Determine the solution to the system of equations:
x = 2y - 5
3y + x = 10

In this system of equations, x is equal to one, and y is equal to three.

Step-by-step explanation:

We can solve this by rearranging either equation to define either x or y, and plugging that into the other.  As it happens, the first equation is conveniently arranged to define x, so let's use that:

3y + x = 10

3y + (2y - 5) = 10

5y - 5 = 10

y - 1 = 2

y = 3

Now that we know the value of y, we can plug it in to either of the original equations to find x:

x = 2y - 5

x = 2(3) - 5

x = 6 - 5

x = 1

Now we can check that our answer is correct by plugging that x value into either of the original equations:

3y + x = 10

3y + 1 = 10

3y = 9

y = 3

And because that gives us our original y value, we know that our answer is correct.

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )