Find the Inverse y=-3/5x+7/5

$y = - \frac{3}{5} x + \frac{7}{5}$

Step 1

Interchange the variables.

$x = - \frac{3}{5} y + \frac{7}{5}$

Step 2

Solve for $y$ .

Rewrite the equation as $- \frac{3}{5} y + \frac{7}{5} = x$ .

$- \frac{3}{5} y + \frac{7}{5} = x$

Simplify each term.

Combine $y$ and $\frac{3}{5}$ .

$- \frac{y \cdot 3}{5} + \frac{7}{5} = x$

Move $3$ to the left of $y$ .

$- \frac{3 y}{5} + \frac{7}{5} = x$

Subtract $\frac{7}{5}$ from both sides of the equation.

$- \frac{3 y}{5} = x - \frac{7}{5}$

Multiply both sides of the equation by $- \frac{5}{3}$ .

$- \frac{5}{3} (- \frac{3 y}{5}) = - \frac{5}{3} (x - \frac{7}{5})$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{5}{3} (- \frac{3 y}{5})$ .

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{5}{3}$ into the numerator.

$\frac{- 5}{3} (- \frac{3 y}{5}) = - \frac{5}{3} (x - \frac{7}{5})$

Move the leading negative in $- \frac{3 y}{5}$ into the numerator.

$\frac{- 5}{3} \cdot \frac{- 3 y}{5} = - \frac{5}{3} (x - \frac{7}{5})$

Factor $5$ out of $- 5$ .

$\frac{5 (- 1)}{3} \cdot \frac{- 3 y}{5} = - \frac{5}{3} (x - \frac{7}{5})$

Cancel the common factor.

$\frac{5 \cdot - 1}{3} \cdot \frac{- 3 y}{5} = - \frac{5}{3} (x - \frac{7}{5})$

Rewrite the expression.

$\frac{- 1}{3} (- 3 y) = - \frac{5}{3} (x - \frac{7}{5})$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3 y$ .

$\frac{- 1}{3} (3 (- y)) = - \frac{5}{3} (x - \frac{7}{5})$

Cancel the common factor.

$\frac{- 1}{3} (3 (- y)) = - \frac{5}{3} (x - \frac{7}{5})$

Rewrite the expression.

$- - y = - \frac{5}{3} (x - \frac{7}{5})$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{5}{3} (x - \frac{7}{5})$

Multiply $y$ by $1$ .

$y = - \frac{5}{3} (x - \frac{7}{5})$

Simplify the right side.

Simplify $- \frac{5}{3} (x - \frac{7}{5})$ .

Simplify terms.

Apply the distributive property.

$y = - \frac{5}{3} x - \frac{5}{3} (- \frac{7}{5})$

Combine $x$ and $\frac{5}{3}$ .

$y = - \frac{x \cdot 5}{3} - \frac{5}{3} (- \frac{7}{5})$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{5}{3}$ into the numerator.

$y = - \frac{x \cdot 5}{3} + \frac{- 5}{3} (- \frac{7}{5})$

Move the leading negative in $- \frac{7}{5}$ into the numerator.

$y = - \frac{x \cdot 5}{3} + \frac{- 5}{3} \cdot \frac{- 7}{5}$

Factor $5$ out of $- 5$ .

$y = - \frac{x \cdot 5}{3} + \frac{5 (- 1)}{3} \cdot \frac{- 7}{5}$

Cancel the common factor.

$y = - \frac{x \cdot 5}{3} + \frac{5 \cdot - 1}{3} \cdot \frac{- 7}{5}$

Rewrite the expression.

$y = - \frac{x \cdot 5}{3} + \frac{- 1}{3} \cdot - 7$

Combine $\frac{- 1}{3}$ and $- 7$ .

$y = - \frac{x \cdot 5}{3} + \frac{- - 7}{3}$

Multiply $- 1$ by $- 7$ .

$y = - \frac{x \cdot 5}{3} + \frac{7}{3}$

Move $5$ to the left of $x$ .

$y = - \frac{5 x}{3} + \frac{7}{3}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = - \frac{5 x}{3} + \frac{7}{3}$

Step 4

Verify if $f^{- 1} (x) = - \frac{5 x}{3} + \frac{7}{3}$ is the inverse of $f (x) = - \frac{3}{5} x + \frac{7}{5}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (- \frac{3}{5} x + \frac{7}{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = - \frac{5 (- \frac{3}{5} x + \frac{7}{5})}{3} + \frac{7}{3}$

Combine the numerators over the common denominator.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 5 (- \frac{3}{5} x + \frac{7}{5}) + 7}{3}$

Simplify each term.

Combine $x$ and $\frac{3}{5}$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 5 (- \frac{x \cdot 3}{5} + \frac{7}{5}) + 7}{3}$

Move $3$ to the left of $x$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 5 (- \frac{3 x}{5} + \frac{7}{5}) + 7}{3}$

Apply the distributive property.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 5 (- \frac{3 x}{5}) - 5 (\frac{7}{5}) + 7}{3}$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{3 x}{5}$ into the numerator.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 5 \frac{- 3 x}{5} - 5 (\frac{7}{5}) + 7}{3}$

Factor $5$ out of $- 5$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{5 (- 1) (\frac{- 3 x}{5}) - 5 (\frac{7}{5}) + 7}{3}$

Cancel the common factor.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{5 \cdot (- 1 \frac{- 3 x}{5}) - 5 (\frac{7}{5}) + 7}{3}$

Rewrite the expression.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{- 1 (- 3 x) - 5 (\frac{7}{5}) + 7}{3}$

Multiply $- 3$ by $- 1$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x - 5 (\frac{7}{5}) + 7}{3}$

Cancel the common factor of $5$ .

Factor $5$ out of $- 5$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x + 5 (- 1) (\frac{7}{5}) + 7}{3}$

Cancel the common factor.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x + 5 \cdot (- 1 (\frac{7}{5})) + 7}{3}$

Rewrite the expression.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x - 1 \cdot 7 + 7}{3}$

Multiply $- 1$ by $7$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x - 7 + 7}{3}$

Simplify terms.

Combine the opposite terms in $3 x - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$f^{- 1} (- \frac{3}{5} x + \frac{7}{5}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \frac{5 x}{3} + \frac{7}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = - \frac{3}{5} \cdot (- \frac{5 x}{3} + \frac{7}{3}) + \frac{7}{5}$

Simplify each term.

Apply the distributive property.

$f (- \frac{5 x}{3} + \frac{7}{3}) = - \frac{3}{5} \cdot (- \frac{5 x}{3}) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{5}$ into the numerator.

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{- 3}{5} \cdot (- \frac{5 x}{3}) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Move the leading negative in $- \frac{5 x}{3}$ into the numerator.

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{- 3}{5} \cdot \frac{- 5 x}{3} - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Factor $3$ out of $- 3$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{3 (- 1)}{5} \cdot \frac{- 5 x}{3} - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor.

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{3 \cdot - 1}{5} \cdot \frac{- 5 x}{3} - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Rewrite the expression.

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{- 1}{5} \cdot (- 5 x) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor of $5$ .

Factor $5$ out of $- 5 x$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{- 1}{5} \cdot (5 (- x)) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor.

$f (- \frac{5 x}{3} + \frac{7}{3}) = \frac{- 1}{5} \cdot (5 (- x)) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Rewrite the expression.

$f (- \frac{5 x}{3} + \frac{7}{3}) = - 1 (- x) - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Multiply $- 1$ by $- 1$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = 1 x - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Multiply $x$ by $1$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x - \frac{3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{5}$ into the numerator.

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{- 3}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Factor $3$ out of $- 3$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{3 (- 1)}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Cancel the common factor.

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{3 \cdot - 1}{5} \cdot \frac{7}{3} + \frac{7}{5}$

Rewrite the expression.

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{- 1}{5} \cdot 7 + \frac{7}{5}$

Combine $\frac{- 1}{5}$ and $7$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{- 1 \cdot 7}{5} + \frac{7}{5}$

Multiply $- 1$ by $7$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + \frac{- 7}{5} + \frac{7}{5}$

Move the negative in front of the fraction.

$f (- \frac{5 x}{3} + \frac{7}{3}) = x - \frac{7}{5} + \frac{7}{5}$

Combine the opposite terms in $x - \frac{7}{5} + \frac{7}{5}$ .

Add $- \frac{7}{5}$ and $\frac{7}{5}$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x + 0$

Add $x$ and $0$ .

$f (- \frac{5 x}{3} + \frac{7}{3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{5 x}{3} + \frac{7}{3}$ is the inverse of $f (x) = - \frac{3}{5} x + \frac{7}{5}$ .

$f^{- 1} (x) = - \frac{5 x}{3} + \frac{7}{3}$

Find the Inverse y=-3/5x+7/5

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