Question

1. solve each equation for the specific variable.

expression	answer
b. $\frac{(j + p)}{e} = d + c$ for $j$

Explanation:

Part a: Solve \( a - q = a + sx \) for \( x \)

Step 1: Subtract \( a \) from both sides

To isolate the term with \( x \), we subtract \( a \) from both sides of the equation.
\( a - q - a = a + sx - a \)
Simplifying both sides, the \( a \) terms cancel out on both sides:
\( -q = sx \)

Step 2: Divide both sides by \( s \) (assuming \( s

eq 0 \))
To solve for \( x \), we divide both sides of the equation by \( s \).
\( \frac{-q}{s} = \frac{sx}{s} \)
Simplifying the right - hand side (the \( s \) terms cancel), we get:
\( x = -\frac{q}{s} \)

Part b: Solve \( \frac{j + p}{e}=d + c \) for \( j \)

Step 1: Multiply both sides by \( e \)

To eliminate the denominator on the left - hand side, we multiply both sides of the equation by \( e \).
\( e\times\frac{j + p}{e}=e\times(d + c) \)
Simplifying the left - hand side (the \( e \) terms cancel), we have:
\( j + p=e(d + c) \)

Step 2: Subtract \( p \) from both sides

To solve for \( j \), we subtract \( p \) from both sides of the equation.
\( j + p - p=e(d + c)-p \)
Simplifying the left - hand side (the \( p \) terms cancel), we get:
\( j = e(d + c)-p \)
We can also expand the right - hand side: \( j=ed + ec - p \)

Answer:

a. \( x = \boldsymbol{-\frac{q}{s}} \) (for \( s
eq0 \))
b. \( j=\boldsymbol{e(d + c)-p} \) (or \( j = ed+ec - p \))