Question

How do you solve the system of equations `2x-y=-5` and `y=-x+1`?

Answer 1

`(x,y)=(-4/3,7/3)`

Explanation:

Given
[1]`color(white)("XXX")2x-color(red)(y)=-5`
[2]`color(white)("XXX")color(red)(y)=color(red)(-x+1)`

Using [2] to substitute for `color(red)(y)` in [1]
[3]`color(white)("XXX")2x-color(red)((-x+1))=-5`

[4]`color(white)("XXX")3x-1=-5`

[5]`color(white)("XXX")3x=-4`

[6]`color(white)("XXX")color(blue)(x)=color(blue)(-4/3)`

Using [6] to substitute for `color(blue)(x)` in [2]`color(white)("XXX")y=-color(blue)(x)+1`

[7]`color(white)("XXX")y=-color(blue)((-4/3))+1`

[8]`color(white)("XXX")y=7/3`

Answer 2

`x = -4/3 and y = 7/3`

In this case, fraction form is better than decimal form because the decimals recur and require rounding, so the answer is not accurate.

Explanation:

This is a nice example to compare 3 different methods of solving this system of equations - also called "simultaneous equations"
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Elimination Method : If you have `color(red)" additive inverses"` they are easily eliminated by adding the two equations together.

`2x - y = -5" and "color(blue)(y = -x +1)" "larr` re-arrange
`color(white)(xxxxxxxxxvvvvvvxxxx) color(blue)(x+y = 1)`

`color(white)(xxxxxxx)2x color(red)(- y) = -5` ...........................................A
`color(white)(xxxxxxx,)x color(red)(+ y) = " "1` ............................................B

`A+B:color(white)(xx)3x" " = -4" "larr` y terms are eliminated
`color(white)(xxxxxxxx)x " "= -4/3" "larr` solve for x

Substitute `x= -4/3` into `y = -x +1`

`rarr y = -(-4/3)+1 " "rarr y = 7/3`
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Substitution method
This is indicated by a 'single variable' in one of the equations.
In this case there is a single variable in each.. choose one.

`2x-color(deeppink)(y) = -5" and "color(deeppink)(y = (-x+1))`

Replace `color(deeppink)(y) " by " color(deeppink)((-x+1))`

`2x-color(deeppink)((-x+1)) = -5" "larr` no y, solve for x

`2x+x-1= -5`
`3x = -4 " " rarr x = -4/3`
`y = -x+1rarr y = -(-4/3)+1 " "rarr y = 7/3`
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Equating method
Note that both equations have a `y` term.

`color(blue)(y = 2x + 5)" and " color(red)(y = -x +1)" "larr` make 'y' the subject

Now, as `color(white)(xxxxxxx)color(blue)(y)= color(red)(y)" "` it follows that

`color(white)(xxxx.x.xx)color(blue)(2x+5) = color(red)(-x+1) " "larr` solve for x

`color(white)(xxxxxxx.x.xx)3x = -4`

`color(white)(xxxx.xx.x.xxx)x = -4/3`

`color(blue)(y = 2x + 5) = 2(-4/3) +5" "rarr y = 7/3`
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