Is this problem right? (X, Y)?

Question

2x - 2y = 12
2x - 5y = 27

I came up with y = 5, and x = 11, I know x is not right can someone help me figure it out? Thank you

Here's what I did

I subtract the two x which cancels out, then I subtract 2 - 5 and got -3y then subtract 12 - 27 and got -15 which is y = 5

Answer 1

2x-5y=27
2x-2y=12
- . + . . -
_________
-3y=15
y=-5

2x-2(-5)=12
2x=2
x=1

Answer 2

The key to solving these simultaneous equations is to make sure you keep the + and - signs correct.

(1) 2x - 2y = 12

(2) 2x - 5y = 27

Subtract (2) from (1) gives

(3) 2x - 2x = 0; and -2y -(-5y) = -2y + 5y = 3y; and 12 - 27 = -15

Gives 3y = -15 so x = -15/3 = -5

Now substitute -5 for x in equation (1)

2x - (2*-5) = 12
2x - (-10) = 12
2x + 10 = 12
2x = 12 - 10
2x = 2
x = 2/2 = 1

Now check with equation (2)

2*1 - (5*-5) =27
2 - (-25) =27
2 + 25 =27 which is correct

At school my maths teacher used to insist that we wright out all these steps. It helped her to see where we had gone wrong and helped us not to go wrong.

Hope this has helped you

Answer 3

My son is working on quadratic equations in Algebra right now so this has been a walk down memory lane for me (I hadn't done this in 25 years until he asked for my help). Anyway, what I recall is that you have to solve for one of the variables and then use that solution to define the other variable and solve for it.

I think "The Potter Boy" above got the right answers. What I would do on the first equation is solve for x, which would be 2x = 12 + 2y, so x = y + 6. Then put this into the second equation, defining it as 2(y + 6) - 5y = 27; 2y + 12 - 5y = 27. Solving for y, you get 2y - 5y = 15; -3y = 15, so y = -5.

Plug your solution for y back into either equation and in either case, you will see that x is 1.

Answer 4

Elimination Method

2x - 2y = 12- - - - - -- Original Equation 1
2x - 5y = 27- - - - - - -Original Equation 2
- - - - - - - - -

Multiply equation 2 by - 1

2x - 5y = 27

- 1(2x) -1(-5y) = - 1(27)

- 2x + 5y = - 27. . . . New equation 2

- - - - - - - - - - - - - -

Elimination of x original equation 1 and new equation 2

2x - 2y = 12
-2x + 5y = - 27
- - - - - - - - - -
3y = - 15
3y/3 = - 15/3

y = - 5

The answer is y = - 5

insert the y value into original equation 2

- - - - - - - - - - - - - - - - - - - - - -

Solving for x

2x - 5y = 27

2x -5(-5) = 27

2x + 25 = 27

2x + 25 - 25 = 27 - 25

2x = 2

2x/2 = 2/2

x = 1

The answer is x = 1

Insert the x value into original equation 2

- - - - - - - - - - - - - - - - - - - - - -

Check for equation 2

2x - 5y = 27

2(1) - 5(-5) = 27

2 + 25 = 27

27 = 27

- - - - - - - - - - - - -

Check for equation 1

2x - 2y = 12

2(1) - 2(- 5) - 12

2 + 10 = 12

12 = 12

- - - - - - - - - - - - - -

The solution set is { 1, - 5 }

- - - - - s-

Answer 5

2x - 2y = 12.....Eq 1
2x - 5y = 27.....Eq 2
Subtract Eq 2 from Eq 1
2x - 2x - 2y -(- 5y) = 12 - 27
-2y + 5y = -15
3y = -15
y = -15/3
y = -5
Substitute the value of y in Eq 1
2x - 2y = 12
2x -2(-5) = 12
2x + 10 = 12
2x = 12 - 10
2x = 2
x = 2/2
x = 1
(x = 1, y = -5) is the solution

Answer 6

subtracting 2 from 1
3y=-15
dividing by 3
y=-5
sub in 1
2+10=12
adding -10
2x=2
dividing by 2
x=1
so soln.set {1,-5}

Answer 7

x=1
y=-5

Answer 8

x=1
y=-5

Answer 9

no there seems to be a problem there just ask your math prof. i think the question is wrong

Answer 10

theres a good site to cheak for your awnser www.maths.com