Question 673701


given:

One number {{{x}}} is {{{8}}}{{{ more}}} than {{{3}}}{{{ times}}} another number {{{y}}} ...=>... {{{x=3y+8}}}.......1

their sum is {{{x+y=68}}}........2

solve the system:

{{{x=3y+8}}}.......1....plug {{{x}}} value in eq. 2

{{{x+y=68}}}........2
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{{{3y+8+y=68}}}........2...solve for {{{y}}}

{{{4y=68-8}}}

{{{4y=60}}}

{{{y=60/4}}}

{{{highlight(y=15)}}}

now find {{{x}}}

{{{x=3y+8}}}.......1

{{{x=3*15+8}}}

{{{x=45+8}}}

{{{highlight(x=53)}}}