Also need help with these two! Text eq's again...!?

Question

1.) The sum of two integers is 10. three times the larger integer is three less than eight times the smaller integer. Find the integer.

For this one, I tried to set it up as:
i=interger

3i-(3(8i)=10

this is way off.. couldn't get the right answer. Larger and smaller part got me confused.

Another one:

2.) One integer is eight lesss than another integer. The sum of the two integer is fifty. Find the integer.

In these two problems, what I am wondering is that are they talking about two different integers? and therefore, should have two different variables?

Any help with these two problems will be appriciated!

Thanks!

Answer 1

problem 1: 7 and 3
7+3=10, (3*8)-3=7*3
simple guess and check, 2 unknown variables to start
problem 2:21 and 29, think 50 divided by 2 is 25 add and subtract four from 25 to get 21 and 29 (4 and 4 is eight)
Its just whether or not you see the pattern, still is two unknown variables to start.

Answer 2

the guy on the third answer has it right. you have to set the equation up with two variables. he has the bigger integer as x and the smaller integer as y. you have to read the problem in math words. "the sum" means addition and "is" means equals. so "the sum" (x+y) "is" = 10. "three times" is just three times the larger integer (3x) "is" = "three less than eight times" ( 8y-3) the "less" means to subtract and of course you use y becuz it is the smaller integer.


i would explain the second problem too but i'm assuming you get it...i hope. if not i'll still help if you want. math comes easy to me so its no problem

Answer 3

1. Start with two simultaneous equations.
x+y=10
3x = 8y-3
isolate and define one of the unknowns

x = 10-y...then replace it in the next equation
3(10-y) = 8y-3 ...then simplify
30-3y = 8y-3
33 = 11y
y=3, x=7
(check 3*7 = 8*3 - 3)

2.
x = y+8
x+y =50

y+8+y=50
2y=42
y=21; x=29
(a quicker way to this one is to start at 25 (50/2), go up 4 and down 4 for the difference of 8, giving 21 and 29.

Answer 4

For the second one; you could simplify it to 'Two numbers with a difference of 8 sum to 50'
You know that 25+25=50, so with a difference of 8 between them the two numbers will be 4 above, and 4 below 25. 25+4=29, 25-4=21. &)

Answer 5

well, we have two integers so you have to start with that

s=smaller integer
b=bigger integer

the first sentence gives us the first equation (sum of integers is 10)

s+b=10

the next sentence gives us the next equation (3 times larger is 3 less than 8 times smaller)

3b+3=8s

two equations, two unknowns, several ways to solve it, substitution is probably simplest

s+b=10 ---> b=10-s

3b+3=8s
3(10-s)+3=8s
30-3s+3=8s
33=11s
3=s

I'm sure you can use s=3 and one of the original equations to get b, the bigger number (I didn't use l for larger because with this type it looks like a one or the pronoun "I")

-----------------------------------------------------------

the second problem is just like the first

i=one integer
a=another integer

i+8=a
i+a=50

substituting

i+(i+8)=50
2i=42
i=21

again, you put the 21 in for i, in either original equation and you get a=29

it is a good idea to go back and check the answers by plugging them both in

it is also a good idea to check my math and arithmetic to make sure you understand and to make sure I didn't screw up

good luck
math is power

Answer 6

2X + 8 = 50 :: 2X = 50 - 8

X = (50 - 8)/2 :: X = 42/2 :: X = 21

21 + 8 = 29 The two numbers are 21 and 29

Answer 7

problem #1:
x+y = 10
3*x+3 = 8*y

combining equations,
3*x + 3 = 8*y = 8*(10-x) = 80 - 8*x

so
3*x + 8*x = 80 - 3,
11*x = 77,
x=7, y=3

Try and solve the 2nd one on your own, it's similar to first.

Answer 8

1)
suppose integers : x,y
then we have
x + y = 10
3x = 8y - 3
__________
x + y = 10 ................ (*3)
3x - 8y = -3 ..................... (* -1)
__________
3x + 3y = 30
-3x + 8y = 3
_________by addng the both equations
11y = 33
y = 33/11
y = 3
x = 10 - 3
x = 7
***************************************************************
2) suppose that the 2 numbers are X,y
then
y = x - 8
x + y = 50
x + x - 8 = 50
2x = 58
x =29
y = 21