Last week, Shelly rode her bike a total of 30 miles over a three-day period. On the second day, she rode LaTeX: \frac{4}{5}45 the distance she rode on the first day. On the third day, she rode LaTeX: \frac{3}{2}32 the distance she rode on the second day

Respuesta :

We make expressions for each afirmation

Where X is the first day, Y second day and Z the third

1. the sum of the 3 days gives us 30

[tex]X+Y+Z=30[/tex]

2. Second day is 4/5 of the first day

[tex]Y=\frac{4}{5}X[/tex]

3.Third day is 3/2 of the second day

[tex]Z=\frac{3}{2}Y[/tex]

Whit the expressions I try to represent everything as a function of X

I must represent Z in function of X, for this I can replace Y of the second expression in the third expression

[tex]\begin{gathered} Z=\frac{3}{2}(\frac{4}{5}X) \\ Z=\frac{12}{10}X \\ Z=\frac{6}{5}X \end{gathered}[/tex]

So I have:

[tex]\begin{gathered} Y=\frac{4}{5}X \\ Z=\frac{6}{5}X \\ \end{gathered}[/tex]

And I can replace on the first expression

[tex]\begin{gathered} X+Y+Z=30 \\ X+(\frac{4}{5}X)+(\frac{6}{5}X)=30 \end{gathered}[/tex]

I must find X

[tex]\begin{gathered} (1+\frac{4}{5}+\frac{6}{5})X=30 \\ 3X=30 \\ X=\frac{30}{3} \\ X=10 \end{gathered}[/tex]

So, if I have X I can replace on this expressions to find de value:

[tex]\begin{gathered} Y=\frac{4}{5}X \\ Z=\frac{6}{5}X \end{gathered}[/tex]

Where X is 10

[tex]\begin{gathered} Y=\frac{4}{5}\times10 \\ Y=\frac{40}{5}=8 \\ \\ Z=\frac{6}{5}\times10 \\ Z=\frac{60}{5}=12 \end{gathered}[/tex]

To check:

[tex]\begin{gathered} X+Y+Z=30 \\ (10)+(8)+(12)=30 \\ 30=30 \\ \end{gathered}[/tex]

The result is correct, therefore:

[tex]\begin{gathered} X=10 \\ Y=8 \\ Z=12 \end{gathered}[/tex]