Question 199118
Let x=# of boys and y=# of girls



Since there's only a choice between boys or girls, this means that there is a total of {{{x+y}}} students. Also, because "on the first day of school, 60% of the class were boys", this means that {{{0.6(x+y)=x}}} (ie take 60% of the total and you get "x")



Also, since "During the school year six girls moved away and 6 boys replaced them." and "At the end of the year, 75% of the class were boys", we know that {{{0.75((x+6)+(y-6))=x+6}}} where "x+6" is the new count for the boys and "y-6" is the new count for the girls.



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{{{0.6(x+y)=x}}} Start with the first equation.



{{{0.6x+0.6y=x}}} Distribute



{{{10(0.6x)+10(0.6y)=10x}}} Multiply EVERY term by 10 to make every number a whole number.



{{{6x+6y=10x}}} Multiply



{{{6x+6y-10x=0}}} Subtract 10x from both sides.



{{{-4x+6y=0}}} Combine like terms. So this is the new equation #1.



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{{{0.75((x+6)+(y-6))=x+6}}} Start with the second equation.



{{{0.75(x+6+y-6)=x+6}}} Remove the inner parenthesis.



{{{0.75(x+y)=x+6}}} Combine like terms.



{{{0.75x+0.75y=x+6}}} Distribute



{{{100(0.75x)+100(0.75y)=100(x)+100(6)}}} Multiply EVERY term by 100 to make every number a whole number.



{{{75x+75y=100x+600}}} Multiply. 



{{{75x+75y-100x=600}}} Subtract 100x from both sides.



{{{-25x+75y=600}}} Combine like terms. So this the new equation #2.








So we have the system of equations:


{{{system(-4x+6y=0,-25x+75y=600)}}}



Let's solve this system by use of substitution.



In order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.





So let's isolate y in the first equation


{{{-4x+6y=0}}} Start with the first equation



{{{6y=4x}}} Add {{{4x}}} to both sides



{{{y=(4x)/(6)}}} Divide both sides by {{{6}}}



{{{y=(2/3)x}}} Reduce




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Since {{{y=(2/3)x}}}, we can now replace each {{{y}}} in the second equation with {{{(2/3)x+0}}} to solve for {{{x}}}




{{{-25x+75highlight(((2/3)x))=600}}} Plug in {{{y=(2/3)x+0}}} into the second equation. In other words, replace each {{{y}}} with {{{(2/3)x+0}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{-25x+(150/3)x=600}}} Multiply.



{{{3(-25x)+cross(3)((150/cross(3))x)=3(600)}}} Multiply EVERY term by the LCD 3. This will eliminate the fractions.



{{{-75x+150x=1800}}} Multiply.




{{{75x=1800}}} Combine like terms on the left side



{{{x=(1800)/(75)}}} Divide both sides by 75 to isolate x



{{{x=24}}} Divide



-----------------First Answer------------------------------



So the first part of our answer is: {{{x=24}}}. 


This means that there are 24 boys (at the beginning of the school year)






Since we know that {{{x=24}}} we can plug it into the equation {{{y=(2/3)x}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(2/3)x}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(2/3)(24)}}} Plug in {{{x=24}}}



{{{y=48/3}}} Multiply



{{{y=16}}} Reduce.  




-----------------Second Answer------------------------------



So the second part of our answer is: {{{y=16}}} This means that there are 16 girls (at the beginning of the school year)






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Summary:



So our answers are: {{{x=24}}} and {{{y=16}}}



This means that at the beginning of the school year, there are 24 boys and 16 girls. At the end of the school year, there are 30 boys and 10 girls.