Solving Simultaneous Equations: y - 2r - 1 0 and 2yx - 8 0

In algebra, solving simultaneous equations is a fundamental skill used in various fields of mathematics and engineering. This article will guide you through the process of solving the specific simultaneous equations:

Introduction to Simultaneous Equations

Simultaneous equations are a set of equations containing multiple variables that are solved simultaneously. The solution is the set of values for the variables that satisfies all of the equations in the set. Understanding and solving simultaneous equations is relevant in various applications, including physics, engineering, and economics.

The Equations Provided

Let's consider the following two simultaneous equations:

y - 2r - 1 0 2yx - 8 0

Step-by-Step Solution

To solve these simultaneous equations, we will follow these steps:

Isolate one of the variables from one of the equations. Substitute the isolated variable into the other equation. Solve the resulting equation for the remaining variable. Substitute the solution back into one of the original equations to find the other variable.

Step 1: Isolate y in the First Equation

Let's start by isolating y in the first equation:

y - 2r - 1 0

Add 2r 1 to both sides:

y 2r 1

Step 2: Substitute y in the Second Equation

Now, we substitute y 2r 1 into the second equation:

2yx - 8 0

This becomes:

2(2r 1)x - 8 0

Simplify the expression:

4rx 2x - 8 0

Factor out 2x:

2x(2r 1) 8

Solve for x:

x 4 / (2r 1)

Step 3: Substitute x Back into One of the Original Equations

Now, we substitute x 4 / (2r 1) back into the first equation:

y - 2r - 1 0

Substitute x 4 / (2r 1) into the expression for y from Step 1:

y 2r 1

Substitute:

y 2r 1

Now substitute x 4 / (2r 1):

(2r 1) - 2r - 1 0

This simplifies to:

0 0

Since this is always true, it confirms that the relationship between y and r is consistent.

Conclusion

We have successfully solved the simultaneous equations:

y 2r 1 x 4 / (2r 1)

The solutions are dependent on the value of r. For example, if r 0, then:

y 1 x 4

For any other value of r, the values of x and y can be calculated using the equations derived.

Complex Solutions

From the equations, we also derived:

4.75x^2 - 14x - 24 0

Using the quadratic formula, x [-b ± sqrt(b^2 - 4ac)] / (2a), where a 4.75, b -14, and c -24, we get:

x [14 ± sqrt(14^2 - 4*4.75*(-24))] / (2*4.75)

Calculating further, we get:

x [14 ± sqrt(196 456)] / 9.5

x [14 ± sqrt(652)] / 9.5

x [14 ± 25.53] / 9.5

This yields two solutions:

x (14 25.53) / 9.5 4.79 x (14 - 25.53) / 9.5 -1.27

Substituting these values back into the equation y 4 - 0.5x, we get:

For x 4.79, y 2.055 For x -1.27, y 5.195

Conclusion and Further Exploration

In summary, we have solved the given simultaneous equations using algebraic methods. The solutions provide insight into the relationships between the variables and can be used to further explore more complex scenarios. Understanding these concepts is crucial for students and professionals alike in fields such as engineering, physics, and mathematics.

Key Takeaways

Simultaneous equations involve solving multiple equations with multiple variables. Algebraic manipulation and substitution are key techniques in solving simultaneous equations. Quadratic equations can be derived from simultaneous equations and solved using the quadratic formula.