Let's define the variables:

• Let $t$ represent the number of days after the current time when both groups will have painted the same amount.
• Let $x_1(t)$ represent the total kilometers painted by the first group after $t$ days.
• Let $x_2(t)$ represent the total kilometers painted by the second group after $t$ days.

Equations

For the first group, the total kilometers painted after $t$ days will be the initial 2 kilometers plus 3 kilometers per day: $x_1(t) = 2 + 3t$

For the second group, the total kilometers painted after $t$ days will be the initial 5 kilometers plus 2 kilometers per day: $x_2(t) = 5 + 2t$

Since we want to find when both groups will have painted the same amount of the bridge, we can set $x_1(t) = x_2(t)$.

Set up the equation

$2 + 3t = 5 + 2t$

Solve for $t$

Subtract $2t$ from both sides: $2 + t = 5$

Subtract 2 from both sides: $t = 3$

Solution for $t$

It will take 3 days for both groups to have painted the same amount of the bridge.

Find the total kilometers painted by both groups after 3 days

Substitute $t = 3$ into either equation (both will give the same result):

Using the first group’s equation: $x_1(3) = 2 + 3(3) = 2 + 9 = 11$

Using the second group’s equation: $x_2(3) = 5 + 2(3) = 5 + 6 = 11$

Both groups will have painted 11 kilometers after 3 days.

Final Answer:

In 3 days, both groups of workers will have painted 11 kilometers of the bridge.

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Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What if the first group painted 4 kilometers per day instead of 3?
2. How long would it take if the second group painted 1 kilometer per day?
3. What if the first group started with 1 kilometer already painted instead of 2?
4. How much would the first group have painted if they worked for 6 days?
5. What if both groups started painting at the same time with no initial progress?

Tip: When solving a system of linear equations, substitution works well when one equation is easy to solve for a variable, but elimination may be more efficient for more complex systems.