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lesya [120]
3 years ago
5

Which is the solution to this pair of linear equations?

Mathematics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

(0,6)

Step-by-step explanation:

So let's start with the top problem.

<u>You have:</u> 3x + y = 6

First, you need to move the number with the X to the right. You will need to subtract the 3x over to the 6

<u>Ex:</u>

3x + y = 6

-3 -3

<u>This now gives you:</u> y = -3x + 6

<u>This</u><u> </u><u>is</u><u> </u><u>bec</u><u>ause</u><u>:</u> When you subtract the 3x, it cannot make the 6 a 3 because the 3 has an x along side it. Because of this, you have to make them separate. The 3x becomes a negative due to subtraction. When the 3x cannot go into anything, it will keep the minus with it and become negative. The y stays the same because it's not accompanying any number.

Now that your problem is in Slope Intercept Form, we can move onto the second problem

<u>Here</u><u> </u><u>you</u><u> </u><u>have</u><u>:</u> -x + 2y = 12

First, just like last time, you have to move the -1x to the right. When a variable doesn't have a number next to it, it's always a 1.

For this one, you have to add the -x

<u>Ex</u><u>:</u>

-x + 2y = 12

+x +x

<u>This</u><u> </u><u>now</u><u> </u><u>gives</u><u> </u><u>you</u><u>:</u> 2y = x + 12

<u>This is because</u> The -x is now positive because you added it to the 12, but since the -1 is accompanying an x, it cannot go into the 12 and make it 13. It stays x. The 2y stays here as well.

For your second step, you have to take the 2 from the 2y and <u>Divide</u><u> </u><u>it</u><u> </u><u>to</u><u> </u><u>ALL</u><u> </u><u>the</u><u> </u><u>numbers </u><u>on</u><u> </u><u>the</u><u> </u><u>right</u><u> </u><u>side</u><u>.</u>

<u>Ex</u><u>:</u>

2y = x + 12

÷2 ÷2 ÷2

<u>This</u><u> </u><u>now</u><u> </u><u>gives</u><u> </u><u>you</u><u>:</u> y = 1/2x + 6

<u>This</u><u> </u><u>is</u><u> </u><u>bec</u><u>ause</u><u>:</u> Since you have to convert the problem to slope intercept form, the y stays on the left, while the 2 divides all the numbers. Since the 2 cannot divide with the 1x since the 1 is accompanying an x, it becomes a fraction.

<u>Now</u><u> </u><u>we</u><u> </u><u>have</u><u>:</u> y = -3x + 6

y = 1/2 + 6

<u>What</u><u> </u><u>you</u><u> </u><u>do</u><u> </u><u>next</u><u>:</u> Personally, I do not have graphing paper on me, so I used Desmos Online Calculator and typed in both problems to get (0,6), but if you're graphing on normal graph paper, you just have to graph the problems like you did in <u>System</u><u> </u><u>of</u><u> </u><u>Linear</u><u> </u><u>Equations</u><u>.</u>

If you have any more questions about anything, I'm always here to help :)

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Answer:

D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

  • Now if we consider the defect in sixth car to be 3 then our data will look like:    3,4,6,7,9,10

So the mean of above data would be \frac{3+4+6+7+9+10}{6} = 6.5

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{6+7}{2} = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

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So the mean of above data would be \frac{4+6+7+9+10+12}{6} = 8

and Median of the above data would be = \frac{3^{rd} obs + 4^{th}  obs}{2} = \frac{7+9}{2} = 8

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  • But Now if we consider the defect in sixth car to be 7 then our data will look like:    4,6,7,7,9,10        

So the mean of above data would be \frac{4+6+7+7+9+10}{6} = 7.167    

and Median of the above data would be =  \frac{3^{rd} obs + 4^{th}  obs}{2} = 7

   Hence mean and median number of defects are not same if the sixth car has 7 defects.      

Therefore option D is correct.                                                                                  

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