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Sagot :
SOLUTION
Given the question, the following are the solution steps to answer the question.
STEP 1: Represent the two daughters with a variable
Let the first daughter represented with y
Let the second daughter be represented with z
STEP 2: Interpret the statements in the question
[tex]\begin{gathered} \text{From second statement:} \\ y+z=21----\text{equation 1} \\ \text{From third statement:} \\ y\times z=110----\text{equation 2} \end{gathered}[/tex]STEP 2: Write out the two gotten equations
[tex]\begin{gathered} y+z=21----(1) \\ yz=110----(2) \\ \end{gathered}[/tex]STEP 3: Make y the subject of the equation 1
[tex]\begin{gathered} \text{Subtract z from both sides} \\ y+z-z=21-z \\ y=21-z \end{gathered}[/tex]STEP 4: Substitute the value for y in equation 2
[tex]\begin{gathered} yz=110 \\ (21-z)z=110 \\ 21z-z^2=110 \\ \text{Subtract 110 from both sides} \\ 21z-z^2-110=110-110 \\ 21z-z^2-110=0 \end{gathered}[/tex]STEP 5: We solve the quadratic equation to get the values of z
[tex]\begin{gathered} 21z-z^2-110=0 \\ \mathrm{Write\: in\: the\: standard\: form}\: ax^2+bx+c=0 \\ z^2+21z-110=0 \\ \text{ Using quadratic formula;} \\ z_{1,\: 2}=\frac{-21\pm\sqrt{21^2-4\left(-1\right)\left(-110\right)}}{2\left(-1\right)} \\ 21^2-4(-1)(-110)=\sqrt[]{1}=1 \\ z_{1,\: 2}=\frac{-21\pm\:1}{2\left(-1\right)} \\ \mathrm{Separate\: the\: solutions} \\ z_1=\frac{-21+1}{2\left(-1\right)},\: z_2=\frac{-21-1}{2\left(-1\right)} \\ z_1=\frac{-20}{-2}=10 \\ z_2=\frac{-22}{-2}=11 \\ z=10,\: z=11 \end{gathered}[/tex]STEP 6: Get the age of the other daughter
[tex]\begin{gathered} y+z=21 \\ \text{when }z=10 \\ y=21-10=11 \\ \text{When z=11} \\ y=21-11=10 \\ \\ \therefore ages\text{ of the daughters are: }10\text{ years and 11 years} \end{gathered}[/tex]Hence, the age of Mr. Thaxton's youngest daughter is 10 years
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