```Question 118890

the general rule for solving this kind of equations is:

if you have to find {{{one}}}{{{ variable }}} you need {{{one}}}{{{ equation}}}
if you have to find {{{two}}}{{{ variables }}} you need {{{two}}}{{{ equations}}}
if you have to find {{{three}}}{{{ variables}}} you need {{{three equations}}}

:and so on
:
:
if you have to find {{{n}}}{{{ variables}}} you need {{{n}}}{{{ equations}}}

If you compare of equations with{{{ one }}} and {{{two}}} variables, they have similarities such as:

both of them have {{{an}}}{{{ unknown}}} number, which you need to get

in order to do it, you will need to use algebraic {{{formulas}}}, {{{multiplication}}}, {{{division}}}, {{{addition}}}, or {{{subtraction}}}

The {{{difference}}} between them is making {{{the }}}{{{ways}}} in which they are treated to get the unknown number.

Many problems can be solved quickly and easily using {{{one }}}{{{equation}}} with {{{one }}}{{{variable}}}.

Other problems that might be difficult to solve in terms of {{{one }}}{{{variable}}} can easily be solved using {{{two}}}{{{ equations}}} and {{{two}}}{{{ variables}}}.

the following example shows the difference in the {{{two}}}{{{ methods}}}; solved first by using one variable and then using two:

1. method

Find the two numbers such that half the first equals a third of the second and twice their sum exceeds three times the second by {{{4}}}.

{{{x}}} is first number

Then

{{{x/2=1/3}}} of the second number=> {{{x=2/3}}}

{{{x=2/3}}}.multiply both sides by {{{3/2}}}

{{{(3/2)x=(3/2)(2/3)}}}.

{{{3x/2=1}}}.

{{{3x/2}}}.is second number

twice their sum exceeds three times the second by {{{4}}}

{{{2(x  + 3x/2)  = 3(3x/2) + 4 }}}

{{{2x  + 3x = 9x/2 + 4 }}}

{{{5x  = 9x/2 + 4 }}}multiply both sides by {{{2}}}

{{{10x  = 9x + 8 }}}

{{{10x -  9x = 8}}}

{{{x = 8}}}

{{{8}}}is first number

{{{3x/2=3*8/2=3*4=12}}}.

{{{12}}} is second number

2. method

If we let {{{x}}} and {{{y}}} be the first and second numbers, respectively, we can write {{{two}}} equations:

{{{x/2=y/3}}}.(1)

{{{2(x+y) = 3*y + 4}}}.(2)

now solve for {{{x}}} first equation, and substitute this value in the second:

{{{x/2=y/3}}}

{{{x =2(y/3)}}}

{{{x = (2y/3) }}}

{{{2(2y/3 + y) = 3y + 4}}}.(2)

{{{(4y/3) +2y = 3y + 4}}}multiply both sides by {{{3}}}

{{{4y + 6y = 9y + 12}}}

{{{ 10y = 9y + 12}}}

{{{ 10y - 9y = 12}}}

{{{ y = 12}}}

{{{x =2(y/3)}}}

{{{x =2(12/3)}}}

{{{x =2(4)}}}

{{{x = 8 }}}

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