Problem 40 - Calculus: One and Several Variables

Question

Find the point where the lines intersect. $$l_{1}: 3 x \cdot y-5=0, \quad l_{2}: 7 x-10 y+27=0$$

Step-by-Step Solution

Answer

The intersection point of the two lines is \((x,y)\) where 'x' and 'y' are the solutions obtained from steps 3 and 4 respectively.

Step 1: Rewrite the equations in standard form

We can rewrite the given equations to express 'y' in terms of 'x'. The first equation will become \(y=\frac{5}{3x}\), while the second equation will be altered to \(y=\frac{7x+27}{10}\).

Step 2: Equate the expressions for y

The y-coordinate of the intersection point of the two lines would be the same, which means the two expressions 'y' must be equal. So we set \(\frac{5}{3x}=\frac{7x+27}{10}\).

Step 3: Solve for x

To solve for 'x', we cross-multiply and rearrange the terms to have the quadratic equation: \(35x^2 - 10x - 15 = 0\) and then solve this quadratic equation to get the value of 'x'.

Step 4: Solve for y

Next, substitute the 'x' value obtained into any of the originally given equations to obtain the corresponding 'y' value.

Key Concepts

Solving Systems of Equations

When we need to find a common solution for two or more equations, we are essentially solving a system of equations. This task involves determining where two or more graphs intersect, or, in algebraic terms, finding the set of values that satisfy all equations simultaneously.

In the context of our exercise, we are looking at two equations representing lines, and we need to find their point of intersection. This is addressed by equating the 'y' values derived from each equation and solving for 'x', since the 'y' values will be the same at the point of intersection. Once 'x' is determined, it can be substituted back into either equation to find the corresponding 'y' value.

This process can handle both linear and non-linear systems, and varying methods such as the substitution method, elimination method, and graphical method can be used depending on the specific problem at hand.

Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can be represented graphically by a straight line and algebraically in the standard form Ax + By = C, where A, B, and C are constants.

In our exercise, the equations given are linear when considered separately. To find L1's intersection with L2, we manipulated L1 to isolate 'y'. In a 'y = mx + b' format, 'm' represents the slope, and 'b' represents the y-intercept. For linear equations, solving systems typically involves lining up like terms and combining them to find the values of the unknowns.

Quadratic Equations

Quadratic equations are a step up in complexity from linear equations. They are polynomial equations of degree two, typically taking the form Ax^2 + Bx + C = 0, where A, B, and C represent constants with A not equal to zero. The graph of a quadratic equation is a curve called a parabola.

In our exercise, once the two expressions for 'y' were set equal, cross-multiplication led to a quadratic equation. Solving this type of equation often involves factoring, completing the square, or using the quadratic formula to find the roots of the equation, which are the x-values where the graphed parabola crosses the x-axis. In this case, finding the root(s) of the equation gave us the 'x' value(s) at the point(s) of intersection between the two lines.

It's important for students to understand how to manipulate equations and solve for unknowns in both linear and quadratic contexts, as these skills form the basis for much of algebra and calculus.