Question

How do you solve the system of equations `3x + y = 16` and `2x + y = 11`?

Answer 1

`x=5`
`y=1`

Explanation:

Solving simultaneously :
`color(lightblue)((1)` `color(lightblue)rarr` `3x+y=16`
`color(lightgreen)((2)` `color(lightgreen)rarr` `2x+y=11`

Multiply `color(lightgreen)((2)` by `-1`:
`color(white)(xxxxx)3xcancel(+y)=+16`
`color(white)(xxx)-2xcancel(-y)=-11`

Add `color(lightblue)((1)` & `color(lightgreen)((2)`:
`color(white)(xxxxx)color(red)(x=5)`

*Substitute the value of `color(red)x` in `color(lightgreen)((2)`:
`2(5)+y=11`

`color(red)(y=1)`

Answer 2

`x = 5 and y=1`

Explanation:

The methods shown by other contributors are the elimination method and the graphical method.

Another option is the method of equating variables.

This is particularly useful for finding the point of intersection of two straight lines algebraically.

In the two equations, notice that they have the SAME y-TERM

`3x color(red)(+y) = 16" and "2x color(red)(+y) = 11`

Make `color(red)(y)` the subject in each case by re-arranging the equations.

`color(red)(y) = 16-3x " and "color(red)(y) = 11-2x`

We know that the value of `" "color(red)(y)" "` will be the same in each equation.

`color(white)(xxxxxxxxxxxx)color(red)(y = y)`

`color(white)(xxx xxx):.11-2x = 16-3x" "larr` only x terms.

`color(white)(xxx xxx):.3x-2x = 16-11" "larr` re-arrange terms

`color(white)(xxxxxxxxxx):.x = 5`

There are now 2 equations to use to find `y`

Substitute `x = 5` into both to check that you get the same answer.

`y= 16-3(5) = 16-15 =1`

`y = 11-2(5) = 11-10 = 1`
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Another option:

Often we forget how useful it is to just look at equations first and we set off doing calculations instead of thinking first.

In `3x +y = 16" and " 2x+y = 11`,

the left hand sides only differ by `1x`
the right hand sides differ by 5.

`x` must be be 5.

If you know the value of `x` you can work out `y` by inspection.