solve each system of equations?

Question

s + 2t =5
7r - 35 + t = 20
2t = 8

Answer 1

3rd Equation:
if 2t = 8
divide by 2
t = 4

substitute 4 to both problems and solve
1st Equation:
s + 2 * 4 = 5
s + 8 = 5
subtract 8 from both sides
s = -3

2nd Equation
7r -35 + 4 = 20
7r - 31 = 20
add 31 to both sides
7r = 51
divide by 7 to both sides
r = 51/7

s = -3
r = 51/7
t = 4

Answer 2

First: take the last equation and solve for "t" > divide both sides by 2...

2t/2 = 8/2
t = 4

Sec: replace 4 with "t" in the 2nd equation...

7r - 35 + 4 = 20
7r - 31 = 20

*Add 31 to both sides (the opposite side means opposite signt)...

7r - 31 +31 = 20 + 31
7r = 51

*Solve for "t" by isolating it on one side > divide both sides by 7...

7r/7 = 51/7
r = 51/7

Third: replace 4 with "t" in the 1st equation...

s + 2(4) = 5
s + 8 = 5

*Subtract 8 from both sides...

s + 8 - 8 = 5 - 8
s = - 3

Answer 3

OK, first use the last equation to say that t = 4

then, plug this into the first equation: s+8 = 5, so s = -3

Then, finally plug t into the second equation:
7r - 35 + 4 = 20
or 7r = 20+31 = 51, so r = 51/7 which is not a whole number, but maybe that's OK.

Answer 4

I would solve the third equation first:
2t=8
t=4

Now, substitute it into the first equation:
s+2(4)=5
s+8=5
s=-3

7r-35+4=20
7r-31=20
7r=51
r=51/7

The solution set is (51/7,-3,4)

Check:
-3+2(4)=5
-3+8=5
5=5

7(51/7)-35+4=20
51-31=20
20=20

2(4)=8
8=8

I hope this helps!

Answer 5

s + 2t =5
7r - 35 + t = 20
2t = 8 ----------------------------> t = 4

s + 2.4 = 5 ---------------------> s = -3

7r - 35 + 4 = 20 ---------------> r = 51/7 = 7.29