Question

arning goals from lesson 1.5
✓ i can solve mathematic and scientific formulas, and other
literal equations, for a specified variable. (standard:
a.ced.4)

solve ( c = \frac{8}{13}z + 16 ) for ( z ).

a. ( z = \frac{8}{13}c - 15 )
b. ( z = \frac{13}{8}c - 26 )
c. ( z = -\frac{8}{13}c + 15 )
d. ( z = -\frac{13}{8}c + 26 )

Explanation:

Step1: Subtract 16 from both sides

We start with the equation \( c=\frac{8}{13}z + 16 \). To isolate the term with \( z \), we subtract 16 from both sides. This gives us \( c-16=\frac{8}{13}z \).

Step2: Multiply both sides by \( \frac{13}{8} \)

To solve for \( z \), we need to get rid of the coefficient \( \frac{8}{13} \) in front of \( z \). We do this by multiplying both sides of the equation \( c - 16=\frac{8}{13}z \) by the reciprocal of \( \frac{8}{13} \), which is \( \frac{13}{8} \).

So, \( z=\frac{13}{8}(c - 16) \). We distribute the \( \frac{13}{8} \) to both terms inside the parentheses: \( z=\frac{13}{8}c-\frac{13\times16}{8} \).

Simplify \( \frac{13\times16}{8} \): \( 16\div8 = 2 \), so \( 13\times2=26 \). Thus, \( z=\frac{13}{8}c - 26 \).

Answer:

b. \( z=\frac{13}{8}c - 26 \)