Problem 272 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Solve \(2 x+8 y+5 z=5\) \(x+y+z=-2\) \(x+2 y-z=2\)

Step-by-Step Solution

Answer

The solution to the system is \(x = -1\), \(y = 2\), and \(z = -2\).

Step 1: Elimination of one variable

Let's choose to eliminate \(x\) from the second and third equations by subtracting the third equation from the second. This gives: \(x+y+z-(x+2y-z) = -2-2\) which simplifies to \(-y + 2z = -4\). Now we have two equations involving two variables \(y\) and \(z\).

Step 2: Solving the system of two equations

The two equation system is now: \(2y + 5z = 5 - 2x\) and \(-y+ 2z = -4\). We can solve for \(y\) in the first equation: \(2y = 5 - 2x - 5z\). Simplifying this will give \(y = 2.5 - x - 2.5z\). Substitute \(y\) into the second equation to solve for \(z\). This gives: \(-(2.5 - x - 2.5z) +2z = -4\). Solving this will give \(z = -2\).

Step 3: Substitute \(z\) into the equations to solve for \(x\) and \(y\)

Now we substitute \(z = -2\) into the above equations to get \(x\) and \(y\). Substituting \(z = -2\) to the equation \(y = 2.5 - x - 2.5z\) will give \(y = 1 - x\). And substituting \(z = -2\) and \(y = 1 - x\) to the first original equation \(2x + 8y + 5z = 5\) will give \(2x + 8(1-x) +5(-2) = 5\). Solving this will give \(x = -1\). Then substituting \(x = -1\) into \(y = 1 - x\) will give \(y = 2\).

Key Concepts

Elimination Method

The elimination method is a systematic technique used in Linear Algebra for solving systems of linear equations. The goal is to simplify the system by removing one variable to make solving the system more manageable. This is accomplished by strategically adding or subtracting equations to cancel out a variable.
For instance, in our exercise, the variable 'x' was eliminated from the second and third equations. By subtracting the third equation from the second equation, we successfully removed 'x'.

• This yielded a new equation: \( -y + 2z = -4 \).

Now, with one less variable, the system becomes easier to solve. Remember, the key steps involve manipulating the equations:

• Aligning equations in terms of variables and constants.
• Adding or subtracting equations to eliminate a variable.
• Simplifying the results.

Overall, the elimination method is extremely useful and efficient in finding solutions to linear systems when combining and simplifying equations.

Linear Algebra

Linear Algebra is the mathematical study of vectors, vector spaces, linear transformations, and systems of linear equations. It is a fundamental area of mathematics, particularly in solving equations involving multiple variables.

• Linear equations form the base of linear algebra, depicting relationships between variables in straight-line forms.
• The beauty of linear algebra lies in its ability to represent complex systems concisely and solve multiple equations simultaneously.

In the exercise provided, linear algebra comes into play to analyze and find solutions for the set of equations involving three variables. By using proven methods like elimination, equations are manipulated:

• Representing each equation as a plane in 3D space.
• Finding intersection points of these planes, which represent the solutions.

Methods of linear algebra, such as elimination, help succinctly solve and interpret these multivariable equations in a logical progression, leading to the solution.

Solving Equations

Solving equations is a core concept in mathematics that involves finding the values of variables that satisfy given equations. Different methods can be employed depending on the type of equations or systems involved.

• Simple equations often require basic operations such as addition, subtraction, multiplication, or division.

For systems of linear equations like in our specific exercise, solving requires integrating methods like elimination to simplify the equations first:

• Identify and isolate variables by eliminating others, as demonstrated in the step-by-step solution.
• Substitute back to verify solutions once variables are found.

In our solved problem, once 'z' was determined to be \(-2\), values could be substituted back to solve for 'y' and 'x'. The approach ensures all solutions satisfy the original set of equations. Keep in mind that no matter the complexity, breaking down the problem into smaller, manageable parts is the key to successfully solving any system of equations.