So first, you want to write down the equations you know. We know the relationship for kinetic energy (KE) is as follows:

KE = (1/2)*m*v^{2} , where m is mass and v is velocity. This relationship of KE, mass, and velocity will be VERY important to solving this problem.

Based off reading the first sentence, we know that the first car has 2 times the mass of the second and half the kinetic energy or:

m_{1}=2m_{2} (relationship 1)

KE_{1}=(1/2)KE_{2} (relationship 2)

In the second sentence we find out that if the velocity of both cars increases by 6, their kinetic energies are equal, or:

when V_{1} = V_{1}+6 and V_{2}=V_{2}+6:

KE_{1}=(1/2)*m_{1}*(V_{1}+6)^{2}

KE_{2}=(1/2)*m_{2}*(V2+6)^{2} and

there kinetic energies are equal at this point or KE_{1}=KE_{2}, therefore, substituting for KE_{1} and KE_{2}, we get our next relationship:

(1/2)*m_{1}*(V_{1}+6)^{2}=(1/2)*m_{2}*(V_{2}+6)^{2} (relationship 3)

Now we have all the relationships we need to answer the question "What were the original speeds of the two cars?"

First we can simplify relationship 2 a bit by substituting KE's for their (1/2)mv^{2} equivalent as follows:

- KE
_{1}=(1/2)KE_{2}(relationship 2) expands to: - (1/2)m
_{1}V_{1}^{2}=(1/2)((1/2)m_{2}V_{2}^{2}), then we can use relationship 1 to get rid of the m_{1}variable by substituting for m_{1}using the relationship m_{1}=2m_{2}: - (1/2)2m
_{2}V_{1}^{2}=(1/2)*((1/2)m_{2}V_{2}^{2}), then we can cancel out the m_{2}term and multiple the numbers out on both sides and this now simplifies to a velocity relationship between car one (V_{1}) and car two (V_{2}): - V
_{1}^{2}=(1/4)V_{2}^{2}or 4V_{1}^{2}=V_{2}^{2}. Take the square root of both sides and we now have a relationship for V_{1}and V_{2}: - 2V
_{1}=V_{2}

Now let's simplify relationship 3 by substituting for m_{1} using the relationship m_{1}=2m_{2}:

- (1/2)*2m
_{2}*(V_{1}+6)^{2}=(1/2)*m_{2}*(V_{2}+6)^{2}. Now we can substitute for V_{2}using the relationship we found above V_{2}=2V_{1}. The equation now simplifies to: - m
_{2}*(V_{1}+6)^{2}=(1/2)*m_{2}*(2V_{1}+6)^{2}

The m_{2} term on both sides cancels out and now we only have one variable in the equation, V_{1}, that we can solve for. If we expand out the squared terms, we now get:

- V
_{1}^{2}+12V_{1}+36=(1/2)*(4V_{1}^{2}+24V_{1}+36), which simplifies to - V
_{1}^{2}+12V_{1}+36=2V_{1}^{2}+12V_{1}+18

If we move the variables to one side and the numbers to the other, the V_{1} terms cancel out, leaving us with:

- V
_{1}^{2}=18.

Taking the square root of this this simplifies to the velocity of car 1 which is:

V_{1}^{2}=3*(2)^{(1/2)} = 4.24 m/s

Since we know from earlier that 2V_{1}=V_{2}, we can find the velocity of car two to be:

V_{2}= 2*(4.24) = 8.48 m/s

These are the original speeds of the two cars!