Question

How do you solve the system of equations: `3x-4y=13` and `2x+3y=3` ?

Answer 1

`(x,y)=(3,-1)`

Explanation:

`E1: 3x-4y=13`
`E2: 2x+3y=3`

I'll multiply `E1` by 2 and `E2` by 3, then subtract the two (to eliminate the `x` terms):

`2E1: 6x-8y=26`
`3E2: ul(6x+9y=9)`
`color(white)(0E3) color(white)(0x)-17y=17`

`:. color(blue)(ul(abs(bar(color(black)(y=-1`

Substitute this into one of the original equations:

`E1: 3x-4y=13`

`E1: 3x-4(-1)=13`

`E1: 3x+4=13`

`E1: 3x=9`

`:. color(blue)(bar(ul(abs(color(black)(x=3`

And here's the graph to show it:

graph{(3x-4y-13)(2x+3y-3)=0}

Answer 2

`x=3" "` and `" "y = -1`

Explanation:

Given:

`{ (3x-4y=13), (2x+3y=3) :}`

Multiplying the first equation by `3` and the second by `4`, we get:

`{ (9x-12y=39), (8x+12y=12) :}`

Adding these two equations together, we get:

`17x=51`

Divide both sides by `17` to get:

`x=3`

Substituting `x=3` into the original second equation, we get:

`6+3y=3`

Subtract `6` from both sides to get:

`3y=-3`

Divide both sides by `3` to get:

`y = -1`