QUESTION ONE:
x+3y=4
2x-y=1
x=? y=?

QUESTION TWO:
On one day, 4 plumbers and 5 helpers earned $182. Working the same number of hours and at the same rate of pay, 5 plumbers and 6 helpers earned $224. How much does a plumber and how much does a helper earn each day?

4p+5h=182
5p+6h=224
p+h=42

QUESTION THREE:
A costumer bought 3 cans of corn and 5 cans of tomatoes for $1.75. The next costumer bought 2 cans of corn and 3 cans of tomatoes for $1.10/ Find the cost of one can of each.

3c+5t=1.75
2c+3t=1.10

QUESTION FOUR:
(same kind of question) 10 pounds of butter and 2 pounds of lard cost $28 while 6 pounds of butter and 3 pounds of lard costs $24. Find the cost of one pound of each.

10b+2c=28
6b+3c=24

Have you tried any of these?

You need to show some of your own work :)

Y = 1
Now Calculate Yourself The Value Of X.

there, is that enough work? Because that's all I got up to, the rest I can't do like my other questions cause I can't do (an example) 2(3x+y=12) it won't leave one variable

Okay, you have a basic problem with how to solve simultaneous equations, that's fine, I'm more than happy to help you with that.

Let's solve your first question:

x+3y=4
2x-y=1

There are 2 ways to solve this kind of equations, You'll want to know both. The first way is substitution, it works like this:

Take one of your equations: x+3y=4
Seperate one of the variables (I'll choose x): x=4-3y
Now take the second equation: 2x-y=1
Plug in the value for x into this equation: 2(4-3y)-y=1
Expand and find a value for y: 8-6y-y=1
8-7y=1
-7y=-7
y= 7/7 = 1

Now you can plug this value back into either of the first equations: x+3y=4
x+3(1)=4
x=4-3
x=1

Now check your numbers with the second equation: 2x-y=1
2(1)-(1)=1

This is correct! So the answer is definately x=1, y=1

Now, there's a second way which is a little more complicated, but is also more powerful, it is called elimination:

x+3y=4
2x-y=1

Here we want to change one of the equations so that we can add the 2 equations together or subtract them to eliminate a variable. In this case I would multiply the bottom equation by 3 so that +3y cancels with -3y when I add them. You could also multiply the top equation by 2 and subtract them to cancel the x term. Let's do my first suggestion by multiplying the bottom equation by 3, which leaves us with:

x+3y=4
6x-3y=3

Now adding them:

x+3y+6x+(-3y)=4+3
(x+6x)+(3y-3y)=(4+3)
7x+0=7
x = 1

Now we do the same as in the substitution method, take one of the initial equations and plug in this value for x:

x+3y=4
1+3y=4
3y=4-1
3y=3
y=1

Now we can check it with the other equation again, but i'll leave that out. We get x=1, y=1 again.

I suggest that you try BOTH methods of solving the questions when you're doing these problems. Elimination especially is very powerful so practice how to do it.
Please ask if you have any questions. These are powerful techniques that you need to be able to perform.

thanks I got the answer already. I found out it didn't say distinct numbers so x and y could be equal. Now I need help with the other questions. I've tried every way I could but couldn't get it. :confused:

Follow the steps that I have outlined in my post, they will solve all of these questions.