# solve algebraically 2x +y =2 3x + 2y=5

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Solve the system 2x+y=2,3x+2y=5 algebraically:

(1) Method I; substitution:

Solve the first equation for y : 2x+y=2

Subtract 2x from each side to get y=-2x+2

Now substitute this expression for y into the second equation:

3x+2y=5

3x+2(-2x+2)=5 solve for x:

3x-4x+4=5

-x=1 ==> x=-1

**If x=-1 then y=-2(-1)+2=4**

**So the solution is x=-1,y=4 or the point (-1,4).**

Check: 2(-1)+4=2 and 3(-1)+2(4)=5.

(2) Method II; elimination:

2x+y=2

3x+2y=5 Choose to eliminate y; multiply the first equation by -2 :

-4x-2y=-4

3x+2y=5 Now add the two equations together:

x+0=-1 ==> x=-1 and proceed as above.

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The solution to the system is (-1,4)

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**Sources:**

2x+y=2(multiply by 2)

4x+2y=4(equation 1)

3x+2y=5(equation 2)

subtract equation 2 from equation 1

x=-1

put value of x in equation 2

3(-1)+2y=5

-3+2y=5

2y=5+3

2y=8

y=8/2

y=4

(x,y)=(-1,4)

2x+y=2

3x+2y=5

In order to solve it algebraically the **method of substitution** can be used.

First the equation 2x+y=2 is to be considered.

2x+y=2

y=2-2x

Insert the above equation in the other one,

3x+2y=5

3x+2(2-2x)=5

Remove brackets and multiply both by 2

3x+4-4x=5

-x+4=5

-x=5-4

-x=1

**x=-1**

Than input the value of x in any one of the above equations to find the value of y:

2x+y=2

2(-1)+y=2

-2+y=2

y=2+2

**y=4**

**Therefore, (-1,4)**

this can be proved by inserting both into any one of the equations:

3x+2y=5

3(-1)+2(4)=5

-3+8=5

5=5 , Therefore proved.

**2x +y =2** and **3x + 2y=5**

2x + y = 2

y = 2 - 2x

3x + 2y = 5

3x + 2(2-2x) =5

3x + 4 - 4x = 5

3x - 4x = 5 - 4

-x = 1

**x = -1 **

y= 2 - 2x

y= 2 - 2(-1)

y = 2 - (-2)

y = 2 + 2

**y = 4**

**Answer: (-1, 4)**

2x + y = 2

3x + 2y = 5

To solve this equation , first multiply everything in the first equation by 2

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

**4x + 2y = 4**

**3x + 2y = 5 **now subtract equation 1 with equation two

By subtracting the two equation , you should get

**x = -1 **this is your answer for " x " , now we have to plug x into one of the equation . Let's plug it into 2x + y = 2

By plugging " x " into the equation , you should get

**2 ( -1 ) + y = 2 **now multiply 2 with -1

By multiplying , you should get

**-2 + y = 2 **now add 2 on both sides

By adding 2 on both sides , you should get

**y = 4 **which is your answer for " y "

So your answer is , **( -1 , 4 ) **

2x +y =2

y = 2 - 2x

now plug in this solution as y into the other problem:

3x + 2(2-2x)=5

take care of the parentheses by distributing the 2 to the numbers inside of the parentheses:

3x + 4 - 4x = 5

move like terms to the same side:

3x - 4x = 5 - 4

combine like terms:

-x = 1

**x = -1**

now plug in x into any one of the equations to get y:

2(-1) +y =2

-2 + y = 2

move like terms:

y = 2 + 2

**y = 4**

so (-1, 4)

To solve the set 2x +y =2 and 3x + 2y=5, write y in terms of x from 2x +y =2 and substitute in 3x + 2y=5. 2x + y = 2 gives y = 2 - 2x

Putting this in 3x + 2y=5 gives 3x + 2(2 - 2x) = 5

=> 3x + 4 - 4x = 5

=> -x = 1

=> x = -1

As y = 2 - 2x, y = 4

Simplify

2x+y=2 3x+2y=5

-y -y -2y -2y

2x=-y+2 3x=-2y+5

Divide both sides by 2 Divide both sides by 3

x=-1/2y+1 x=-2/3y+5/3

Then Substitution