Question 148736
1.{{{x+y=60}}} 
2.{{{.03x=.60}}}
Your equations are correct. 
You could have solved eq. 2 directly to get x=..., then substitute back into 1 to solve for y.
{{{0.03x=0.60}}}
{{{x=20}}}
{{{y=60-x}}}
{{{y=40}}}
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.
.
Let's see where you went wrong. 
{{{x=60-y}}}
{{{.03x = .60}}}
{{{.03(60-y)=.60}}}
{{{1.80-highlight(y)=.60}}}<---- Here's where you went wrong.
You didn't distribute 0.03 through the entire expression.
Here's what it should have been.
{{{.03(60-y)=.60}}}
{{{.03(60)-0.03y=.60}}}
{{{1.8-0.03y=.60}}}
{{{-0.03y=-1.2}}}
{{{highlight(y=40)}}}
{{{x+y=60}}}
{{{highlight(x=20)}}}
.
.
.

You could have also done it this way to simplify.
Divide by 0.03 first, no need to distribute.
{{{.03(60-y)=.60}}}
{{{60-y=20}}} Divide both sides by .03.
{{{y=40}}}