract 2r from each side
(8r-2r) + 4 = (10+2r-2r)
6r + 4 = 10
subract 4 from each side
6r +4 -4 = (10-4)
6r=6
divide by 6
6r/6 = 6/6
r=1

28-0.3y=0.7y-12
first add 12 to each side
(28+12) - 0.3y = 0.7y - 12 +12
16 - 0.3y = 0.7y
now add 0.3y to each side
16 = 1y
y=16

3c-4c+1 = 5c +2+3
-1c +1 = 5c +5
1= 6c + 5
-4 = 6c
-4/6 = c
-2/3 = c

5-(t-3) = 5c +2(t-3)
My Dear Aunt Sally (multiply, divide, add, subtract)
Solve for t
5 - t-3 = 5c +2t -6
ADd 6
11-t-3 = 5c+2t
8-t = 5c +2t
Add t
8= 5c +3t
3t = 5c - 8
t = 5c-8/3

Answer 4

You shouldn't ask people to do your homework for you but...
1)8r+4=10+2r first subtract 2r from both sides and get 6r+4=10 then subtract 4 from both sides and get 6r=6 then divide both sides by 6 and get r=1
2)28-0.3y=0.7y-12 First add 0.3y to both sides and get 28=y-12 then add 12 to both sides and get y=40

That is all i will show you because this is cheating but as long as i am not referenced as a source it is ok.

Answer 5

8r+4=10+2r Subtract 2r from each side
6r+4=10 Subtract 4 from each side
6r=6 Divide each side by 6
r=6

20-0.3r=0.7-12 Add 0.3r to each side
20= 1r-12 Subtract 12 from each side
8=1r which is also written
r=8

3c-4c+1= 5c+2+3 do the math to add common terms on each side

-1c+1= 5c+5 Add 1c to each side
1=6c+5 Subtract 5 from each side
6c=-4 Divide each side by 6
c= -4/6 Reduce the fraction
c= -2/3

You get to finish your homework. Do the last one yourself.

Answer 6

8r+4=10+2r
6r+4=10
6r=6 /
r =1

Answer 7

8r+4=10+2r
-2r -2r
6r+4=10
-4 -4
6r=6
/6 /6
r=1


28-0.3=0.7y-12
+12 +12
40-0.3=0.7y
+0.3 +0.3
40=y

3c-4c+1=5c+2+3
-2 -2
3c-4c-1=5c+3
-3 -3
3c-4c-4=5c
/c
3-4c-4=5
-3 -3
4c-4=2
+4 +4
4c=6
/4 /4
c=1.5


5-(t+3)=5c+2(t-3)
cant do this one


vote best answer i worked really hardd.

Answer 8

8r+4=10+2r
or 8r - 2r = 10 -4
or 6r = 6
or = 6/6 = 1

Follow the same steps for rest of the problems.

Answer 9

8r+4=10+2r
8r=6+2r
6r=6
r=1

28-0.3y=0.7y-12
.3y=40+.7y
-.4y=40
y=10

im too lazy to do the rest..but you can just google 'how to use substitution method'

Answer 10

8r+4=10+2r, subtract 2r from each side so
6r+4=10 then subtrat 4 from each side so
6r=6 then divide by 6 so
r=1
28-.3y=.7y-12 first add 12 to each side so
40-.3y=.7y then add .3y to each side so
40=1y then divide by 1 so
40=y
3c-4c+1=5c+2+3 first add the numbers up so
c+1=5c+5 then subtract 1 from each side so
c=5c+4 then subtract 5c so
-4c=4 then divide by 4 so
-c=1 then divide by - no
c=-1
5-(t+3)=5c+2(t-3) first distribute by multiplying t-3 by 2 so
5-(t+3)=5c+2t-6 then add each side by t+3 so
5=5c+2t-6+t+3 then put the t's and numbers together so
5=5c+3t-3 then add 3 to each side so
8=5c+3t then subtract 3t from each side so
8-3t=5c then subtract 8 from each side so
-3t=5c-8 then divide each side by -3 so
t=-5/3c+8/3