x+ y = 5    2x-y = 4   solve x and y

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hala718's profile pic

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

Posted on

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

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

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

jess1999's profile pic

jess1999 | Student, Grade 9 | (Level 1) Valedictorian

Posted on

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

nisarg's profile pic

nisarg | Student, Grade 11 | (Level 1) Valedictorian

Posted on

x+ y = 5   

2x-y = 4

3x=9

x=3

x+y=5

3+y=5

y=2

jeyaram's profile pic

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)}

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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