Question

How do you solve the following system: `3x – 5y = 53 ,5x - 7y = 12`?

Answer 1

Why on earth do they chose such awful numbers????

`x=-77.75`
`y=-57.25`

Explanation:

In solving equations for variable values the method is to manipulate so that you have a load of values and just one variable. It is then solvable. Variables in this case being `x and y`.

`3x-5y=53" "....................Equation(1)`
`5x-7y=12" "....................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for "x)`

Consider equation(1)

To make the -5y positive multiply everything by -1

`-3x+5y=-53`

Add `3x` to both sides

`color(green)(-3xcolor(red)(+3x)+5y=-53color(red)(+3x))`

But `-3x+3x=0`

`0+5y=3x-53`

Divide both sides by 5 ( this is the same as `color(red)(xx1/5)" "`)

`color(green)(5/(color(red)(5))y=3/(color(red)(5))x-53/(color(red)(5))`

But `5/5=1`

`y=3/5x-53/5" ".....................Equation(1_a)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using `Equation(1_a)` substitute for y in `Equation(2)`

`5x-7y=12" "->" "5x-7(3/5x-53/5)=12`

`" "25/5x" "-21/5x+ 371/5=12`

`" "4/5x=12-371/5`

`" "4x=60-371`

`" "color(blue)(x= -311/4)`

`color(blue)("This is the same as "x=-77.75)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving for "y)`

Substitute for `x` in `Equation(2)`
It does not matter which of the 2 equations you use.

`5x-7y=12" "->" "5(-311/4)-7y=12`

`" "-1555/4-7y=12`

`" "-1555/4-28/4y=48/4`

`" "28y=-1555-48`

`" "y= -229/4`

`color(blue)("This is the same as "y=-57.25)`

Answer 2

`x = -77.75 and y = -57.25`

Explanation:

There are several methods for solving a system of equations, also known as simultaneous equations. The method you choose will depend on the format of the equations you are given.

• Elimination - [make additive inverses and add the equations]
• Substitution - [write one variable in terms of the other]
• Equating - [one variable is expressed in two different ways]
• Matrices - [ multiply a matrix by its inverse to get the unit matrix]
• Graphically - [ graph each equation and find the intersection]

In this example I will use the elimination method.

`color(white)(.....................)3x -5y = 53color(white)(.............)A`
`color(white)(.....................)5x -7y = 12color(white)(.............)B`

`A xx 5:" "color(red)(15x)-25y = 265color(white)(..........)C`
`B xx"-3: "color(red)(-15x)+21y = -36color(white)(.......)D`

`C+D:color(white)(.............)-4y = 229" "` there is no x-term!
`color(white)(...............................)color(blue)(y = -57.25)`

Now that you have value for `color(blue)(y)`, substitute it into any of the equations above to find `x`

`color(white)(.....................)5x -7color(blue)(y) = 12color(white)(.............)B`

`color(white)(.............)5x -7color(blue)(("-57.25")) = 12`
`color(white)(...................)5x color(blue)("+400.75") = 12`
`color(white)(....................)5x " "= -388.75`
`color(white)(......................)x " "= -77.75`
.

[Note that: `color(red)(15x) and color(red)(-15x)` are additive inverses. They therefore ADD to give 0.]

This was done by multiplying B by `-3` to change the sign.