Question 12590
This problem isn't really a Rate-of-Work Problem, it is actually a system of linear equations. 
<P>
First Define your Variables, Let C be the Cost/Day for a Carpenter, and P be the Cost/Day for a Painter
<P>

Equation 1: 6c + 2p = 970 
<BR>
Equation 2: 3c + 4p = 770
<P>
Now you can solve this by grpahing, substitution or elimination, but for this answer, I will solve by elimination.
<P>
Choose one of your variables to get to be the same, I am going to choose C, so I need to multiply the Bottom equation by 2
<P>
6c + 2p = 970 
<BR>
2( 3c + 4p = 770 )
<P>
Now I get <BR>
6c + 2p = 970 <B>
6c + 8p = 1540

<P>
Subtract your two equations and you get -6p = -570, divide by -6 and P = 95.
<BR>
That is, it cost $95.00/Day to hire a painter.
<P>
Now substitute P = 95 into one of your original equations <BR>
6c + 2(95) = 970 <BR>
Simplfy 6c + 190 = 970
Subtract 190 from both sides 6c = 780
Dive by 6 on both sices c = 130
<P>

So it costs 130/day for a carpenter and 95/day for a painter.