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# what is x+y=2 2x=y=-1

I don't know how to solve this linear equation

### 4 Answers by Expert Tutors

Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
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Because upper case for "=" button is "+" , I might assume that system of linear equations is
x + y = 2     ......(1)
2x + y = -1    .....(2)
Let's use method of illumination, and subtract (2) from (1)
x + y =  2
–
2x + y = - 1
- x       =  3

x = – 3

Now, let's plug in value of x into (1)
– 3 + y = 2
y = 2 + 3

y = 5

The solution is pair of numbers  (– 3, 5)
(1) ...  - 3 + 5 = 2
2 = 2

(2) ...  2 · (- 3) + 5 = - 1
- 1 = - 1

Jaison N. | A PhD to Teach You Math and PhysicsA PhD to Teach You Math and Physics
4.9 4.9 (150 lesson ratings) (150)
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Hi Elica,

I'd like to give you a solution while pointing out something to look for in the future. We have the system

x + y = 2    2x - y = -1

Let me move the y in the second equation over to the right and the -1 to the left.

x + y = 2    2x + 1 = y

Now, I'll use the symmetric property on the second equation, which tells me that I can flip the equation without changing it.

x + y = 2   y = 2x + 1

Now, in the first equation, move the x to the right by subtracting it from both sides.

y = 2 - x   y = 2x + 1

Now, each equation represents a line and you're trying to figure out if they intersect. I'll assume that they do intersect. If they don't intersect, I'll get some nonsense like 0 = 2 because its a bad assumption. If they do, I'll get the solution. And if I look at my equations, each has a y on the left. So, I can equate them.

2 - x = 2x + 1

2 = 3x + 1

Subtract 1 from both sides

1 = 3x

Divide both sides by 3.

x = 1/3

Plug this guy into one of the equations to get y = 5/3.

What I want to point out is that if you look at the original equations, the numbers in front of the y's differ only by a sign, which you can change by moving one to the other side. If the y's are on one side of the equation and they have the same number and sign in front, then you can equate them! This would work even if each equation had 5.3y instead of y because of the substitution property.

Hope this helps.

Thomas T. | Thomas - Professional TutorThomas - Professional Tutor
4.5 4.5 (78 lesson ratings) (78)
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Set it up to solve the equations simultaneously: align the variables vertically, then add to solve for x.

x+y=2                x+y=2

2x=y-1           +  2x-y=-1

3x=1        x=1/3    so substitute x=1/3

x+y=2     1/3+y=2    y=1and2/3       Check: 2x=y-1         2(1/3)=5/3-1        2/3=2/3!

Jorge L. | Spanish / Italian / Mathematics TutorSpanish / Italian / Mathematics Tutor
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I'm assuming your system looks like this (since you have two = signs on the right) :

x + y = 2

and

2x - y = - 1

To solve this, you need to find the point where both equations meet.

SOLVING THROUGH ELIMINATION
When solving through elimination, you want to make sure the variables all appear in the same side of the equation. On this case, they're already in that form so you're good to combine (add/subtract the equations).

x    + y    = 2

2x  -y       = -1

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

3x  0y     = 1          Add all the variables as if they were a single sum.

3x = 1.

x = 1/3                 Solve for x.

Now you can find the value of y by plugging (1/3) in place of x in either equation.

x + y = 2                                   OR                     2x - y = -1

(1/3) + y = 2                                                      2(1/3) - y = -1

-(1/3)         -(1/3)                                              (2/3) - y = -1

y =       5/3                                                       (-2/3)       (-2/3)

-y = -(5/3)

y = (5/3)

The solution to this system is the ordered pair ( 1/3, 5/3).