Algebra help...?

Question

I'm in high school, but I still can't pass Pre Algebra. I got a worksheet today, but I don't understand it. No, I'm not asking you to do my homework, I'm just asking for help.
The worksheet says:
For this task you will use deductive reasoning to determine which of the digits from 0 to 9 each of the letters represents. Each of the digits from 0 to 9 is used exactly once.

1. g + g + g = d

2. j + e = j

3. g times g= d

4. b + g = d

5. f - b = c

6. a x c = a

I kind of understand the instuctions, from what I understand you have to figure out what the letters are and what number they are. But I can't figure out how to do it. =[

Thanks in advance.

My bad, I did forget one. I forgot

i/h = a; h is less than a

I really don't get that one.

Answer 1

First off, 1 and 3 give us the value of g. Since they both equal d, set them equal to each other!

3g = g^2, now divide both sides by g.
3 = g. There is your first answer.
This also reveals that, since

d = 3g,
d = 9.

Part 2 tells us that j + e = j. Subtract j from both sides to reveal that:

e = 0.

Part 4 says to add an unknown value, b, to g, and that equals d. We know g and d, so subtract g on both sides to find out that the value of b is...

b = 9 - 3 = 6.

Let's skip to part 6. a * c = a. Divide a on both sides to reveal that:

c = 1.

Now solve part 5, the last piece of the puzzle. we don't know f, but we know b and c. Add b to both sides to determine that...

f = 1 + 6 = 7

And there you go. =D

Answer 2

Let's try to understand based on what was given and deduce the answers as instructed --

What we know: each letter can be any unique number 0 to 9 and can only be that number, e.g. if you assign c = 1 then c will always be one in the whole set of questions.

We take a look at #1 and #3

g+g+g = d
g(g) = d

or

3g = d
g^2 = d, so

3g = g^2 or 3g = g*g thus by cancellation g = 3
by substitution, d = 9

Using #4, b+g = d, so

b+3=9 thus b=9-3 or b=6

Using #2, j + e = j which is the definition of the additive identity so e=0

So far, we recognize:

g = 3
d = 9
e = 0
b = 6 but we still don't know j, f, c, & a

Using #6, a(c) = a which is the definition of the multiplicative identity so c = 1

Using #5, f - 6 = 1 thus f = 7

Now, j and a are still undefined and the remaining unused numbers are 2, 4, 5, and 8.

But you say, i/h = a so we take a look at the remaining four and check which ones fit -- there is 2, 4, & 8 which are factors of each other so

8/2 = 4 thus i = 8, h = 2, & a = 4

and indeed h < a is true so we are left with j and 5 so

j = 5

ta da!

Answer 3

Challenging qn ^^

1. g + g + g = d

2. j + e = j

3. g times g= d

4. b + g = d

5. f - b = c

6. a x c = a

eq 1 and eq 3 states that when u add 3 g together or let g * g, both will work out to d. Hence, d have to have a sq no. either 1,4,9. However 1 and 4 is out so d have to be 9

since d = 9 = g*g = g+g+g,
3g = 9
g = 3

now with d and g solve, next will be still be eq 4. b + g = d
b+3=9
b=6

eq 2. j+e = j, so e have to be 0 for sure
eq6. a*c = a, so c definately have to 1.

eq 5. f - b = c
f-6=1
f=7

so u have
e = 0, c = 1, g = 3, b = 6, f = 7, d = 9. which leaves 2, 4, 5, 8 left.

so only left a and j which can be either 2, 4, 5, 8.

Hope my explaining is clear enough

Answer 4

The key is to use deductive reasoning - think differently than you would with a normal problem.

Looking at the problems, there are two that look similar, and thus may be a good place to start.

1) g + g + g =d
3) g x g = d

From 1), d must be greater than 3 (as g cannot be less than 1). From 3), d must also be a square number.

There are only 3 square numbers less than 10: 1, 4 and 9. As g must be a whole number, d cannot be 4. Because, from problem 3), g would then equal the square root of 4 (not a whole number). So therefore d = 9, and g = 3.

Now we can work out problem 4):
b + g = d
b + 3 = 9
b = 9 - 3 = 6.

I'll leave the rest up to you (try looking at problem 6 before problem 5. Then you should be able to work out problem 2 from the numbers that are left.

Good luck!

Answer 5

1. g + g + g = d
g could be 1,2, or 3.
d could be 3,6, or 9

2. j + e = j
e can only be 0, because any number can only be added to zero to equal itself.

3. g times g= d
refer to question 1.Which of the g's and d's will satisfy both questions?
Hint: 1x1=?... 2x2=?... 3x3 = ?

4. b + g = d
By now, we know d=9, e=0, and g=3 (or did you skip ahead to this part? LOL), so let us plug in some known values.
b + 3 = 9
subtract 3 from both sides of the equation.
b + 0 = 6 or b=6.

5. f - b = c
f - 6 = c
Since we don't know f or c yet, we'll come back to it.

6. a x c = a
c can only be 1, because any number can only be multiplied by 1 to equal itself.
Going back to question 5: f - 6 = 1, so f = 7.

7. i/h = a; h is less than a
This is a tough one, as we don't have a clear answer for any of the parts. Let's experiment with plugging in some of the left over values.
4/2 = 2... oops that doesn't work, because h would = a.
8/2 = 4 2<4, so this works.

What we know so far: a=4, b=6, c=1, d=9, e=0, f=7, g=3, h=2, i=8.
j has to be 5, because it's the only number left!

Answer 6

This is one of those silly-αssed pseudo-math problems that some Dork with a PhD in Education thought looked pretty good on a math apptitude test.

But a lot of them are just 'horse sense' kind of questions.

Look at question # 2. The only thing in the World that e can be is 0 (since adding it to something didn't change the value). And
a x c = a looks as if c = 1. Then g+g+g=d along with g*g = d makes me think that g is 3 and d is 9. The b+g=d leads to b=6 (since we already know what g and d are). Just kinda 'step around' them trying different values

HTH

Doug

Answer 7

Forget numbers at first, and simplify.
g+g+g=g*g, so g=3
d=g*g, so d=9
b+g=d, so b=6
and I now see the person above me is doing the same thing, so read that one.

Answer 8

I have no clue...im in pre-al too....and am in 9th grade..but ive never seen this before.