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

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### 5 Answers

First, rearrange one of the equations to solve for a single variable.

(2x + y = 2)

rearranges to:

(y = 2-2x)

THen we can substitute the value (2-2x) for y in the second equation.

(3x + 2y = 5)

substitute (2-2x) for y:

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

multiply and simplify:

(3x + 4 - 4x = 5)

(-x + 4 = 5)

(-x = 1)

therefore **x = -1**

Now we just apply the X value to either equation.

(-2 + y = 2)

**y = 4**

We can check by applying these values to the second equation.

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

(-3 + 8 = 5)

Another way of solving the system of equations

2x + y = 2 ...(1)

3x +2y = 5 ...(2)

is using the elimination method.

First, let's determine x by eliminating y.

2*(1) - (2)

4x + 2y - 3x - 2y = 4 - 5

x = -1

Now, determine y by eliminating x.

3*(1) - 2*(2)

6x + 3y - 6x - 4y = 6 - 10

-y = -4

y = 4

The solution of the given system of equations is x = -1 and y = 4

2x + y = 2 ------eq(i)

3x + 2y = 5 ------eq(ii)

We can solve this through the method of substitution, let us consider eq(i)

2x + y = 2 ------eq(i)

y = 2 - 2x

Input this value in eq(ii)

3x + 2y = 5 ------eq(ii)

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

3x + 4 - 4x = 5

-x + 4 = 5

-x + 4 - 4 = 5 - 4

-x = 1

x = -1

To find the value of y input the value of x in eq(i)

2x + y = 2 ------eq(i)

2(-1) + y = 2

-2 + y = 2

-2 + 2 + y = 2 + 2

y = 4

Therefore,

**x = -1**

**y = 4 Answer.**

**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)**

Solve for a single variable first. You can choose which, but y is easier to do.

2x+y=2

y=2-2x

No input the 2-2x into the second equation for y.

3x+2y=5

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

Distribute

3x+4-4x=5

Combine like terms and solve

-1x=1

**x=-1**

Now input x back into the first equation that you rearranged to solve for y.

y=2-2x

y=2-2(-1)

**y=4**