Q. \(\frac{1}{5}\left(20y – 13\right)\).

Q: What is the simplest form of \frac{1}{5}\left(20y-13\right)?

Multiply through: \frac{1}{5}\cdot 20y = 4y and \frac{1}{5}\cdot (-13) = -\frac{13}{5}, so the simplified form is 4y-\frac{13}{5}.

Q: How do you distribute \frac{1}{5} across the parentheses?

Multiply \frac{1}{5} by each term: \frac{1}{5}\left(20y\right)=4y and \frac{1}{5}\left(-13\right)=-\frac{13}{5} ; combine to get 4y-\frac{13}{5} .

Q: Can you simplify before distributing?.

Yes. Divide 20y by 5 first to get 4y, leaving -\frac{13}{5} for the constant. So \frac{1}{5}\left(20y-13\right)=4y-\frac{13}{5}..

Q: Is this expression linear in y, and what are its slope and intercept?

Yes. As a linear function f(y) = 4y - \frac{13}{5}, the slope is 4 and the y-intercept (constant term) is -\frac{13}{5}.

Q: For which y does \frac{1}{5}\left(20y-13\right)=0?

Solve 4y-\frac{13}{5}=0. Then 4y=\frac{13}{5}, so y=\frac{13}{20}.

Q: How do you evaluate \frac{1}{5}\left(20y-13\right) at y=2 ?

How do you evaluate \frac{1}{5}\left(20y-13\right) at y=2 ?

Q: How can you express 4y-\frac{13}{5} with a mixed number?

Write \frac{13}{5} as 2\frac{3}{5} , so the expression is 4y-2\frac{3}{5} ..

Q: What is the inverse function of f(y)=\frac{1}{5}\left(20y-13\right)?

Start with x=\;4y-\frac{13}{5}. Solve for y: 4y=x+\frac{13}{5}, so y=\frac{x}{4}+\frac{13}{20}.

Answer

\( \frac{1}{5}(20y-13)=\frac{1}{5}\cdot20y-\frac{1}{5}\cdot13=4y-\frac{13}{5} \)

Detailed Explanation

Original expression: \( \frac{1}{5}\left(20y-13\right) \)

Apply the distributive property:The factor \( \frac{1}{5} \) multiplies each term inside the parentheses. Write the product separately for each term:\( \frac{1}{5}\left(20y-13\right) = \frac{1}{5}\cdot 20y \;-\; \frac{1}{5}\cdot 13 \)
Simplify the first product \( \frac{1}{5}\cdot 20y \):Multiply the numerical coefficients \( \frac{1}{5} \) and \(20\). Treat \(y\) as a factor that is carried along.\( \frac{1}{5}\cdot 20y = \left(\frac{20}{5}\right)y \)
Simplify the fraction \( \frac{20}{5} \) by dividing numerator and denominator by 5:

\( \frac{20}{5} = 4 \), so \( \left(\frac{20}{5}\right)y = 4y \)
Simplify the second product \( \frac{1}{5}\cdot 13 \):Multiply the numerical coefficients. There is no variable here, so the result is a fraction:\( \frac{1}{5}\cdot 13 = \frac{13}{5} \)
Combine the simplified parts:Substitute the simplified first and second products back into the expression:\( \frac{1}{5}\left(20y-13\right) = 4y – \frac{13}{5} \)

Final simplified result: \( 4y – \frac{13}{5} \)

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Algebra FAQs

What is the simplest form of \(\frac{1}{5}\left(20y-13\right)\)?

Multiply through: \(\frac{1}{5}\cdot 20y = 4y\) and \(\frac{1}{5}\cdot (-13) = -\frac{13}{5}\), so the simplified form is \(4y-\frac{13}{5}\).

How do you distribute \( \frac{1}{5} \) across the parentheses?

Multiply \( \frac{1}{5} \) by each term: \( \frac{1}{5}\left(20y\right)=4y \) and \( \frac{1}{5}\left(-13\right)=-\frac{13}{5} \); combine to get \( 4y-\frac{13}{5} \).

Can you simplify before distributing?.

Yes. Divide \(20y\) by \(5\) first to get \(4y\), leaving \(-\frac{13}{5}\) for the constant. So \(\frac{1}{5}\left(20y-13\right)=4y-\frac{13}{5}\)..

\( \)Is this expression linear in \(y\), and what are its slope and intercept?\( \)

Yes. As a linear function \(f(y) = 4y - \frac{13}{5}\), the slope is 4 and the \(y\)-intercept (constant term) is \(-\frac{13}{5}\).

For which \(y\) does \(\frac{1}{5}\left(20y-13\right)=0\)?

Solve \(4y-\frac{13}{5}=0\). Then \(4y=\frac{13}{5}\), so \(y=\frac{13}{20}\).

How do you evaluate \( \frac{1}{5}\left(20y-13\right) \) at \( y=2 \) ?

How can you express \(4y-\frac{13}{5}\) with a mixed number?

Write \( \frac{13}{5} \) as \( 2\frac{3}{5} \), so the expression is \( 4y-2\frac{3}{5} \)..

What is the inverse function of \(f(y)=\frac{1}{5}\left(20y-13\right)\)?

Start with \(x=\;4y-\frac{13}{5}\). Solve for \(y\): \(4y=x+\frac{13}{5}\), so \(y=\frac{x}{4}+\frac{13}{20}\).

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