`i. 4x-y = 5`

`ii. 4x+4y = -4`

Multiply equation i. by -1 making the x-coefficients opposites.

`i. -4x+y=-5`

`ii. 4x+4y = -4`

Add the 2 equations together.

`5y = -9`

`y = -9/5`

Now substitute `-9/5` for y in either of the original equations to solve for x.

`4x...

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`i. 4x-y = 5`

`ii. 4x+4y = -4`

Multiply equation i. by -1 making the x-coefficients opposites.

`i. -4x+y=-5`

`ii. 4x+4y = -4`

Add the 2 equations together.

`5y = -9`

`y = -9/5`

Now substitute `-9/5` for y in either of the original equations to solve for x.

`4x - (-9/5) = 5`

`4x+ (9/5) = 5`

`4x = 5 - 9/5`

`4x = 25/5 - 9/5`

`4x = 16/5`

`x = 16/20 = 4/5`

**The solution to the system is `(4/5, -9/5).`**

Given the system of two equations with two unknowns:

`4x-y=5` (1)

`4x+4y=-4` (2)

Solve by elimination.

Multiply (1) by -1 and add it to (2).

`-4x+y=-5`

`4x+4y=-4`

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

`5y=-9` (3)

Divide (3) by 5 to solve for y.

`y=-9/5`

Substitute `-9/5` for y in (1).

`4x-(-9/5)=5`

`4x+9/5=5` (4)

Solve (4) for x: subtract 9/5 to both sides.

`4x=5-9/5=16/5`

Divide both sides by 4.

`x=4/5`

Verify by substituting 4/5 for x and -9/5 for y in (2).

`4(4/5)+4(-9/5)=-4`

`16/5-36/5=-4`

`-20/5=-4`

`-4=-4`

**Thus the solutions are x=4/5 or x=0.8 and y=-9/5 or y=-1.8**

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