Question

How do you solve the system of equations `16x - 5y = - 33` and `16x + y = - 51`?

Answer 1

`x=-3` and `y=-3`

Explanation:

We have two equations `16x−5y=−33` and `16x+y=−51`,

now let us subtract first equation from second. We get

`16x+y-(16x-5y)=−51-(-33)`

or `16x+y-16x+5y=−51+33=-18`

i.e. `6y=-18` or `y=-3`

Putting this in second equation

`16x-3=-51` or `16x=3-51=-48`

or `x=-48/16=-3`

Hence, `x=-3` and `y=-3`

Answer 2

`y=-18/5`
`color(white)(.)`

`x= -237/80`

Explanation:

There are at least 2 ways of doing this. The thing is to always try and spot the most efficient approach (less work).

Notice that both equations have `16x` so I am going to use the time tested method of subtraction. As they both have `16x` there is no need to make any adjustments. Thus we have:

`color(blue)("Determine the value of "y)`

`16x+color(white)(.)y=-51`
`ul(16x-5y=-33) larr" Subtract"`
`color(white)(1)0x+5y=-18`

Divide both sides by 5

`y=-18/5`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determine the value of "y)`

Substitute `y=-18/5" into "16x+y=-51` giving:

`16x+(-18/5)=-51`

`=>16x" "=" "-51+18/5" " =" " -47 2/5" " =" "-237/5`

`=>x" "=" "-237/(5xx16)" "=" "2 77/80 = -237/80`

Answer 3

`x=-3 and y= -3`

Explanation:

Solving the system of two unknowns is determined by eliminating
`" "`
one unknown then find the second.
`" "`

Name the equations:

`16x - 5y = -33` `" "` `color(red)(EQ1)`
`" "`
`16x + y = -51` `" "` `color(red)(EQ2)`

Method:
`" "`
`" "`

First:
`" "`
Multiply one of the equations by an integer to obtain opposite
`" "`
coefficients in both equations of the same unknown.

in the given system we will multiply `" "color(red)(EQ1)" "` by`" "-1" "` it is:
`" "`
`-16x + 5y = +33` `" "` `color(green)(EQ1)`
`" "`
`" "`

Second:
`" "`
Add the two equations to find the first unknown :

`color(green)(EQ1)+ color(red)(EQ2)`
`" "`
`rArr-16x + 5y +16x + y = +33 -51`
`" "`
`rArrcancel(-16x )+ 5ycancel( +16x) + y = +33 -51`
`" "`
`rArr6y=-18`
`" "`
`rArry=-18/6`
`" "`
`rArry=-3`
`" "`
`" "`
Third:
`" "`
Substitute the computed unknown in the second equation to find
`" "`
the second unknown;
`" "`
In the given system we will substitute`" "y=- 3" "` to find the
`" "`
value of `" "x" "`.
`" "`
`" "`
`16x + y = -51` `" "` `color(red)(EQ2)`

`rArr 16x +(-3) = - 51`
`" "`
`rArr 16x -3 = -51`
`" "`
`rArr 16x = -51+3`
`" "`
`rArr 16x = -48`
`" "`
`rArr x = -48/16`
`" "`
`rArr x = -3`
`" "`
`" "`
Therefore;
`" "`
`x = -3 and y=-3`