Question 639570


let's numbers be {{{x}}} (the first), {{{y}}} (the second) and {{{z}}} (the third)

given:

The sum of three numbers is {{{90}}} .

{{{x+y+z=90}}} ..............1

The first number is {{{6}}} less than the second .

{{{x+6=y}}}.............2

 The third number is {{{2}}} times the second.
 
{{{z=2y}}}..............3

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solve this system:

{{{x+y+z=90}}} .........1
{{{x+6=y}}}.............2
{{{z=2y}}}..............3.....plug in 1
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{{{x+y+2y=90}}} .........1...solve for {{{x}}}

{{{x+3y=90}}}

{{{x=90-3y}}}...........plug in 2
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{{{90-3y+6=y}}}.............2..solve for {{{y}}}

{{{96=y+3y}}}

{{{96=4y}}}

{{{96/4=y}}}

{{{highlight(24=y)}}}....now find {{{x}}}

{{{x=90-3y}}}

{{{x=90-3*24}}}

{{{x=90-72}}}

{{{highlight(x=18)}}}...now find {{{z}}}

{{{z=2y}}}..............3

{{{z=2*24}}}

{{{highlight(z=48)}}}