I Need Help With Algebra?

Note: I will help, but you have too many problems in one question. Even half of these would be a bit much.
1.) the sides of a triangle can be represented by three consecutive even numbers. if the first side is increased by 1 and the second side is decreased by 1, the triangle becomes isosceles and has a perimeter of 30 in. find the length of each side of the original triangle.
Let x = 1st side
x + 2 = 2nd side
x + 4 = 3rd side
Can you take if from there?
2.) mark is 10 years younger than larry. larry’s age in 8 years will exceed twice marks age 3 years ago by 4 years. how old is each now.
Let x = Larry’s age now
x – 10 = Mark’s age now
x + 8 = Larry’s age in 8 yrs
(x – 10) – 3 = Mark’s age 3 yrs ago
x + 8 = 2[(x - 10) - 3] + 4
3.) mr. west drove his car from home to chicago at the rate of 40 mph. and returned at a rate of 45 mph. if his time going exceeded his time returning by 30 minutes, find his time going and his time returning.
d = r x t
Distance is the same both ways, so r(going) x t(going) = r(returning) x t(returning).
Let t = time returning
t + 30 = time going
r(going) = 40 mph
r(returning) = 45 mph
40(t + 30) = 45t
4.) tom, dick, and harry weigh 50, 85, and 120 pounds. respectively and use a 12 foot plank for a teeterboard. tom sits 5 feet, dick sits 4 feet, and harry sits 3 feet from the fulcrum on the same side. how far must a 200 pound friend sit from the fulcrum to balance them?
I’m assuming the fulcrum is in the center.
x = distance from fulcrum of the 200-lb friend
50(5) + 85(4) + 120(3) = 200x
I think. Haven’t done this in awhile.
5.) find 4 consecutive even integers such that the sum of the third and fourth exceeds 1/3 of the first by the sum of the first and third.
Let x = 1st integer
x + 2 = 2nd integer; x + 4 = 3rd integer; x + 6 = 4th integer.
(x + 4) + (x + 6) = 1/3(x) + x + (x + 4)
6.) a sum of $3,500 is invested in two parts. one part brings a return of 5% and the other a return of 8%. the total annual return is $250. find the amount invested at each rate.
P = principle of 1st part
3500 – P = principle of 2nd part
(5% of P) + (8% of (3500 – P)) = 250
7.) how many pounds of seed worth 60 cents a pound must be mixed with 300 pounds of seed worth 35 cents a pound to produce a mixture worth 50 cents per pound?
x = lbs seed worth 60 cents/lb
total lb of seed = x + 300
60x + 35(300) = 50(x + 300)
Hope this helps!
d

1. Okay letting x be first side, then
you know first side is x, 2nd side is x+2,
and third side is x+4.
So, then you know that
(x+1)+(x+2-1)+(x+4)= 30
2. Let m be marks age and
L be larrys age, then
m=L-10 and
L+8=2(m-3)+4 or
L+8=2(L-10-3)+4.
3 Let t be time returning , so
the time going is t+.5, so
40*(t+.5)=45*t
4. Im not sure about this but I think that
distance the 200 friend is from fulcrum is
((3*120)+(4*85)+ (5*50))/(3*200)
5 Let x be first number then
((x+4)+(x+6))=(x/3)+(x+(x+4))
6.Let x be part return of 5% and
y be return of 8%, then
x+y=3500
.05x+.08y=250
7. Let x be number of pounds, the
60x+(300*35)=(x+300)50