Answer :
SOLUTION
Steps1: Define a parameter for the unknow
[tex]\begin{gathered} a=\text{outer ring point } \\ b=\text{bull's eye point} \end{gathered}[/tex]Step2: Write out the equation for Helen
Since Helen landed 3arrows on the outer ring and 1 on the bull's eye and have a total of 174, the equation becomes
[tex]3a+b=174[/tex]Step3: Wite out the equation for Jordan
Since Jordan Landed 3arrows on outer ring and 4arrows in the bull's eye with a total point of 444, the equation becomes
[tex]3a+4b=444[/tex]Step4: Solve the system of equation with elimination method
To solve the equation, we label them eliminate the variables separately
hence
[tex]\begin{gathered} 3a+b=174\ldots Eq1 \\ 3a+4b=444\ldots Eq2 \end{gathered}[/tex]Subtract Eq1 from Eq2 (Eq2-Eq1) to eliminate the variable a.
[tex]\begin{gathered} 3a_{}+4b=444 \\ 3a+b=174 \\ \text{Then} \\ 3b=270 \end{gathered}[/tex]From the equation in the last line above divide both sides by 3, we have
[tex]\begin{gathered} \frac{3b}{3}=\frac{270}{3} \\ \text{then} \\ b=90 \end{gathered}[/tex]To eliminate the variable b, Multiply Eq1 by 4 and subtract from Eq 2, we have
[tex]\begin{gathered} 4\times Eq1\rightarrow4(3a+b=174)=12a+4b=696 \\ \text{Then subtract from Eq2, we have } \\ 12a+4b=696 \\ 3a+4b=444 \\ \text{hence} \\ 9a=252 \end{gathered}[/tex]From the equation in the last line above, divide both sides by 9, we obtain
[tex]\begin{gathered} \frac{9a}{9}=\frac{252}{9} \\ \text{Then} \\ a=28 \end{gathered}[/tex]Hence
a=28,b=90.
Therefore
The outer ring worth 28 point
The bull's eye worth 90 point