Question 1151666


Let '{{{a}}}' be the amount of solution {{{A}}} (in liters)

Let '{{{b}}}' be the amount of solution {{{B}}} (in liters)

{{{Solution}}} | {{{Concentration}}} | {{{Acetic}}}{{{ Acid}}}

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... {{{a}}} ...... | ...... {{{8}}}% ......... | {{{0.08a}}}

... {{{b}}} ...... | ...... {{{20}}}% ....... | {{{0.20b}}}

Final solution:

... {{{4}}} ..... | ...... {{{11}}}% ........ | {{{4(0.11) = 0.44}}}

Write two equations:

{{{a + b = 4}}}

{{{0.08a + 0.20b = 0.44}}}

Multiply the last equation by {{{100 }}}to get rid of decimals:

{{{8a + 20b = 44}}}

Divide everything by {{{4}}} to simplify:

{{{2a + 5b = 11}}}

Solve the first equation for one variable, say {{{a}}}:

{{{a = 4 - b}}}

Substitute into the last equation:

{{{2(4 - b) + 5b = 11}}}

{{{8 - 2b + 5b = 11}}}

{{{3b = 11 - 8}}}

{{{3b = 3}}}

{{{b = 3/3}}}

{{{b = 1}}}

Then solve for{{{ a}}}:

{{{a = 4 - 1}}}

{{{a = 3}}}