Question

1. determine the values of a, b, and c when $y - 7 = 3(x + 4)$ is written in standard form, $ax + by = c$.

choices:
-19 -1 1 3 5 11
____x + __y = ____

1. select the inverse of $f(x) = \frac{3}{4}x - \frac{1}{2}$.

a. $f^{-1}(x) = \frac{4}{3}x + \frac{2}{3}$
b. $f^{-1}(x) = \frac{4}{3}x - 2$
c. $f^{-1}(x) = -\frac{4}{3}x + \frac{1}{2}$
d. $f^{-1}(x) = \frac{3}{4}x + \frac{1}{2}$
advanced

1. if $f(x) = 5x + a$ and $f^{-1}(10) = -1$, find $a$.

Explanation:

Question 7

Step1: Expand the right side

Given \( y - 7 = 3(x + 4) \), expand the right side: \( y - 7 = 3x + 12 \)

Step2: Rearrange to standard form

Subtract \( 3x \) from both sides and add 7 to both sides: \( -3x + y = 12 + 7 \)
Simplify the right side: \( -3x + y = 19 \)
We can also write it as \( 3x - y = -19 \) (multiplying both sides by -1 to make \( A \) positive, though standard form allows \( A \) to be negative, but often \( A \) is positive). So comparing with \( Ax + By = C \), we have \( A = -3 \) (or \( 3 \)), \( B = 1 \) (or \( -1 \)), \( C = 19 \) (or \( -19 \)). From the choices, the numbers are -19, -1, 1, 3, 5, 11. So using \( -3x + y = 19 \) is not matching, but \( 3x - y = -19 \) gives \( A = 3 \), \( B = -1 \), \( C = -19 \). Wait, maybe I made a mistake. Let's do it again.

Original equation: \( y - 7 = 3(x + 4) \)
Expand: \( y - 7 = 3x + 12 \)
Bring all terms to left: \( -3x + y - 7 - 12 = 0 \) → \( -3x + y - 19 = 0 \) → \( -3x + y = 19 \). So \( A = -3 \), \( B = 1 \), \( C = 19 \). But the choices have -19, -1, 1, 3, 5, 11. Wait, maybe the standard form is written as \( 3x - y = -19 \), so \( A = 3 \), \( B = -1 \), \( C = -19 \). So the blanks would be 3, -1, -19? But the choices have -19, -1, 1, 3, 5, 11. Let's check the choices again. The choices are -19, -1, 1, 3, 5, 11. So maybe the equation is written as \( -3x + y = 19 \), but that's not matching. Wait, maybe I messed up the sign. Let's do it step by step.

Start with \( y - 7 = 3(x + 4) \)
Expand: \( y - 7 = 3x + 12 \)
Subtract \( 3x \) from both sides: \( -3x + y - 7 = 12 \)
Add 7 to both sides: \( -3x + y = 19 \)
Now, standard form is \( Ax + By = C \), where \( A \), \( B \), \( C \) are integers, and \( A \) is non-negative (usually). So multiply both sides by -1: \( 3x - y = -19 \)
So \( A = 3 \), \( B = -1 \), \( C = -19 \)
From the choices, the numbers are -19, -1, 1, 3, 5, 11. So the first blank (x coefficient) is 3, second (y coefficient) is -1, third (C) is -19. So the blanks: 3, -1, -19. But let's check the choices. The choices are -19, -1, 1, 3, 5, 11. So 3 (A), -1 (B), -19 (C).

Question 8

To find the inverse of \( f(x) = \frac{3}{4}x - \frac{1}{2} \), follow these steps:

Step1: Replace \( f(x) \) with \( y \)

\( y = \frac{3}{4}x - \frac{1}{2} \)

Step2: Swap \( x \) and \( y \)

\( x = \frac{3}{4}y - \frac{1}{2} \)

Step3: Solve for \( y \)

Add \( \frac{1}{2} \) to both sides: \( x + \frac{1}{2} = \frac{3}{4}y \)
Multiply both sides by \( \frac{4}{3} \): \( y = \frac{4}{3}x + \frac{4}{3} \times \frac{1}{2} \)
Simplify \( \frac{4}{3} \times \frac{1}{2} = \frac{2}{3} \), so \( y = \frac{4}{3}x + \frac{2}{3} \)
So \( f^{-1}(x) = \frac{4}{3}x + \frac{2}{3} \), which is option a.

Question 9

If \( f(x) = 5x + a \) and \( f^{-1}(10) = -1 \), we know that if \( f^{-1}(b) = a \), then \( f(a) = b \). So here, \( f^{-1}(10) = -1 \) means \( f(-1) = 10 \).

Step1: Substitute \( x = -1 \) into \( f(x) \)

\( f(-1) = 5(-1) + a \)

Step2: Set equal to 10 and solve for \( a \)

\( 5(-1) + a = 10 \)
\( -5 + a = 10 \)
Add 5 to both sides: \( a = 10 + 5 = 15 \)

Question 7 Answer (assuming the blanks are filled with the choices):

From the choices, the numbers are -19, -1, 1, 3, 5, 11. So the equation \( 3x - y = -19 \) gives \( A = 3 \), \( B = -1 \), \( C = -19 \). So the blanks: \( \boldsymbol{3}x + \boldsymbol{-1}y = \boldsymbol{-19} \) (or \( 3x - y = -19 \))

Question 8 Answer:

a. \( f^{-1}(x) = \frac{4}{3}x + \frac{2}{3} \)

Question 9 Answer:

\( a = \boldsymbol{15} \)

Answer:

If \( f(x) = 5x + a \) and \( f^{-1}(10) = -1 \), we know that if \( f^{-1}(b) = a \), then \( f(a) = b \). So here, \( f^{-1}(10) = -1 \) means \( f(-1) = 10 \).

Step1: Substitute \( x = -1 \) into \( f(x) \)

\( f(-1) = 5(-1) + a \)

Step2: Set equal to 10 and solve for \( a \)

\( 5(-1) + a = 10 \)
\( -5 + a = 10 \)
Add 5 to both sides: \( a = 10 + 5 = 15 \)

Question 7 Answer (assuming the blanks are filled with the choices):

From the choices, the numbers are -19, -1, 1, 3, 5, 11. So the equation \( 3x - y = -19 \) gives \( A = 3 \), \( B = -1 \), \( C = -19 \). So the blanks: \( \boldsymbol{3}x + \boldsymbol{-1}y = \boldsymbol{-19} \) (or \( 3x - y = -19 \))

Question 8 Answer:

a. \( f^{-1}(x) = \frac{4}{3}x + \frac{2}{3} \)

Question 9 Answer:

\( a = \boldsymbol{15} \)