Linear modelling

In the linear model $y=15x+30$, we first recall what each symbol means. The problem states that $x$ is the number of sessions and $y$ is the total monthly fee. The $y$-intercept is the value of $y$ when the input ($x$) is zero.

To get the $y$-intercept using algebra, substitute $x=0$ into the equation. This matches the case where zero sessions are booked.

$\begin{array}{rl}y& =15(0)+30\\ y& =0+30\\ y& =30\end{array}$

The calculation shows that when $x=0$ (no sessions), the monthly fee is $y=30$. So even with no sessions booked, the customer still pays $30. In everyday terms, this is a fixed starting charge (a base subscription fee).

Use what we found to check each option:

• Option A is incorrect because the per-session cost is the gradient ($15$), not the intercept.
• Option B is incorrect because a $y$-intercept does not state any maximum number of sessions.
• Option C is incorrect because the equation models one person’s fee, not how many users exist.

The $y$-intercept is the amount charged when no sessions are booked, so it represents the base monthly fee.

We are given the linear model:

$y=15x+30$

Here, $y$ is the total monthly fee and $x$ is the number of personal training sessions booked. This is a standard idea in Coordinate Geometry and Linear Modelling.

Compare the model to the slope-intercept form:

$y=mx+c$

Where:

• $m$ is the gradient (how much $y$ changes when $x$ increases by 1).
• $c$ is the $y$-intercept (the starting value).

So in $y=15x+30$, the $y$-intercept $c$ is 30.

The $y$-intercept happens when $x=0$.

In this setting, $x=0$ means zero personal training sessions are booked. Substituting gives:

$\begin{array}{rl}y& =15(0)+30\\ y& =30\end{array}$

So there is still a fixed charge of 30 even with no sessions, which is a base/subscription fee.

• Option A: This describes the gradient ($m=15$), the fee added per session.
• Option B: Nothing in the equation sets a maximum; it only calculates cost from $x$.
• Option C: The equation gives a cost for one customer, not the number of customers.

Because the $y$-intercept is the value of $y$ when $x=0$, it must be the fixed monthly amount charged even when no sessions are booked.

The model $y=15x+30$ gives total monthly fee $y$ from the number of sessions $x$. It follows:

$y=mx+c$

Where:

• $m$ is the slope (rate of change).
• $c$ is the $y$-intercept (the value of $y$ when $x=0$).

This is part of Coordinate Geometry and Linear Modelling. Now eliminate options that cannot describe $c=30$.

Option A talks about the cost per session. In the equation, every increase of 1 in $x$ increases $y$ by 15, which is the slope $m$, not the intercept.

$\text{Slope }m=15=\text{Cost per session}$

Option B suggests a maximum number of sessions. A maximum would be a restriction on possible $x$ values, but the constant term $+30$ does not set any limit on $x$.

Option C refers to the total number of users. But this equation is about one user’s monthly fee based on their sessions, so user population is not represented by the intercept.

The intercept is found when $x=0$. Compute $y$:

$\begin{array}{rl}y& =15(0)+30\\ y& =30\end{array}$

This 30 is the amount paid even when zero sessions are booked, so it is the base subscription fee.

After eliminating A, B, and C, the only option that matches the starting cost (the $y$-intercept) is Option D.

We are told the relationship is $y=4x+12$ and we want the input $x$ when the output is $y=32$. So we put $32$ into the formula wherever $y$ appears.

Substituting $y=32$ gives the equation:

$32=4x+12$

To get the $x$ term by itself, subtract $12$ from both sides:

$\begin{array}{rl}32-12& =4x\\ 20& =4x\end{array}$

This kind of rearranging is a key skill in the Coordinate Geometry and Linear Modelling chapter.

Now divide both sides by $4$ to solve for $x$:

$\begin{array}{rl}x& =\frac{20}{4}\\ x& =5\end{array}$

To be sure, check that the other options do not make $32=4x+12$ true:

• For Option B: $4(9)+12=36+12=48\ne 32$
• For Option C: $4(7)+12=28+12=40\ne 32$
• For Option D: $4(12)+12=48+12=60\ne 32$

Our solution is $x=5$, and the remaining choices do not work, so Option A is correct.

The model is $y=4x+12$ and we need the option for $x$ that makes the output $y=32$. With substitution of options, we plug each proposed $x$ into the rule until we hit 32.

Substitute $x=9$:

$\begin{array}{rl}y& =4(9)+12\\ y& =36+12\\ y& =48\end{array}$

Because $48\ne 32$, Option B is not correct.

Substitute $x=7$:

$\begin{array}{rl}y& =4(7)+12\\ y& =28+12\\ y& =40\end{array}$

Because $40\ne 32$, Option C is not correct.

Substitute $x=12$:

$\begin{array}{rl}y& =4(12)+12\\ y& =48+12\\ y& =60\end{array}$

Because $60\ne 32$, Option D is not correct.

Now substitute $x=5$:

$\begin{array}{rl}y& =4(5)+12\\ y& =20+12\\ y& =32\end{array}$

This equals the required value $y=32$.

Graphically, we want the $x$-value where the line gives an output of $32$. Treat each side as its own function:

$y_1=4x+12$ $y_2=32$

On a graphing tool, draw the line $y=4x+12$ and the horizontal line $y=32$. Choose a viewing window that shows where they cross; since $y=32$, a $y$-maximum around $40$ helps.

The solution is the intersection point of the two lines. The graph shows they meet at $(5,32)$, so when $y=32$, the corresponding $x$ is $5$.

Since the intersection gives $x=5$, the other listed $x$-values are not correct and can be crossed out.

The intersection has $x$-coordinate $5$, so the answer is Option A.

The problem gives a linear equation for the total cost:

$y=25x+150$

This matches the familiar slope-intercept format from Coordinate Geometry and Linear Modelling:

$y=mx+c$

Comparing $y=25x+150$ with $y=mx+c$ lets us identify the important parts:

1. The slope ($m=25$): how much the total increases for each extra 1 chair.
2. The $y$-intercept ($c=150$): the constant amount, meaning the value of $y$ when $x=0$.

If you graph the line, it crosses the vertical axis at $150$. That means when $x$ is zero, the graph begins at $y=150$.

Set $x=0$ (no chairs rented) and see what the model gives:

$\begin{array}{rl}y& =25(0)+150\\ y& =150\end{array}$

So even with zero chairs, the cost is still $150$. That shows there is a fixed charge you pay no matter what, like a booking or delivery/setup fee.

Check why the other answers don’t match:

• Option A: The equation describes cost, not profit, so it can’t tell us a maximum profit.
• Option B: The per-chair cost is the slope ($m=25$), not the constant $150$.
• Option D: For $25$ chairs, the cost is $25(25)+150=775$, not $150$.

Because $150$ is the cost when $x=0$, it represents the fixed fee charged regardless of how many chairs are rented.

To interpret the equation, first note what the variables represent in this model:

• $x$: the number of chairs rented.
• $y$: the total cost in dollars.

To find the amount charged before any chairs are added, substitute $x=0$ into the equation:

$\begin{array}{rl}y& =25(0)+150\\ y& =0+150\\ y& =150\end{array}$

When $x=0$, the model still gives $y=150$. This is the cost that exists even before renting a single chair. In modelling, that constant charge is called a fixed cost (or setup/base fee), because it does not depend on how many chairs are rented.

The remaining choices can be rejected:

• Option A: This equation is only for cost, not profit, and it gives no revenue information.
• Option B: The per-chair rate is the coefficient of $x$, which is $25$.
• Option D: If $x=25$, then $y=25(25)+150=775$, so $150$ is not the total for 25 chairs.

Since $150$ is the cost when there are zero chairs, it must be the fixed setup/base fee that stays the same regardless of the number rented.

The equation is $y=25x+150$, which has the form:

$y=mx+c$

Here, $y$ is total cost, $x$ is the number of chairs, $m$ is the rate per chair, and $c$ is the constant starting amount. This is part of Coordinate Geometry and Linear Modelling.

The term $25x$ changes when $x$ changes. That means the $25$ is the amount added for each additional chair. So $25$ is the cost per chair, not $150$.

Test $x=25$ to see the total cost for 25 chairs:

$\begin{array}{rl}y& =25(25)+150\\ y& =625+150\\ y& =775\end{array}$

Because the result is $775$ (not $150$), that option is incorrect.

The model states that $y$ is total cost. Profit would require revenue information (Profit = Revenue − Cost), which is not provided here. So it cannot represent profit.

The number $150$ is the constant term ($c$). It equals the cost when $x=0$:

$y=25(0)+150=150$

So, even if no chairs are rented, the company still charges $150$. That is a fixed setup/base fee that does not depend on the quantity.