Question

The monthly payment \(p\) on a mortgage varies directly with the amount borrowed \(B\). If the monthly payment on a 30 -year mortgage is \(\$ 6.49\) for every \(\$ 1000\) borrowed, find a linear equation that relates the monthly payment \(p\) to the amount borrowed \(B\) for a mortgage with these terms. Then find the monthly payment \(p\) when the amount borrowed \(B\) is \(\$ 145,000\).

Short answer

The monthly payment is \(\$941.05\) for \(\$145,000\) borrowed.

Step-by-step solution

Identify the variables

Let the monthly payment be denoted as \(p\) and the amount borrowed be denoted as \(B\).

Understand the direct variation

Since the monthly payment \(p\) varies directly with the amount borrowed \(B\), we have the relationship \(p = kB\) where \(k\) is the constant of proportionality.

Find the constant of proportionality (k)

Given that the monthly payment is \(\$6.49\) for every \(\$1000\) borrowed, we can set up the equation \(6.49 = k \times 1000\). Solving for \(k\), we get \(k = \frac{6.49}{1000} = 0.00649\).

Formulate the linear equation

Substitute the value of \(k\) into the direct variation equation: \(p = 0.00649B\). This is the linear equation that relates the monthly payment \(p\) to the amount borrowed \(B\) for a mortgage with these terms.

Calculate the monthly payment when \(B = \$145,000\)

Using the equation \(p = 0.00649B\), substitute \(B = 145000\). Thus, \(p = 0.00649 \times 145000 = 941.05\). The monthly payment \(p\) is \$941.05.

Key concepts

linear equations

Linear equations are mathematical expressions that show a straight-line relationship between two variables.
They follow the format of usually looking like this: oindent \(y = mx + b\), where \(y\) represents the dependent variable, \(x\) denotes the independent variable, \(m\) serves as the slope (or rate of change), and \(b\) is the y-intercept (where the line crosses the y-axis).

In our exercise, the relation between monthly mortgage payments and the borrowed amount is described by the linear equation: oindent \(p = 0.00649B\) Where: \(p\) represents the monthly payment and \(B\) represents the amount borrowed.

Since there is no additional constant term added to the equation, this line passes through the origin (0,0). This kind of linear equation means that if you borrow more, your monthly payment increases proportionally.

constant of proportionality

A constant of proportionality is a fixed number that simplifies the proportional relationship between two variables.
This can be spotted in directly proportional relationships, such as the one we have in our exercise. It is usually represented as 'k'.
In general terms, the direct variation equation looks like: oindent \(y = kx\).

In our specific case: oindent \(p = kB\), where \(p\) is the monthly payment and \(B\) is the amount borrowed.

We are given that for every \(\text{1000}\text{ USD (B)}\), the payment is \(\text{6.49}\text{ USD (p)}\). Let's find `k`:
oindent \(6.49 = k \times 1000\), solving for `k` we get: oindent \[ k = \frac{6.49}{1000} = 0.00649 \] So the constant of proportionality \(k\) equals oindent \(0.00649\).

monthly mortgage payment

Understanding your monthly mortgage payment is crucial in financial planning. It is a regular payment you make to repay the money you've borrowed from a lender to buy a house or property.
In our example, the monthly payment \(p\) is \(\text{6.49 USD}\) for every \(\text{1000 USD}\) borrowed. Hence, our linear equation was determined to be: oindent \(p = 0.00649B\).

If you want to find out your monthly payment for an amount borrowed, let's say \(\text{145,000 USD}\), you would set \(B = 145000\) in the equation.

Thus, oindent \(p = 0.00649 \times 145000\). After calculating, oindent \(p = 941.05\).

So, if you borrowed oindent \(\text{145,000 USD}\), your monthly mortgage payment would be oindent \(941.05\).

Be sure to always account for other potential fees and interests that might apply to your mortgage. But this calculation gives you a straightforward view of the principal repayment.