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Sagot :
To solve this question, we proceed as follows:
Step 1: Let x be the worth of one of the type of coins Jared has, and let y be the worth of the other type of coin
Thus:
Since 3 of the coins are of a different type, we have that:
[tex]\begin{gathered} 3x+(12-3)y=75 \\ \Rightarrow3x+9y=75 \end{gathered}[/tex]Also, since 3 of the coins are worth twice as much as the rest, we have that:
[tex]x=2y[/tex]Now, substitute for x in the first equation:
[tex]\begin{gathered} 3x+9y=75 \\ \Rightarrow3(2y)+9y=75 \\ \Rightarrow6y+9y=75 \\ \Rightarrow15y=75 \\ \Rightarrow y=\frac{75}{15} \\ \Rightarrow y=5cents \end{gathered}[/tex]Since y = 5 cents, we have that:
[tex]\begin{gathered} x=2y \\ \Rightarrow x=2(5) \\ \Rightarrow x=10cents \end{gathered}[/tex]Now, since x = 10 cents (the equivalent worth of a dime), and y = 5 cents (the equivalent worth of a nickel), we have from the first equation that:
[tex]3x+9y=75\text{cents}[/tex]From the above equation, therefore, we can conclude that Jared has nine 10 cents coins (dimes), and three 5 cents coins (nickels)
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