Problem 35 - A History of Mathematics: An Introduction

Question

If the sum of the two quotients formed by dividing the two parts of a given number by two different known numbers is given, then each of the parts is determined. Namely, solve the system \(x+y=a, x / b+y / c=d\). Jordanus sets \(a=10, b=3, c=2\), and \(d=4\).

Step-by-Step Solution

Answer

Answer: The two parts of the given number are x = 6 and y = 4.

Step 1: First Equation: x+y=10

It's given that \(x+y=10\). This equation represents the sum of the two parts of the given number.

Step 2: Second Equation: \(\frac{x}{3} + \frac{y}{2} = 4\)

It's given that \(\frac{x}{3} + \frac{y}{2} = 4\). This equation represents the sum of the two quotients when dividing the two parts by known numbers 3 and 2.

Step 1: Isolate y from the first equation.

Rearrange the first equation to isolate y in terms of x: \(y=10-x\).

Step 2: Substitute y into the second equation.

Now, substitute the expression for y from Step 1 into the second equation: \(\frac{x}{3} + \frac{10-x}{2}=4\).

Step 3: Solve for x.

Multiply each term by 6 to eliminate the denominators: \(6\left(\frac{x}{3}\right) + 6\left(\frac{10-x}{2}\right)=6(4)\). Simplifying, we get: \(2x + 3(10-x)=24\). Distribute the 3 and then simplify: \(2x + 30 - 3x = 24\). Combine like terms, and solve for x: \(-x = -6\), hence \(x = 6\).

Step 4: Substitute the value of x back into the expression for y.

Substitute the value of x we found in Step 3 back into the expression for y: \(y=10-x\). Evaluating, we have \(y = 10-6 = 4\).

Step 5: State the final values of x and y.

The solution of the system is \(x=6\) and \(y=4\), which correspond to the two parts of the given number.

Key Concepts

Equation Solving

Equation solving is a fundamental aspect of algebra and involves finding the values of variables that satisfy one or more equations. In the given problem, we are tasked with solving a system of two equations, which involve the variables \(x\) and \(y\). One of the equations specifies the sum of these variables: \(x + y = 10\). The other equation is slightly more complex: \(\frac{x}{3} + \frac{y}{2} = 4\). To solve such systems, the first step is usually to express one variable in terms of another. In our case, we isolate \(y\) from the first equation, giving us \(y = 10 - x\). This expression is then substituted into the second equation, allowing us to solve for \(x\). This substitution method simplifies the process of finding the values of both variables.

Quotients

Quotients play an essential role in our system of equations. A quotient is the result of dividing one number by another. In the given exercise, each variable \(x\) and \(y\) is divided by a distinct number, 3 and 2 respectively, forming the quotients \(\frac{x}{3}\) and \(\frac{y}{2}\). The sum of these quotients is another equation: \(\frac{x}{3} + \frac{y}{2} = 4\). Understanding how to manipulate these quotients, such as by clearing denominators through multiplication, is critical in simplifying and solving equations. By multiplying through by 6, which is the least common multiple of 3 and 2, we eliminate the fractions entirely, making it more straightforward to combine and solve for our unknowns.

Linear Algebra

Linear algebra is the study of vectors, vector spaces, and linear transformations. It also deals with systems of linear equations, much like the one we solved in this problem. The equations \(x + y = 10\) and \(\frac{x}{3} + \frac{y}{2} = 4\) are linear because they represent straight-line relationships between \(x\) and \(y\). When plotted, each equation forms a line on the Cartesian plane, and the solution to the system is the point where these lines intersect. This solution provides the values for \(x\) and \(y\), which in this case are \(x = 6\) and \(y = 4\). Linear algebra helps understand and solve such equations systematically, using various methods like substitution, elimination, or matrix operations.