Mrs J - > X years ago,
Her son - > Y years old
X=5∗YX =5*Y\\X=5∗Y
(5Y−3)(Y−3)=185\begin{alignedat}{2} (5&Y-3)&(Y-3) = 185 \end{alignedat}(5Y−3)(Y−3)=185
5∗Y2−18∗Y−176=05*Y^{2}-18*Y-176=0\\5∗Y2−18∗Y−176=0
a=5, b=-18, c=-176=0
D=b2−4∗a∗c,D=b^{2}-4*a*c,D=b2−4∗a∗c,
D=(−18)2−4∗1∗(−176)=622,D=(-18)^{2}-4*1*(-176)=62^2,D=(−18)2−4∗1∗(−176)=622,
Y1=(−b+D)/(2∗a),Y1=(−(−18)+622)/(2∗5)=8,Y2=(−b−D)/(2∗a),Y2=(−(−18)−622)/(2∗5)=−4.4.Y1=(-b+\sqrt{D})/(2*a),\\ Y1=(-(-18)+\sqrt{62^2})/(2*5)=8,\\ Y2=(-b-\sqrt{D})/(2*a), \\ Y2=(-(-18)-\sqrt{62^2})/(2*5)=-4.4. \\Y1=(−b+D)/(2∗a),Y1=(−(−18)+622)/(2∗5)=8,Y2=(−b−D)/(2∗a),Y2=(−(−18)−622)/(2∗5)=−4.4.
The age must be greater than 0, so Y=8,
X=5*Y=5*8=40.
Mrs John is 40 years old, her son is 8 years old,
Checking: (40-3)*(8-3)=185.
Answer: Now Mrs John is 40 years old, her son is 8 years old.
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