Question 335088
Let the first number be A, second number B.
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"three times a first number is decreased by a second number is 1"
1.{{{3A-B=1}}}
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"the first number increased by twice the second number is 12"
2.{{{A+2B=12}}}
From eq. 1,
{{{3A-B=1}}}
{{{B=3A-1}}}
Substitute into eq. 2 for B.
{{{A+2B=12}}}
{{{A+2(3A-1)=12}}}
{{{A+6A-2=12}}}
{{{7A=14}}}
{{{highlight(A=2)}}}
Now go back and solve for B.
{{{B=3A-1}}}
{{{B=3(2)-1}}}
{{{B=6-1}}}
{{{highlight(B=5)}}}