CIE/2020/w/22/No 4
Solve the simultaneous equations.
$\begin{aligned}\log _{3}(x+y) &=2 \\2 \log _{3}(x+1) &=\log _{3}(y+2)\end{aligned}$
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CIE/2020/w/22/No 4
Solve the simultaneous equations.
$\begin{aligned}\log _{3}(x+y) &=2 \\2 \log _{3}(x+1) &=\log _{3}(y+2)\end{aligned}$
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*********math solution*************
$\begin{array}[t]{ll}&x+y=3^{2}=9 \\&y=9-x \quad \cdots \quad(1) \\&(x+1)^{2}=y+2 \\&x^{2}+2 x+1=9-x+2 \\&x^{2}+3 x-10=0 \\&(x+5)(x-2)=0 \\&x=-5 \text { or } x=2 \\&\text { since } x=-5 \text { is imposible, } x=2 . \\&\therefore y=9-2=7\end{array}$
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