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C=0.25(L-1200)
The cost,
C, in dollars, to the holder of a liability insurance account given a total liability of
L dollars is given by the equation as long as
C is positive. By how many dollars does the liability increase when the cost to the holder increases by 1 dollar?

$C=0.25(L-1200)$ $\newline$ The cost, $C$ , in dollars, to the holder of a liability insurance account given a total liability of $L$ dollars is given by the equation as long as $C$ is positive. By how many dollars does the liability increase when the cost to the holder increases by $1$ dollar?

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Write linear functions: word problems

Full solution

Q.

C=0.25(L-1200)

\newline

The cost,

C

, in dollars, to the holder of a liability insurance account given a total liability of

L

dollars is given by the equation as long as

C

is positive. By how many dollars does the liability increase when the cost to the holder increases by

1

dollar?

Isolate L in equation: We are given the equation $C = 0.25(L - 1200)$ , where $C$ is the cost to the holder and $L$ is the total liability in dollars. We need to find out how much $L$ increases when $C$ increases by $1$ dollar. $\newline$ First, let's isolate $L$ in the equation to see how it changes with respect to $C$ .
Add $1200$ to both sides: We add $1200$ to both sides of the equation to get $C + 1200 = 0.25L$ .
Divide by $0$ . $25$ to solve: Next, we divide both sides of the equation by $0.25$ to solve for $L$ . This gives us $L = \frac{C + 1200}{0.25}$ .
Find change in $L$ : Now, we want to find out the change in $L$ when $C$ increases by $1$ dollar. Let's call the increase in $L$ as $\Delta L$ and the increase in $C$ as $\Delta C$ . Since $\Delta C$ is $1$ dollar, we can write the new cost as $L$ $0$ .
Substitute $C + \Delta C$ : Substituting $C + \Delta C$ into the equation for $L$ , we get $L + \Delta L = \frac{C + \Delta C + 1200}{0.25}$ .
Solve for $\Delta L$ : Since $\Delta C$ is $1$ , we can substitute it into the equation to get $L + \Delta L = (C + 1 + 1200) / 0.25$ .
Substitute $L$ into equation: We can now subtract the original $L$ from both sides to find $\Delta L$ . This gives us $\Delta L = \frac{(C + 1 + 1200)}{0.25} - L$ .
Simplify the equation: We know that $L = \frac{(C + 1200)}{0.25}$ from our earlier step, so we can substitute this into our equation for $\Delta L$ . This gives us $\Delta L = \left[\frac{(C + 1 + 1200)}{0.25}\right] - \left[\frac{(C + 1200)}{0.25}\right].$
Calculate $\Delta L$ : Simplifying the right side of the equation, we see that the terms $(C + 1200) / 0.25$ cancel out, leaving us with $\Delta L = (1 / 0.25)$ .
Final result: Calculating $1 / 0.25$ gives us $\Delta L = 4$ . This means that the liability increases by $4$ dollars when the cost to the holder increases by $1$ dollar.

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