# Solve this linear system by elimination: 4x-y=5 4x+4y=-4

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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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4x - y = 5 ------eq(i)

4x + 4y = -4 ------eq(ii)

The first step in the method of elimination is to subtract eq(i) from eq(ii), after doing this we will be left with one unknown variable, with the help of which we can find the value of the other.

(4x - y) - (4x + 4y) = 5 - ( -4)

4x - y - 4x - 4y = 9

-5y = 9

-5y/5 = 9/-5

**y = -9/5**

Now insert this value in either one of the equations to find the value of x,

4x - y = 5

4x - (-9/5) = 5

4x + 9/5 = 5

4x + 9/5 - 9/5 = 5 - 9/5

4x = 25/5 - 9/5

4x = 16/5

4x/4 = (16/5)/4

x = 16/20

**x = 4/5 **

Now insert both values in eq(ii) to verify the answer,

4x + 4y = -4

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

16/5 - 36/5 = -4

-20/5 = -4

-4 = -4

LHS = RHS

Proved.