40 boys and 68 girls
let's start with variables and an equation.
boys: x
girls: 2x - 12
and we know that boys and girls make up the class. so let's add the two expressions.
x + 2x - 12 = 108
3x - 12 = 108
+ 12 + 12
3x = 120
x = 40 boys
BUT x is only the number of BOYS in the class. so we have to find girls now. let's plug in our value of x into the girls' equation.
2x - 12 = # of girls
2(40) -12 = # of girls
80 - 12 = # of girls
68 girls
let's check our answer!
68 + 40 does in fact add to 108, therefore our answer is correct.
This question is worded slightly strangely but I believe I understand it.
To eliminate x, multiply the second equation by 8, so that it becomes 4x + .8y= 96
Then you can subtract 4x on the top by 4x on the bottom to eliminate it.
Given
we have
Squaring both sides, we have
And finally
Note that, when we square both sides, we have to assume that
because we're assuming that this fraction equals a square root, which is positive.
So, if that fraction is positive you'll actually have roots: choose
and you'll have
Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose
and you'll have
Squaring both sides (and here's the mistake!!) you'd have
which is not a solution for the equation, if we plug it in we have
Which is clearly false
Outlier would be 25 because is the number that is the farthest away from the other numbers
