First an informal solution using logical reasoning and simple arithmetic. You can get some good mental exercise solving problems like that.
3 tacos and 3 burritos cost $11.25; 4 tacos and 2 burritos cost $10. The difference is 1 more taco and 1 less burrito, for $1.25 less.
So adjust the order repeatedly the same way, adding one taco and taking away one more burrito, and decreasing the cost by $1.25 each time.
3 tacos, 3 burritos, $11.25
4 tacos, 2 burritos, $10.00
5 tacos, 1 burrito, $8.75
6 tacos, 0 burritos, %7.50
The cost of each taco is $7.50/6 = $1.25.
3 of each cost $11.25, so 1 of each cost $3.75; the taco costs $1.25, so the burrito costs $2.50.
ANSWERS: taco $1.25, burrito $2.50
Then here is one of many ways to work the problem with formal algebra....
3t+3b=11.25
4t+2b=10
With the two equations in that form, a solution by elimination is probably easiest. Multiply each equation by some constant so that the coefficients of b are opposites, so that b disappears when the two equations are added.
-6t-6b=-22.50
12t+6b= 30.00
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6t = 7.50
t = 7.50/6 = 1.25
Plug that into either of the original equations to find b:
4(1.25)+2b=10
5+2b=10
2b=5
b=2.50