# Solve using substitution method. 4x-y=5 4x+4y=-4 solve the linear system from the problem using elimination.

Let,

EQ1: `4x-y=5 `                         EQ2: `4x+4y=-4`

Note that we may re-write EQ2 as:

`4x=-4-4y`

Then, substitute this to EQ1.

`4x-y=5`

`-4-4y-y=5`

`-4-5y=5`

Isolate -5y.

`-5y=5+4`

`-5y=9`

Then, isolate y.

`y=-9/5`

Substitute the value of y to  4x=-4-4y and...

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Let,

EQ1: `4x-y=5 `                         EQ2: `4x+4y=-4`

Note that we may re-write EQ2 as:

`4x=-4-4y`

Then, substitute this to EQ1.

`4x-y=5`

`-4-4y-y=5`

`-4-5y=5`

Isolate -5y.

`-5y=5+4`

`-5y=9`

Then, isolate y.

`y=-9/5`

Substitute the value of y to  4x=-4-4y and solve for x.

`y=-9/5` ,   `4x=-4 -4(-9/5) = -4+36/5= 16/5`

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

Hence, the solution to the given system of equations is `x=4/5` and `y=-9/5` .

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