# Lars Eighner's Homepage

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calumus meretrix et gladio innocentis

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3x+5y=8 and x-2y=1

1) Multiply the second equation through by 3

(x-2y=1) x 3 = (3x-6x = 3)

2) Subtract from the first equation

```  3x+5y=8
-(3x-6y=3)
_________
11y=5
y=5/11
```

3) Substitute in either equation, in this case, we will use the second

x-2y=1

``` x - 2(5/11) = 1
x = 1+ 10/11
x = 21/11
```

Now check by substituting both values in both equations:

``` 3x+5y=8
3(21/11) + 5(5/11) = 8
63/11 + 25/11 = 8
88/11 = 8

x-2y=1
21/11 - 2(5/11) = 1
11/11 = 1
```

So the values x=21/11 and y=5/11 are the solutions to the equations.

All of this is strictly valid. You multiplied an equation through by 3. It is valid to multiply an equation by a number. You subtracted this equation from the other. It is valid to subtract equal quantities from each side of an equation. You then had one equation in one unknown which you can solve. You substituted this value back into one of the equations to get another equation which you can solve. You substituted both values in both of the original equations to prove your solutions were correct.

### March 19, 2019

Lars

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Lars Eighner
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