Question

1. Show the work involved regardless of the method you use to graph, such as your table of values and the work involved in plugging numbers into the equations, or the slope and y-intercept if you use that method to graph.

y = - x + 3
y = x - 2

Solve Using Elimination

2. 4. 2x + 3y = 12
2x – y = 4

3. 5x – 2y = - 19
2x + 3y = 0

It's worth 99 points so uh... Yeah

Answer 1

Using the elimination method:

`\left \{ {{y=-x+3} \atop {y=x-2}} \right.`

1. Add all members of two equations to each other:

`y=-x+3`

`y=x-2`
----------------------

`2y=1`
[/tex]y=1/2=0.5[/tex]

2. Replace
`y` with found value in first equation to find
`x`:

`0.5=-x+3`

`x=3-0.5=2.5`

So, the solution is:
`x=2.5` and
`y=0.5`



`\left \{ {{2x+3y=12} \atop {2x-y=4}} \right.`

1. Subtract members of second equation from members of first:

`2x+3y=12`

`2x-y=4`
----------------------

`4y=8`
[/tex]y=8/4=2[/tex]

2. Replace
`y` with found value in first equation to find
`x`:

`2x+3*2=12`

`2x=12-6`

`2x=6`

`x=6/2=3`

So, the solution is:
`x=3` and
`y=2`



`\left \{ {{5x-2y=-19} \atop {2x+3y=0}} \right.`

1. Multiply every member of first equation by 3 and every member of second equation by 2:

`5x-2y=-19 | * 3` ⇒
`15x-6y=-57`

`2x+3y=0 | * 2` ⇒
`4x+6y=0`


2. Add all members of two equations to each other:

`15x-6y=-57`

`4x+6y=0`
----------------------

`19x=-57`

`x=-57/19=-3`

3. Replace
`x` with found value in second equation to find
`y`:

`2*(-3)+3y=0`

`-6+3y=0`

`3y=6`

`y=6/3=2`

So, the solution is:
`x=-3` and
`y=2`