Can you answer this 2 problems using substitution or elimination method?

Question

Prob #1:
Thomas is 4 times as old as his son. In 5 yrs, he'll be 3 times as old as his son. How old is each now?

Prob#2:
There is a number such that 3 times the number diminished by 8 is equAL to 34. Find the number.

Pls answer me coz i rily nid help...and so I can vote for you a best answer.

Answer 1

prob 1
thomas' age = x
his son's age = y

x = 4y and x+5 = 3(y+5)
so 4y +5 = 3y + 15
y = 10
x = 40

prob 2
say the number is x
3x -8 = 34
3x = 42
x = 14

Answer 2

Prob #1:
Thomas is 4 times as old as his son. In 5 yrs, he'll be 3 times as old as his son. How old is each now?
t=thomas; s=son
t=4s
t+5=3(s+5) substitute
4s+5=3s+15 subtract 3s+5
s=10 so t=4s=40
in 5 yrs t+5=45
s+5=15
15*3=45

Prob#2:
There is a number such that 3 times the number diminished by 8 is equAL to 34. Find the number.
3x-8=34 add 8 to each side
3x=42 divide by 3
x=14

Answer 3

Question 1, Thomas is 40, his son is 10
In five years his son will be 15. 15 x 3 =45 and 40 + 5 = 45
Just start with a reasonable figure, like 5 and 20 and work it out until you get the right answer.

Question 2, x times 3, minus 8 equals 34
So: 3x-8=34 (add 8 to both sides)
3x = 42 (Divide by 3)
x = 14
Plug in the values.
14 times 3 equals 42, minus 8 is 34...true

Answer 4

1) T for thomas age and S for the son age :
T = 4S
T+5 = 3 (S+5)

So by substitution :
4S+ 5 = 3S + 15
so
S = 10 years (The son)
T = 40 years (Thomas)

2) Let N the number :
3N -8 = 34
so 3N = 42
so N = 14

Answer 5

1.
Let x = Thomas' age
y = his son's age
x = 4y
x + 5 = 3(y + 5)
using substitution:
4y + 5 = 3 y + 15
y = 10
x = 40

2.
let n = number
3n - 8 = 34
3n = 42
n = 14

Answer 6

Prob #1 Answer is Thomas is 40 and his son is 10.
Prob #2 Answer is 42

Answer 7

1. let x be his son's present age
then 4x+5 = 3(x+5)
4x+5 = 3x+15
4x-3X =15-5
x = 10 Therefore, Thomas is 40 yrs old now while his son is 10 yrs old now
2. let x be the #
3x-8=34
3x=34+8
3x=26
x=8.6666...

Answer 8

36.) multiply both equations by technique of 30 to cancel out the denominaters. so..... 18s-5t=30 6s+25t=330 (you are able to both situations the first equation by technique of 5 or the 2d one by technique of three) so like 90s-25t=one hundred fifty 6s+25t=330 upload mutually to cancel out t and go away s by technique of itself 96s=480 divide by technique of ninety six s=480 plug into 1st or second equation to get t try the 2d one by technique of self! desire this helpd:)