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Sagot :
Let M be the initial amount of men at the party, and W the initial amount of women.
Since there were 80 more men than women, then:
[tex]M=W+80[/tex]If 60% of men leave the party, then 40% remain. If 35% of women leave the party, then 65% remain.
Then, (40/100)M and (65/100)W remain at the party, Now, there are 13 more women than men. Then:
[tex]\frac{65}{100}W=\frac{40}{100}M+13[/tex]We got a 2x2 system of equations:
[tex]\begin{gathered} M=W+80 \\ \frac{65}{100}W=\frac{40}{100}M+13 \end{gathered}[/tex]Solve the system using the substitution method. Since M is isolated in the first equation, replace M for W+80 into the second equation and solve for W:
[tex]\begin{gathered} \Rightarrow\frac{65}{100}W=\frac{40}{100}(W+80)+13 \\ \\ \text{ Multiply both members by 100} \\ \Rightarrow65W=40(W+80)+1300 \\ \\ \text{ Expand the product on the right member} \\ \Rightarrow65W=40W+3200+1300 \\ \\ \text{ Simplify the right member} \\ \Rightarrow65W=40W+4500 \\ \\ \text{ Subtract 40W from both members} \\ \Rightarrow65W-40W=4500 \\ \\ \text{ Simplify the left member} \\ \Rightarrow25W=4500 \\ \\ \text{ Divide both members by 25} \\ \Rightarrow W=\frac{4500}{25} \\ \\ \text{ Simplify} \\ \Rightarrow W=180 \\ \\ \therefore W=180 \end{gathered}[/tex]Therefore, there were 180 women at first.
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