Because 50% of something means half (because 100-50=50), he would have to make the number of girls and boys equal. Since there are 21 girls, he has to make it so there are also 21 boys.
So, if we know that 21-15=6, this means he would have to add 6 boys to the club if he wants the girls and boys equal.
In each case, you can use the second equation to create an expression for y that will substitute into the first equation. Then you can write the result in standard form and use any of several means to find the number of solutions.
System A
x² + (-x/2)² = 17
x² = 17/(5/4) = 13.6
x = ±√13.6 . . . . 2 real solutions
System B
-6x +5 = x² -7x +10
x² -x +5 = 0
The discriminant is ...
D = (-1)²-4(1)(5) = -20 . . . . 0 real solutions
System C
y = 8x +17 = -2x² +9
2x² +8x +8 = 0
2(x+2)² = 0
x = -2 . . . . 1 real solution
4.125 x 14 = 57.75
So 57.75 is the answer and I have to show my work
