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x+ y = 5    2x-y = 4   solve x and y

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seedrasoso | Student, Grade 10 | eNoter

Posted August 15, 2010 at 8:49 AM via web

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x+ y = 5    2x-y = 4   solve x and y

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 15, 2010 at 8:51 AM (Answer #1)

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x+y = 5 ........(1)

2x-y = 4 .........(2)

Using the emimination method, let us add (1) and (2):

==> 3x = 9

==> x= 9/3 = 3

Now using the substituting method substotite in (1):

x+ y = 5

==> 3 + y = 5

==> y= 5-3 = 2

==> y= 2

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pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted August 15, 2010 at 8:53 AM (Answer #2)

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X = 3 and Y = 2.  Here is how we find this.

What we have to do is use the first equation to get a value for y.  This is quite easy because all we must do is say

x + y = 5 so y = 5 -x

Now we plug that value into the second equation.

2x - (5-x) = 4

This becomes

2x - 5 + x = 4 (because you are subtracting negative x, you are in effect adding x)

3x - 5 = 4

3x = 9

x = 3

If x = 3, then 3 + y = 5 and y = 2

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neela | High School Teacher | Valedictorian

Posted August 15, 2010 at 10:32 AM (Answer #3)

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x+y = 5

2x-y = 4.

To solve for x and y:

From the 2nd equation, y = 2x-4. Put this in 1st equation. x+y =5.

x+ 2x- 4 =5.

3x=5+4 =3

x = 9/3 =3.

Put x= 3 and from 1st equation, x+y =5,  we get: 3+y =5. So y = 5-3 =2.

So (x,y) = (3,2)

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giorgiana1976 | College Teacher | Valedictorian

Posted August 15, 2010 at 10:59 PM (Answer #4)

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Because the equations of the system are linear, we'll use the matrix formed by the coefficients of the variables to calculate the determinant of the system:

             1        1

det A =

             2         -1

 

det A = -1-2=-3

det A = -3 different from zero.

Now, we'll calculate x:

x = det x/ det A

             5         3

det x =

             4         -3

 

det x = -15 - 12 = -27

x = -27/-3

x = 9

Now, we'll calculate y:

 

             1         5

det y =

             2         4

det y = 4 - 10

det y = -6

 

y = det y/ det A

y = -6/-3

y = 2

The solution of the system is: {(9 , 2)}

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jeyaram | Student, Undergraduate | Valedictorian

Posted August 16, 2010 at 3:37 PM (Answer #5)

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x+ y = 5............(1)

2x-y = 4.............(2)

(1)+(2) => 3x=9

so                x=3

put  x=3 in (1)   y=2

Answer {x=3;y=2}

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nisarg | TA , Kindergarten | Valedictorian

Posted July 12, 2014 at 2:16 AM (Answer #6)

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x+ y = 5   

2x-y = 4

3x=9

x=3

x+y=5

3+y=5

y=2

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jess1999 | TA , Grade 9 | Valedictorian

Posted July 21, 2014 at 2:25 PM (Answer #7)

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x+ y = 5   

2x - y = 4 

First multiply everything in the first equation by 2 .

By multiplying everything in the first equation by 2 , you should get :

2x + 2y = 10

2x - y = 4 now subtract the two equation .

By subtracting the two equations , you should get :

3y = 6 divide both sides by 3 .

By dividing both sides by 3 , you should get :

y = 2 which is your answer for " y " 

Now plug " y " into one of the equation 

x + 2 = 5 subtract both sides by 2 .

By subtracting both sides by 2 , you should get

x = 3 which is your answer for " x " 

So " x " equals to 3 and " y " equals to 2

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