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Question Number 85461 by oustmuchiya@gmail.com last updated on 22/Mar/20

Commented by jagoll last updated on 22/Mar/20

$(2) the line x = 5 - 2 y$ $intersect the curve y^{2} + xy = 6$ $\Rightarrow y^{2} + y (5 - 2 y) = 6$ $- y^{2} + 5 y = 6 \Rightarrow y^{2} - 5 y + 6 = 0$ $(y - 3) (y - 2) = 0$ $A (- 1, 3) & B (1, 2)$
Commented by jagoll last updated on 22/Mar/20

$(4 a) C_{2}^{4} \times C_{2}^{2} = \frac{4 \times 3}{2 \times 1} \times 1 = 6$ $at least 2 girls = C_{2}^{2} \times C_{2}^{4} = 6$
Commented by jagoll last updated on 22/Mar/20

$(4 b) COMPUTER$ $= 8!$
Commented by mathmax by abdo last updated on 22/Mar/20
$2) we solve { ((x+2y=5)),((xy+y^2 =6 )) :} ⇒ { ((x=−2y+5)),(((−2y+5)y +y^2 −6=0)) :} ⇒ { ((x=−2y +5)),((−2y^2 +5y +y^2 −6 =0)) :} ⇒ { ((x=−2y+5)),((y^2 −5y +6=0)) :} y^2 −5y+6 =0→Δ=25−24=1 ⇒y_1 =((5+1)/2)=3 y_2 =((5−1)/2) =2 y=3⇒x=−1 and y=2 ⇒x=1 so the pnts of intersection are A(−1,3) and B(1,2)$
$2) w e s o l v e {\begin{cases} x + 2 y = 5 \\ x y + y^{2} = 6 \end{cases}$ $\Rightarrow {\begin{cases} x = - 2 y + 5 \\ (- 2 y + 5) y + y^{2} - 6 = 0 \end{cases}$ $\Rightarrow {\begin{cases} x = - 2 y + 5 \\ - 2 y^{2} + 5 y + y^{2} - 6 = 0 \end{cases}$ $\Rightarrow {\begin{cases} x = - 2 y + 5 \\ y^{2} - 5 y + 6 = 0 \end{cases}$ $y^{2} - 5 y + 6 = 0 \to Δ = 25 - 24 = 1 \Rightarrow y_{1} = \frac{5 + 1}{2} = 3$ $y_{2} = \frac{5 - 1}{2} = 2$ $y = 3 \Rightarrow x = - 1 a n d y = 2 \Rightarrow x = 1 s o t h e p n t s o f i n t e r s e c t i o n a r e$ $A (- 1, 3) a n d B (1, 2)$
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