Isaac has 21 green marbles and 7 blue marbles. He wants to place them in identical groups without any marbles left over. What is
2 answers:
Answer:
7 groups.
Step-by-step explanation:
We have been given that Isaac has 21 green marbles and 7 blue marbles. He wants to place them in identical groups without any marbles left over.
To find the greatest number of groups Issac can make we will find the GCF of 7 and 21.
Factors of 7 are: 1, 7.
Factors of 21 are: 1, 7, 3, 21.
We can see that 7 is the greatest common factor of 7 and 21, therefore, Issac can make 7 groups from 21 green marbles and 7 blue marbles.
Let us find number of each type of marble is each group.
Therefore, there will be 3 green marbles in each group.
Therefore, there will be 1 blue marble in each group.
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Answer:
A. (1, -2)
B. the lines intersect at the solution point: (1, -2).
Step-by-step explanation:
A. The equations can be solve by substitution by using the y-expression provided by one of them to substitute for y in the other.
This gives ...
3x -5 = 6x -8
Adding 8-3x to both sides, we get ...
3 = 3x
Dividing both sides by 3 gives ...
1 = x
Substituting this value into the first equation, we can find y:
y = 3(1) -5 = -2
The solution is (x, y) = (1, -2).
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B. The lines intersect at the solution point, the point that satisfies both equations simultaneously. That point is (1, -2).
