Question

How do you solve the following system: `3x+y=8, 2x +5y = 5`?

Answer 1

The more formal layout approach.

This `ul("clearly")` shows the thinking processes and is easy to follow for teacher marking.

I have determined the value of `x` to start you off.

Explanation:

Given:
`3x+y=8" "......................Equation(1)`
`2x+5y=5" "...................Equation(2)`

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`color(blue)("This bit is not part of the formal approach.")`

We need one equation with just one unknown.

I am using method by example: known that `color(green)(2=2) color(red)(larr" True"`

Multiply both sides by `color(purple)(3)`

`color(green)(ubrace([2color(purple)(xx3)]) = ubrace([2color(purple)(xx3)])`
`color(green)( color(white)(".d") darr color(white)("dddd.d")darr )`
`color(green)(color(white)("dd") 6color(white)("dd") =color(white)("dd") 6 color(red)(larr" Still true"))`
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`color(blue)("Back to the solution.")`

Multiply both sides of `Eqn(1)` by 5 `color(white)("d")` ( 'to get rid' of the `y`)

`15x+5y=40" "........Equation(1_a)`

`Eqn(1_a) - Eqn(2)`

`15x+5y=40`
`ul(color(white)(1)2x+5y=color(white)(4)5larr" Subtract")`
`13x+0color(white)("d") =35`

Divide both sides by 13

`(13x)/13=35/13`

`x=35/13`

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`color(blue)("Your bit")`

Substitute this into either `Eqn(1) or Eqn(2)` to determine the value of `y`

I will let you do that