Solve the system of linear equations by substitution.
1 answer:
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2 ways: make a little divide tree based on how many times you can divide it. Like you have sqrt88 you know it can divide by 2. So its 2 and 44, and 44 can be broken to 2 and 22, then 2 and 11. 11 can’t be divided anymore.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.