Question 323496
Let T be the amount of 30% solution and E be the amount of 80% solution,
1.{{{T+E=10}}}
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{{{30T+80E=50(10)}}}
2.{{{3T+8E=50}}}
From eq. 1,
{{{T=10-E}}}
Substitute into eq. 2,
{{{3(10-E)+8E=50}}}
{{{30-3E+8E=50}}}
{{{5E=20}}}
{{{E=4}}}
Then from eq. 1,
{{{T+4=10}}}
{{{T=6}}}
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6 ounces of 30% solution and 4 ounces of 80% solution.