Answer: [tex]\frac{10}{7}[/tex] meters
Step-by-step explanation:
Let x be the length of Kaya's string and y be the length of Ethan's string
Then According to the question, we have the following equations
[tex]\frac{1}{2}y=\frac{2}{3}x\\\Rightarrow\ y=\frac{4}{3}x......(1)\\x+y=10....(2)[/tex]
Substitute the value of y from (1) in (2), we get
[tex]\frac{4}{3}x+x=10\\\Rightarrow\frac{4x+3x}{3}=10\\\Rightarrow\ \frac{7x}{3}=10\\\Rightarrow\ x=\frac{30}{7}[/tex]
The length of Kaya's string =[tex]\frac{30}{7}[/tex] feet
The length of Ethan's string =[tex]\frac{4}{3}\times\frac{30}{7}=\frac{40}{7}[/tex] meters
The difference in their lengths=[tex]\frac{40}{7}-\frac{30}{7}=\frac{10}{7}[/tex]
Hence, Ethan's string is [tex]\frac{10}{7}[/tex] meters longer than the Kaya's string.
Answer:
Ethan's string is [tex]=\frac{10}{7}[/tex] feet longer than Kayla's string.
Step-by-step explanation:
Let Ethan's string = x feet
and Kayla's string = y feet
According to question,
Half of Ethan's string is equal to 2/3 of Kayla's string that is,
[tex]\frac{1}{2}x=\frac{2}{3}y[/tex]
[tex]\Rightarrow x=\frac{4}{3}y[/tex] ..............(1)
Also,The total length of their strings is 10 feet that is,
[tex]x+y=10[/tex]
Put value of x from (1),
[tex]\frac{4}{3}y+y=10[/tex]
Solving for y, we get,
[tex]\Rightarrow y(\frac{4}{3}+1)=10[/tex]
[tex]\Rightarrow y(\frac{4+3}{3})=10[/tex]
[tex]\Rightarrow y(\frac{7}{3})=10[/tex]
[tex]\Rightarrow y=\frac{10 \times 3}{7}[/tex]
[tex]\Rightarrow y=\frac{30}{7}[/tex]
Thus, Length of Kayla's string is [tex]\frac{30}{7}[/tex] feet.
and Put value of y in (1) to get value of x,
[tex]\Rightarrow x=\frac{4}{3} \times \frac{30}{7}[/tex]
[tex]\Rightarrow x=\frac{40}{7}[/tex]
Thus, Length of Ethan's string is [tex]\frac{40}{7}[/tex] feet.
Length of Ethan's string is longer than Kayla's string = Length of Ethan's string-Length of Kayla's string.
[tex]=\frac{40}{7}-\frac{30}{7}[/tex]
[tex]=\frac{10}{7}[/tex]
Thus, Ethan's string is [tex]=\frac{10}{7}[/tex] feet longer than Kayla's string.
