How do you solve the following system: `5x+8y= -29 , x+3y=3`?

`color(green)(5x)+8y =-29`.............equation `(1)`

`x+3y=3`, multiplying by `5`
`color(green)(5x)+15y=15`.................equation `(2)`

Solving by elimination:
Subtracting equation `2` from `1` would eliminate `color(green)(5x)`

`cancelcolor(green)(5x)+8y =-29`
`-cancelcolor(green)(5x)-15y=-15`

`-7y =-34`

`y =(-44)/(-7)`

`color(green)(y =44/7`

Finding `x` from equation `2`:

`color(green)(5x)+15y=15`

`5x=15 -15y`

`5x=15 -15 xx(44/7)`

`5x=15 -(660/7)`

`5x=105/7 -(660/7)`

`5x=(-555/7)`

`x=(-555/(7xx5))`

`color(green)(x=-111/7`

From the given equations
`5x+8y=-29" "`first equation
`x+3y=3" "`second equation

start with the second equation
Let `x=3-3y` substitute in the first equation

`5x+8y=-29" "`first equation
`5(3-3y)+8y=-29" "`
`15-15y+8y=-29`
`-7y=-29-15`
`-7y=-44`
`y=44/7`

Use `y=44/7` in the second equation

`x=3-3y`
`x=3-3(44/7)`
`x=(21-132)/7`
`x=-111/7`

checking
`5x+8y=-29" "`first equation
`5(-111/7)+8(44/7)=-29" "`

`-555/7+352/7=-29`
`-203/7=-29`
`-29=-29`

checking
`x+3y=3" "`second equation
`(-111/7)+3(44/7)=3" "`
`-111/7+132/7=3`
`21/7=3`
`3=3`

God bless....I hope the explanation is useful..