Question 732496

Two numbers, {{{x}}} and {{{y}}} have the ratio {{{3 : 4}}}. 

{{{x:y=3:4}}}......eq. 1

If {{{4}}} is added to each of the numbers the resulting ratio is {{{4 : 5}}}.

{{{(x+4):(y+4)=4:5}}}.....eq. 2

solve the system:

{{{x:y=3:4}}}......eq. 1

{{{(x+4):(y+4)=4:5}}}.....eq. 2
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{{{x:y=3:4}}}......eq. 1...solve for {{{x}}}


{{{4x=3y}}}

{{{x=3y/4}}}.........substitute in eq. 2


{{{(3y/4+4):(y+4)=4:5}}}.....eq. 2..solve for {{{y}}}

{{{5(3y/4+4)=4(y+4)}}}


{{{5((3y+4*4)/4)=4(y+4)}}}

{{{5((3y+16)/4)=4y+16}}}

{{{(15y+80)/4=4y+16}}}

{{{(15y+80)=4*4y+4*16}}}

{{{15y+80=16y+64}}}

{{{15y-15y+80=16y-15y+64}}}

{{{80=y+64}}}


{{{80-64=y+64-64}}}

{{{highlight(16=y)}}}

go back to {{{x=3y/4}}} and find {{{x}}}

{{{x=3*16/4}}}

{{{x=3*cross(16)4/cross(4)}}}

{{{x=3*4}}}

{{{highlight(x=12)}}}


check:

{{{x:y=3:4}}}......eq. 1
{{{12:16=3:4}}}
{{{0.75=0.75}}}

{{{(x+4):(y+4)=4:5}}}.....eq. 2
{{{(12+4):(16+4)=4:5}}}
{{{16:20=4:5}}}
{{{0.8=0.8}}}